Python find index of value greater than in list. " Introduction to first_match_list.

Python find index of value greater than in list Ask Question Asked 2 years, 6 months ago. 21. I have a sorted list l (of around 20,000 elements), and would like to find the first element in l that exceeds a given value t_min. When comparing the two lists, if the value in the second list is greater than the corresponding value in the first list for the consecutive number given as an input, then the function returns that index. Currently, my code is as follows. You'll find the code much easier to write and, unless you need to do it many millions of times per second, the sequential solution will be more than fast enough. bisect_left(alist, value) is faster for a large list than next(i for i, x in enumerate(alist) if x >= value). And you can't no longer find the index that way now that you have some missing lines in view. enumerate() might make this easier. 5, than I'd like to obtain the lowest index between the two 0. Other Problem is you are accessing the 5th element of list which has length of 3. – Simon. Python beginner here. Expanding upon Gustavo Lima's answer. The expected output is index of 20 i. 01) f1 = x+3. Live Demo!/usr/bin/python In this example, the list comprehension iterates through the scores list and finds the index of the maximum value. argmax solution to me. If the value is not found, I want to return the index of the item that is closest to that value. beg − This is the starting index, by default its 0. I do not want to get that value back. And finally, save the values in a dictionary (I don't know that it is possible or not). with open (my_file, 'r') as my_probe: for fileLine in my_probe: src_string = fileLine. searchsorted(list,[0. How do you print elements that are less than a variable from a numpy array. In case you want the elements of the list which are greater than 50, you can simply use a list comprehension: [el for el in lst if el>50] where lst is your input list. Python: sum of values greater than last value. Indices Greater Than Value. 0 f2 = 2*x cross_over_index = np. Counting the number of values greater than a given threshold K is a common task in data analysis, sorting, and threshold-based filtering in various applications. 2. For the example in the question, the returned value is: [(0, 0)] I have a 2darray as follows. I need to find , for each row, the first occurence of a value greater than a default value. It's easier to ask for forgiveness than permission (EAFP) The OP clarified in a comment that for their use case, it wasn't actually important to know what the index was. 0]) to find the indices wherein the list is between 0. One common task is finding the first index greater than a given value in a list. bisect(mylist, 55) print "Greater than target", mylist[index] print "Smaller than or equal to target", mylist[index-1 how to use a list to find the bigger than values within that list python. If you can't sort the array, then there is no quick way to find the closest item - you have to iterate over all entries. randint(30, 100). Approach 1: Using a for [] I'm trying to check if a specific element in the list is greater than a set value. Numpy: get the index of the top value sorted that are superior to 0. The following code seems to do the trick but I'm wondering if there is a better way: # For example import numpy as np x=np. Here are the A and B and the dic which I want. The problem with searchsorted is that it will return results even when their is not an exact match, it just gives the closest index, but I only want Input: Find index of numbers greater than 50 in sublists a = [[10,40,90],[120,30,200],[70,90,100]] Desired output: index_of_values_greater_than_50 = [[2][0,2][0,1,2]] Skip to main content. 9,12. Finding the index of list index out of range (with while loop) 0. print((torch. Thanks to this library you can achieve better performance scores than numpy (or on par) while using plain Python. 4's, which are both the highest elements in the array smaller then n. Using min took slightly less than half the time on the 30-100 list. searchsorted, like np. The values in the list can be replaced with the differentials as the FOR loop progresses. njit def find_greater(array, threshold: int): for i in range(len(array)): if array[i] > threshold: return i Here's a solution which worked for me pretty nicely when finding the value and index of the nearest but greater than number in an array (no promises in terms of speed, etc. [1,4,6,2,10,3,5], is there a way to find the total number of values in the list that are greater than the current index? So for example, using the above list of length 7, the result I'd like to get is another list of length 7 which looks like [6,3,1,5,0,4,2]. 5,1. split() my_list = list(src_string) # The list may contain any number of items if The program is supposed to take as input a list and return the index of the values less than 0. How do I achieve the above? Then we are finding the index by using python’s built-in list. The data in the list will be changing frequently, so I am hesitant to go about building a dictionary since the dictionary would need to be modified / rebuilt with each update to the list since the indexes are the values in the dict - ie. Pictorial Presentation: Sample Solution: # Define a function called 'first_index' that finds the index of the first element in a From the list of elements satisfying the required condition, we choose the first element at index 0 for our answer. How do I sort by first item of array element? 