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问题 I've been struggling with the workings of foreach loops in R. To speed up my code I'm trying to change my for loop into a foreach loop with %dopar%. My goal is to end up with three lists of the same length, each filled with data frames that represent scores between two users (I'm comparing three different calculation methods). My code used to be (very basic representation): for (a in 1:5) { #Just creating some sample data resultA <- data.frame(matrix(nrow = 40, ncol = 3)) resultB <- data

问题 I've been struggling with the workings of foreach loops in R. To speed up my code I'm trying to change my for loop into a foreach loop with %dopar%. My goal is to end up with three lists of the same length, each filled with data frames that represent scores between two users (I'm comparing three different calculation methods). My code used to be (very basic representation): for (a in 1:5) { #Just creating some sample data resultA <- data.frame(matrix(nrow = 40, ncol = 3)) resultB <- data

问题 This question already has answers here : Closed 9 years ago . Possible Duplicate: When should I use a List vs a LinkedList If I expect not to use the access by index for my data structure how much do I save by using LinkedList over List ? If if am not 100% sure I will never use access by index, I would like to know the difference. Suppose I have N instances. inserting and removing in a LinkedList will only be a o(1) op , where as in List it may me be O(n), but since it it optimized, it would

问题 I have class CatalogProduct(id: String, name: String) to declare a product I have two list below: val newestCatalogProductList = mutableListOf<CatalogProduct>() newestCatalogProductList.add(CatalogProduct("A1", "Apple")) newestCatalogProductList.add(CatalogProduct("A2", "Banana")) newestCatalogProductList.add(CatalogProduct("A3", "Orange")) newestCatalogProductList.add(CatalogProduct("A4", "Pineapple")) val popularCatalogProductList = mutableListOf<CatalogProduct>() popularCatalogProductList

问题 Hello I have a quick question I cant seem to solve. I have a list: a = [item1, item2, item3, item4, item5, item6] And I want to split this list into two seperate ones by everything other item such that: b = [item1, item3, item5] c = [item2, item4, item6] 回答1: Use slicing, specifying a step: b,c = a[::2], a[1::2] 回答2: Using filter is one option: a = [item1, item2, item3, item4, item5, item6] b = filter(lambda x: a.index(x) % 2 == 0, a) c = filter(lambda x: a.index(x) % 2 != 0, a) EDIT: This

问题 I hope you are great! I want to search through a VBA ArrayList and get the index number, the problem is, with For loop, you can only get the exact matches' index. I have the most of my search element (highlighted in the red box) and I want to get the elements which highlighted in the blue box, is there any way to do this in VBA? 回答1: You can use the in-built function InStr to find an occurrence of one string inside another. In your case change this: If list(j) = search_element Then To: If

问题 I'm a bit mentally stuck at something, that seems really simple at first glance. I'm grabbing a list of ids to be selected and scores to sort them based on. My current solution is the following: ids = [1, 2, 3, 4, 5] items = Item.objects.filter(pk__in=ids) Now I need to add a score based ordering somehow so I'll build the following list: scores = [ {'id': 1, 'score': 15}, {'id': 2, 'score': 7}, {'id': 3, 'score': 17}, {'id': 4, 'score': 11}, {'id': 5, 'score': 9}, ] ids = [score['id'] for

问题 I'm a bit mentally stuck at something, that seems really simple at first glance. I'm grabbing a list of ids to be selected and scores to sort them based on. My current solution is the following: ids = [1, 2, 3, 4, 5] items = Item.objects.filter(pk__in=ids) Now I need to add a score based ordering somehow so I'll build the following list: scores = [ {'id': 1, 'score': 15}, {'id': 2, 'score': 7}, {'id': 3, 'score': 17}, {'id': 4, 'score': 11}, {'id': 5, 'score': 9}, ] ids = [score['id'] for

问题 This is my list a=[ ['a','b','a','a'], ['c','c','c','d','d','d']] I wanna find most common elemments. I have tried this from collections import Counter words = ['hello', 'hell', 'owl', 'hello', 'world', 'war', 'hello', 'war','aa','aa','aa','aa'] counter_obj = Counter(words) counter_obj.most_common() but it works just for simple list. my output should be like this [('a', 3), ('c', 3), ('d', 3), ('b', 1)] 回答1: Apply Counter().update() option on the elements of your list, Based on suggestion

问题 I know this question has been asked before, and my solution is the same as many of the answers but I have a special test case that won't work correctly with the common solution to this problem. The solution that I have found for the zip problem like many others is (define (zip l1 l2)(map list l1 l2)) . . .which works great with given arguments such as (zip '(a b c) '(1 2 3)) => ((a 1) (b 2) (c 3)) but I also want the zip function to work for cases where my arguments do not match length like