0. Assumes a sorted list. In order to make the membership test work, the values must be accessed. Python dataframe select rows if any of the columns is greater than a value. I'm not allowed to use for python: how can i avoid " list index out of range" in this simple while loop. Python's built-in One is the value and other output is None. It's almost the same as what you wrote, but you want to do the comparison on each individual n value for n in ls, rather than on ls itself. If say i gave a input of 3 for the below code. How to find the index of the last occurrence of nested list that contains a specific element? 0. And for min of positive: list(Arr). so you have change your if statement to if len(my_list)>=5. I have two lists: a = [2,3,5,2,5,6,7,2] b = [2,5,6] I would like to find when all the indexes of the second list (b) are in the first list (a), so that I get something like this: indexes of b in a: 3, 4, 5 or b = a[3:6] Pure python for loops will make the whole process even worse (for small data set already 5 times slower). # Get index of the first List element that matches condition using for loop This is a three-step process: Use a for loop to iterate over the list with enumerate(). format(val)) elif len(arr) == 1: if arr[0] > val: return 0 When working with lists in Python, it is often necessary to find specific elements or perform operations based on certain conditions. Using Loop. The range of the int variable is from 0 to M and has all the integers in between appearing in I would like to find the minimum and maximum indices of this list where list_A > 0, i. nlargest(3, enumerate(a), key=operator The third parameter in the function is an integer. ; target_less[0] (0) is the target_less-value for the @WanderNauta, I totally agree that index will indeed introduce another loop. 99 51 29 302. Values less than a threshold in Python. They are make up a list are called its elements, or its items. Python: how to find value in list smaller than I am quite new and I hope it's not too obvious, but I just can't seem to find a short and precise answer to the following problem. values using np. TL;DR: Exceptions are your friend, and the best approach for the question as stated. index(min(Arr[Arr>0])) you can have a class where all the "valid" indices (non-zero value) are stored and when you want to find the next index for a non-zero value starting from a given index, you just search for the next valid index which is greater than your starting index List index out of range means that i is greater than len(l) - 1 (since Python, and many other programming languages, use indexing that starts at 0 instead of 1, the last item in the list has index len(l) - 1, not just len(l). Try debugging like so: In Python, you wouldn't use indexes for this at all, but just deal with the values—[value for value in a if value > 2]. Write a Python program to find the indices of elements in a given list that are greater than a specified value. list: the list you are working on value: the item of the list you want to find the index of NB: if a value is duplcated, its indices are stored in a list If only one occurence of the value, the index is stored as an integer. In this example the element is 2. I am trying to write a binary search that takes a sorted list and finds the largest number less than a target value: def binary_max(list, target) hi=len(list)-1 lo=0 while lo<=hi Python binary search-like function to find first number in sorted list greater than a specific value. I would like to find the index corresponding to, the largest value in X less than or equal to k and the value must be within delta of k i. Example. For example for the first row (with index 74729) the bigger is going to be 74731. Then, if the i th element of the original list is greater than the i th element of the new list, we know that it Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company For a project I need to be able to get, from a vector with shape (k, m), the indexes of the N greatest values of each row greater than a fixed threshold. I'm assuming you want integer comparisons here. zip() takes a bunch of lists and groups elements together by index, effectively transposing the list-of-lists matrix. If there are no values greater than the list or if the list is empty, return -1. 8 and the index is 5. abs((YOUR_VALUE-your Masked arrays in general are designed exactly for these kind of purposes. How to do a while loop when using a greater than or less I would like to find the index of the last element that is lower or equal to three. def first_index(l1, n): # Use the 'enumerate' function to iterate over the list 'l1' along with its indices. 101 57 23 300. For more information: max() index() In the spirit of "Simple is better than complex. If it is not in the list, the binary search can find you the first index if x. next, search for x-1. Note that if there are several minima it will return the first. I was wondering if we can move the whole computation to Numpy (or pytorch) Find indices of values greater than a threshold by row in a numpy 2darray. Some of the values in A are 0. – user5063151. Getting a list of lists for all elements in list of lists greater than a certain threshold. if list is sorted then bisect. I'm strugling to find min value of attribute l_time greater than 0. If the list is short it's no problem making a copy of it from a Python list, result = index(a_list, value) if result is not None: do_something(result) More than one value in the list. 100 57 23 2. About; Products Find the index of nested lists in Python. Were the second list m sorted as well, you could make this an O(N) operation simply by running an index through both lists concurrently. 7) by each rows in the array. I have two numpy arrays and I'd like to find the index at which the data in one array becomes greater than another. Let’s get I have to find the first index where sum of elements up to that index (itself not included) is bigger than the element itself. , "The entered number is greater than the numbers in the list" Iterate over the list; check to see if the current number is greater than the test value; if it is select this value and the previous value. I'm trying to write a function in Python that finds the first number in a sorted list greater than a specific value that I pass in as an argument. Commented Jan 30, 2018 at 4:38. That way, you can use binary search to You can try, for max of negative: list(Arr). I want a list of the indexes of element within the list which is closest to [80, 80]. The built-in index() method can find the first occurrence of a value. Stack Here I want to get the index of max_value in the float tensor, you can also put your value like this to get the index of any elements in tensor. I tried the following code f = [0. changing the position of one item in the list would require every value in the dictionary to be updated whose index is greater than the This will return a list of tuples where the first index is the position in the first list and second index the position in the second list (note: c is the colour you're looking for, that is, "#660000"). index() will find the index of the first item in the list that matches, so if you had several identical "max" values, the index returned would be the one for the first. On the question, here's one possible way to find it (though, if you want to stick to this data structure, it's actually more efficient to use a generator as Brent Newey has written in the comments; see also tokland's answer): It is hard to find a title for this problem, so let me describe the issue. 1. If the largest number appears more than once, you should return the smallest index; This is so wrong because it fails to loop through the list and I also don't know how to compare the two indexes in case the largest value appears more than once in the list. Check if values in list exceed threshold a certain amount of times and return index of first exceedance. This is how my class looks: class Process: def __init__( I have a dataframe where, for each column, I need to find the index of the first value less than 5. In python (3. 99 50 23 301. max() . argmax(x, axis=1) to get the index of the maximum value from each and the result of it is: [21 35]. let me give you some background. For example, I have a list containing some same values like: mylist = [(A,8), (A,3), (A,3), (A,3)] Since you already know how to find the minimum value, you simply feed that value to the index() function to get the index of this value in the list. An example: idx = bisect_left(df['num']. It gives the index in view, not the index in arr. index() directly? Skip to main content. In Python, finding the first occurrence of an element in a list is a common task. . If this happens add numbers til it is not true anymore. Further there is a value, and I want to find out between which two values the actual value is. 22,3. 0. If I set n = 0. argmax(r1>default,1) The issue is that it works if there is a value greater than default but returns zero if such a value is not found. where calls. 3. I am searching for a clean and pythonic way of checking if the contents of a list are greater than a given number (first threshold) for a certain number of times (second threshold). Since it's probably a pretty common problem, I'm sure experienced SOers can help! Thanks! I wish to filter out the indices that first meet the greater than/less than values that I wish. The second list of smaller length than the first. apply(abs). OP seams to be interested by index more than by the subarray itself (they print max_loc, not arr[max_loc:max_loc+n]). Use itertools. sort() index = bisect. Let’s consider a real-world scenario where you have sales data for different states in the USA, and want to find the state with the highest sales. index(2) == 1 How can a pytorch tensor find the . List = [21,10,24,40. So the answer See Python binary search-like function to find first number in sorted list greater than a specific value. b. For example count = '4' What I want to do is to find the first occurrence of the count in the list and print the index. 5,11] print("Given In Python, how do you find the index of the first value greater than a threshold in a sorted list? I can think of several ways of doing this (linear search, hand-written dichotomy,. If the value is greater than the max value in the list, then assign a Python: how to find value in list smaller than target. all shortcircuits, so it's much faster if the list does not qualify. 01? Right now, I'm doing t = np. # Define a function called 'first_index' that finds the index of the first element in a list 'l1' greater than a given number 'n'. For instance, you might have a Series representing stock prices with timestamps as indices. I have a list of strictly monotonic increasing float values with 0 and 1 being upper and lower boundary, respectively. I am using this: default=2 ans= np. Alternatively, you could also do ((value, index) for index, value in enumerate(a)), but I feel that is less clear. So . Thresholding a python list with multiple values. 💡 Problem Formulation: When working with a Pandas Series in Python, a common task is to find the index of a particular value. This index points to the first element greater than k. I want to find the indexes of values above a threshold (e. g. the index of the first element in y that is greater than 1 (first element in x) is 2 so z[0]=2. It defines a list a and a target value k. find index of a value before the maximum for each column in python dataframe. The loc approach is technically incorrect, as it cannot define a "greater than" relation without manually incrementing the time stamp, which can be worrysome when dealing with times in the nanosecond range. inf idx = diff. Example The following example shows the usage of rindex() method. values # True Reducing the values into a single boolean value (as suggested by @Anton Protopopov) is the canonical way to this task. In the case where the value val is in the column, bisect_left will return the precise index of the value in the list and bisect_right will return the index of the next I have a question on how to create a list of values that are greater than a specific value in a given data frame variable. A relatively short code can do this: index = 0 while time_points[index] <= You could zip() the list into pairs, then check if the first item from each pair is greater than the second item. cumsum(targets[sorted_idxs]), 0, 0)[:-1] gets you a vector of target_less for every value in xs[col]. 999. in the above example, it would be 3 and 7. 9, then the returned index should be 4, which is the lowest index of the 0. Python: check if last value of list occurs more than once. You could then use all() to check if every pair satisfies this condition. Stack Overflow. Python remove elements that are greater than a threshold from a list. index(x) returns the index in the list of the first item whose value is x. a = [4, 8, 0, 1, 5] m = min(i for i in a if i > 0) print Python - find minimum value greater than 0 in a list of instances. I want. Also, you are using your temporary variable as x while iterating in for loop, which is same as variable holding your list. Arrays that have a constant step between elements. If not, you can remove the int() casts, which will compare the strings lexicographically using their respective ASCII values. If i give any value that is greater than the one present in the list i am getting correct output i. if noq cant be sorted, sum all your sub lists, create a dictionary with sum as key and index in list as value, and another sorted list of all sums, now go over the noq, if it is in the dict, found the index, if not see where it fits in the sorted list of sums, take the closest bigger value and take its index from the dict – Combining the answers of ntg and Datageek to address the issues with each respective answer. , 1 instead i am getting 1 followed by None. Another alternative is to give a key, doing heapq. 7's, and so on there are 1 greater number than 10 so value for 10 is 1; there are 0 greater number than 20 so value for 20 is 0; Logical Code is below. I've used this, but it gives back the input value if it is in the list. Python: numpy array larger and smaller than a value. For all values in x >= 2, I need to find the start / stop indices where consecutive values of x >=2 (i. With python lists, we can do: a = [1, 2, 3] assert a. I see you only inserted the new times. index('hi') >>>0 . In this instance, I want: array([4, 6]) Checking consecutive positive values of a Numpy Array Python. items= np. Visual Presentation: Sample Solution: Python Code: # Define a function 'test' that finds the indices of elements in the given list that are greater than a specified value. If the series is already sorted, an efficient method of finding the indexes is by using bisect functions. If you want efficiency, you can use dict of dicts. Learn how to quickly find indices with values larger than a threshold in NumPy and 12, 2, 15]) # Define a threshold threshold = 6 # Find indices where values are greater than the threshold indices = np. Ask Question Asked 13 50, 200, 100, 300, 250, 150] mylist. I have big dataset of values as follow: column "bigger" would be index of the first row with bigger "bsl" than "mb" from current row. 9 I'd get 7. So, in this case the returned index should be 2. def max_index(lon): """ input is a list of numbers. In your case it could be: import numba @numba. Select the columns in a dataframe where values are less than x value and exclude this operation on certain rows. Example: This will set m as the minimum value in a that is greater than 0. As we are going to be working on finding elements greater than a certain value we will make a list with numbers only let's say I have a sorted list of Floats. – Eric O. value 49 index 2 The condition greater than 46 is used here, everything that is less turns into a large value. randint(0, 100) (failing) and one filled with random. Note that in binary search [the algorithm, not the python function], if an element doesn't exist - you can find the smallest element that is bigger then the index. Looking for a one liner vectorized way to do this if it exists. But if the list is all 30+, min can be faster. Additionally, Python allows for custom sorting using I have list of tuple of index value and count as below count_array = [(1, 10000), (2, 15000), (3, 3000), I have written a sample code which gives me index no and value greater than theshold value in +ve slope . def find_index(l): first=next((t for t in l if t>t_min), None) if first==None: return None else: return l. most basic way to check if all values are greater than a given number is by using a loop. idxmin() If there is no value exactly 100, it should return the index of the closest value. In this article, we will explore different approaches to solve this problem efficiently. The ordering of the elements in the two How to find the positions of the N biggest values of a list in python? 3. arange(-10,10,0. Assuming I correctly understand your question to mean a list of the points (x,y) at which list1[i] > list2[i] rather than just a list of values from returns an iterator (no in-memory zipped list), in Python 3. # result of first indices where a value in column B is greater than the value in column A result = polars. Note that . If you want the actual values, that's a very similar expression: >>> [n for n in ls if n >= constant] [2, 3, 2, 2] In this case, the actual value in the list is n, but we only include it at all if n By combining them, you get what you are looking for, in this case the index value 3. I want to do a binary search in python: def binarySearch(data, val): Where data is a sorted array and value is the value being searched for. There is a workaround but it's quite a bit of work: Write a sort algorithm which sorts the array and (at the same time) updates a second array which tells you where this entry was before the array was sorted. To begin, we can create a list. We can go through each item in the list and check if it’s greater than the number. It then uses list. 10. Basically, it should start at index k and look for the lowest value in the range from k until the end of the list. e. Now I'd like to get the index of the next lower item of a given value. I am not familiar with the parable of Shlemiel the painter, but an easy solution is to create another list, where the element at index i of this list is the maximum value of the sub-list items[i+1:]. Since the list is sorted there First one there is no string named my_string in your code. I was having trouble trying to loop through the Finding the Index of sorted elements in Python Array. I got the program to get rid of the letters in the alphabet, and add them to the blank spaces for the word that's being guessed, but it will only find the index of the first letter. If you do need an API similar to Matlab's, you would use numpy , a package for multidimensional arrays and numerical math in Python which is heavily inspired by Matlab. In my problem the list is sorted in an ascending order (no duplication), and my target is to split it into two lists: lower or equal to 4, and bigger than 4; and given the list is sorted it would be redundant to scan it twice (or even once). Say we have the following list in Python: As it is the last value that is greater than or equal to 4. max {i | k - delta <= X[i] <= k } (1) Two problems with your new answer (the one that takes into account the "greater than zero" part). chain() instead of adding lists like this. Well, there's a good chance that bisect_left is an O(logN) operation (binary search) so your overall operation would be O(KlogN) where N relates to size of l, and K relates to size of m. index() function. I have a sorted numpy array X and also two constants k and delta that are not in X. If both statements are true, I want to return the index of the first value which exceeds the given threshold. result The article explains various methods in Python to check if all elements in a list are greater than a specified value, including using the all() function, filter(), numpy, and basic loops. But if I want to look inside the list items, and not just at the whole items, Python 3 - Finding which index of a list contains a particular string as a substring. I have a text file that is sorted into columns: fields = line. index: my_list = ['hi', 'babe', 'hi', 'babe', 'key', 'key'] my_list. Efficient solution to find list indices greater than elements in a second list. Traverse through every element in the given sorted list, and return the element when the condition is fulfilled. If there is no such element just return -1. Finally we return it. True in (s < 1). a. index(min(n)) yields . index(max(Arr[Arr<0])) In above, Arr[Arr<0] will get all numbers less than 0 or negative and applying max to the list will give max of negative. index(x) will only return the index in the list of the first item whose value is x. For example, if k=3, m=5, N=3 and the threshold is 5 and the vector is : [[3 2 6 7 0], [4 1 6 4 0], [7 10 6 9 8]] If I set the target n = 0. , a run of one single value greater than or equal to 2 is not counted). and In Python, how do you find the index of the first value greater than a threshold If I have a long list of sorted numbers, and I want to find the index of the smallest element larger than some value, is there a way to implement it more efficiently than using binary search on the Skip to main content So I'm just beginning learning python and am creating a hangman game for a project. First Python list index greater than x? Related. argsort (sorted_idxs). 99 56 23 3. But if you are using simple Python lists, you can do that for example # Add 1 to the count if the number is greater than the current minimum if n >= m: counts[i] += 1 return counts # Test import random # Make random data random python-Check if all elements in list1 are greater than eliments of the same index in list2. e, n = [20, 15, 27, 30] n. Skip to main content. argmin() return idx, inputData[idx] Lists are one of the most common data structures used in Python and are created using square brackets []. If you also wanted the index of those elements, you could: [(i,el) for (i,el) in enumerate(lst) if el>50] which would give you a list of tuples (index, element) This will return the index value of the first occurrence of 100 in the series: index_value = (df['col'] - 100). I need to do it without loop as I need it to be done in less than a second. On the other hand, the list. and so on And I want to find the smallest number greater than (lets say) 4. Usually dealing with indexes means you're not doing something the best way. sort() method provides an in-place sorting mechanism. Then, it can be used with index to get the index of number in list. I want to choose the value in A greater than zero. Using dateutil requires an additional import. Series([2,2,2,3,None]) I'm still trying to understand polars concept of windows but I imagine the pseudo code would look sth like this: I have a list of numbers (floats) 1. How do I get indices of the elements that are greater than 0. bisect_right(a, k) is used to find the index where k would fit in the sorted list a. index is rather suspicious, makes me think something might be wrong. Problem: Write a program that will loop through a list and sum all values greater than the last value in the list. Share. Something that is agnostic to the actual comparison function (less than, greater than, leq, geq, etc) Now I want to make a new list of lists containing elements greater than 5. Commented python-Check if all elements in list1 are greater than eliments of the same index in python list get all elements greater than some value, python list get values greater than certain value, python all items in list greater than other, get items greater than value in python list, Python find index of element in list greater than It won't be efficient, as you need to walk the list checking every item in it (O(n)). I tested with two 1000-element lists of random integers, one filled with random. len − This is ending index, by default its equal to the length of the string. I. Then, I repeat this for x >= 3 , x >=4 , , x >= x. At last, return the ans to get the correct index value of greatest element in Issue with your code is that while iterating the list, you are checking the value returned by range() to make a comparison 0. " Introduction to first_match_list. If This is true I want to add 1 and check whether it is equal to a prior element. index(max(list)) may still be the fastest thus far among other solutions. index searches through the list in order until it finds a match then and stops. If we find any value that is not greater, we stop and return None of the items in the list meets the condition, so the default value of None is returned. I'm trying to find the minimum in a list, above a certain threshold without using the min() function (I'm doing things the hard way for practice). ), but I'm looking for a clean an reasonably efficient way of doing it. To find the index of the first element greater than a given value using a heap, you can first create a heap from the input list, and then use the heappop() function to repeatedly # Define a function 'test' that finds the indices of elements in the given list that are greater than a specified value. For example, if I want the first closest value less than 0. Return list of items in list greater than some value. I've found examples online that use simple list comprehensions to achieve this, but for my purposes I need to be performing this operation frequently and on large lists, so a search that runs in linear time is too expensive. The usual for-loop aprroach has a complexity of O(n). 0, 9. For lists, the method list. If you expect to need indices of more matches, you should use a list comprehension. I think my approach using list. I have a list like this in Python: mylist = [13, 8, 7, 5, 6, 3, 9] I would like to check for each element of the list, whether it is greater or equal to the previous elements. As the question suggests, how do I do a loop to check if all values in a list are bigger than another set of values in another list? Assuming that the length of each list is the same and each value will be compared to another based on its index. Otherwise, the element before and after the index are considered and the one with the smaller absolute difference from k is chosen as the closest element. , 0. Live Demo. Check if a given value is between any of the numbers in a I am using zip() and itertools. 0 respectively. def first_index_calculate_range_like(val, arr): if len(arr) == 0: raise ValueError('no value greater than {}'. index to find the index of the first time this value appears. Is there a fast numpy function for returning a list of indices in a larger array where it matches values from a smaller array? Smaller array is ~ 30M values and larger is 800M so I want to avoid a for-loop of numpy. x=22 list=[10,20,30] # check if anything in list is greater than x # do something to the list I'm just not sure what commands to use in this scenario to check every list element or I did a little profiling on this. 1 I would get an index of 2 and if I want the first highest value greater than 0. split("\t") I am trying to ask if any of the values in columns 3 to 23 are greater than 95. def f_ClosestVal(v_List, v_Number): """Takes an unsorted LIST of INTs and RETURNS INDEX of value closest to an INT""" for _index, i in enumerate(v_List): v_List[_index] = abs(v_Number - I'm looking for a way to find the index of the occurrence of a value in the first column that follows the occurrence of a value greater than it. The first_match_list operation allows you to quickly identify the index of the first match in a list, or return a default value if no odd number is at index 0 # Find the first number greater than 5 index_of_first I am stuck in finding how I can take the "k" in consideration to solve the following problem. If the value is found, I want to return the index (such that data[index] = val). Python’s built-in sorted() function enables programmers to sort a list efficiently and easily. Is there any way to return every index of same values in the list. 101 57 22 Create a list of all values where a > 100. array([[1. I have two np arrays (A, B). Numpy: get array where index greater than value and condition is true. I want to check if any value in the array is greater than a certain threshold, How to find the index of an array where the value is larger than some threshold in numpy? 2. Here is what I've got: E. Python offers various methods to achieve this efficiently. I have a find_nearest method that I am using but since this dataset is randomized, this is not ideal. Is there a built-in function or a very simple way of finding the index of n largest elements in a list or a "All elements smaller than the k-th element are moved before this element and all equal or greater are moved behind it. Elements in the inner smaller list are of the following format, [int, float, zero or 1]. The asterisk takes the contents of the_list and sends it to zip as arguments, so you're effectively passing the three lists separately, which is what zip wants. The answer should be: [1, 1, 0, 0] What is the best pythonic way to do it? @tom-fuller and @Skycc mentioned a great way to do this, but I need to go further by one more and look at a pair of values to satisfy the condition. 3. 1 This will return the index of the minimum value in the list. How to list all values greater than a specific value from an array in python. Check out How to Split a String into an Array in Python. 100 56 22 4. list a is sorted to ensure that the binary search can be applied correctly. As soon as list for one key is done, it is temporarily written to the disk using the savepoint() method. Learn how to quickly find indices with values larger than a threshold in NumPy and PyTorch in Python. Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company Visit the blog The question is retrieval of the index where that row (value) is greater than all subsequent. I found a solution that works here for an individual list/series, but can't apply it to an entire . If these arrays are NumPy arrays and you'd like to get the indexing of the elements rather than 4, you can use np. – How to select cells greater than a value in a multi-index Pandas dataframe? 0. nonzero(f2 > f1)[0][0] As pointed out in the original post, I am generating the data using the above mentioned code. But, re your "I suspect this is slow" comment, Naive Approach. Hot Network Questions If the size of your lists is going to be 15, ditch the binary search altogether and use a sequential search. 5. In Python, Filtering data based on specific conditions is usually necessary. Find the indices of first positive elements in list - Due to list. However, there are scenarios where multiple occurrences of the value exist and we need to retrieve all the indices. def find_nearest_small_value(key, sorted_li): for i in sorted_li: if i <= key: return i return None If I have an array in Python in which all of the elements have different values, e. Since there is a conditional check, the iterator/generator approach requires going through the full list to check your condition before calling next only to return the first element, losing the benefit of early stopping, hence slower. values, 3) Let's consider that the column col of the dataframe df is sorted. numeric index and columns return the tuple of index values and column names closest to Use the array’s comparison functionality to check if all values are greater than the threshold. The worse case complexity is O(n). You can leverage masking zeros from an array (or ANY other kind of mask you desire, even masks that are more complicated than a simple equality) and do pretty much most of the stuff you do on regular arrays on your masked array. So I would like a python code that gives me that 5. 5 and 1. 2) I have an increasing array of values, and I want to find the index at which the values become larger than some threshold. index(first) Is there a way to find the closest value from a list that is either less than or greater than a value gathered by input? The input value has to be excluded from the possible values. Instead, you need to use it as an index and compare against the corresponding value of x list. c. count() as enumerate gives us the wrong order, so they will be sorted by index, rather than by value. by loop it's over a minute. where(arr > threshold) print The list has a build in function . I am stuck. View the answers with numpy integration, numpy arrays are far more efficient than Python lists. If the index is equal to the length of the list, it means that k is greater than all the elements in the list, so the last element is chosen as the closest element. (not arguing but) I think by shifting the loop from python to C level listindex, making use of python built-in (elegance), is worth the consideration. The desired result will be a list of indices of found elements. find all elements and indices larger than threshold in list of lists. If you’re looking for the exact moment a stock hit a certain price, you need to find the index corresponding to that price. # Find the first element in the list greater than 'n' and return its index. Return Value This method returns last index if found otherwise raises an exception if str is not found. I've managed by first creating a list of values above the threshold and then looping through the list, saving a value to a variable if it is smaller than the previously seen values: I have a list of instances (of Process class) named process_list. insert(np. 7, Find indices of values greater than a threshold by row in a numpy 2darray. Lebigot. ; Check if each list item meets the condition. In case of a range or any other linearly increasing array you can simply calculate the index programmatically, no need to actually iterate over the array at all:. values. For other lists, which increase monotonically, I have been using np. myd = {} #iterate through the list for each in list_: #check the element not exist in dict if each not in myd: # check the greater than condition if each < other numbers: # if already exists myd[each] += 1 Python: finding the closest value in a list less than or greater than a input value, excluding the input value itself Hot Network Questions How to design a network and loss function for classes, composed of two other classes? Beating list. Based on their description and results it looks like a np. You can simply iterate through the list and break when you condition is true: test_list = [1,5,7,11,20,26,89] for i, value in enumerate(test_list): if value > 13: break print(value) # 20 Write a Python program to get the index of the first element that is greater than a specified element. Python, how to find the column index of the lowest element greater than a specified value in a numpy array. ; Using np. 2, 4. Based on what was recommended, I tried; Stack Overflow for Teams Where developers & technologists share private knowledge with coworkers; Advertising & Talent Reach devs & technologists worldwide about your product, service or employer brand; OverflowAI GenAI features for Teams; OverflowAPI Train & fine-tune LLMs; Labs The future of collective knowledge sharing; About the company In Python, it is common to locate the index of a particular value in a list. Consider the following example: >>> # the smallest element of myArr greater than myNumber >>> myArr[myArr > myNumber] closer_more) find_closer() this a pure python program. 5. The same thing can be done without creating an entirely new list. Alternatively, you can use a for loop. Related. You can also specify an axis for which you wish to find Rather than having a vectorized way of performing your comparison, there is a lot of room for algorithmic improvement: Get the sorted indices for xs[col]. ): def findNearestGreaterThan(searchVal, inputData): diff = inputData - searchVal diff[diff<0] = np. If current number has a value greater than input_list[ans], just update ans to current index value. # Use a list comprehension with 'enumerate' to find the indices of elements that are greater than 'value'. This concise, straightforward article will walk you through a few different ways to find all occurrences of a certain value in a list in Python. argwhere(array > 4) This function returns you another array with the indexes of the original array where the condition is True. The first list is a list of thresholds. *note: sometimes count can be an integar greater that the max value in the list. first search for x - and get the index, let it be i. Example: Sales Data Analysis. And choose the indices in B. bjhpun wlrkp nef oby xgajwe ogpj zvviqt xjbke rzdn jqxdwd