sublist

问题 I am trying to solve a simple prolog question but I am not able to solve it. From a list a need to create a sublist given the index I and then from I the next elements given as N. If the index is greater than the list lenght I will get the sublist empty. If N (number of elements) is greater than the rest of elements in the list I will get all the elements from I until the end. Here, I got one part of the assignment, I can get from the index I, the next elements N. Now I ask about the other

问题 In Python, what is the time complexity when we create a sublist from an existing list? For example, here data is the name of our existing list and list1 is our sublist created by slicing data. data = [1,2,3,4,5,6..100,...1000....,10^6] list1 = data[101:10^6] What is the running time for creating list1? Is it O(10^6) i.e.O(N), or O(1)? 回答1: Getting a list slice in python is O(M - N) / O(10^6 - 101) Here you can check python list operations time complexity By underneath, python lists are

问题 I need Python to search all the sublists of a given list, but this does not work when I'm searching for an element contained in only one of them. For example, this is my code: data = [[4,5],[4,7]] search = 4 for sublist in data: if search in sublist: print("there", sublist) else: print("not there") print(data) and this works very well if my search is contained in all of the sublists of the lists. However, if my search is, for instance, 5, then I get: there [4,5] #I only want this part. not

问题 Closed. This question is off-topic. It is not currently accepting answers. Want to improve this question? Update the question so it's on-topic for Stack Overflow. Closed 6 years ago . This is a little confusing so I will try my best to explain my goal. In a nutshell i'm trying to look at a sublist within a list. In those sublists, some have the same starting element (sublist[0]) and i want to record the differences between that sublist with other sublists starting with the same element data =

问题 So I have this program that needs to be written in Scheme using Racket that has the following properties and I am stumped. The function is called sublist? with two inputs of S and L which are both lists. It checks whether S is a sublist of L and returns #t or #f . Examples would be similar to: sublist? of (A A) and (A B C) is #f sublist? of (A B C) and (A B D A B C D) is #t sublist? of (A (B)) and (C ((A (B))) (C)) is #t A small function called extractLists needs to be created to extract the

问题 I would like to create an User Event script (Before Submit event), applied on Transactions (Purchase Orders, Sales Orders) that will Sort the Items lines based on a certain field (i.e. Item Name). My approach would be to store all information in an array, sort it and then use the APIs nlapiRemoveLineItem method to remove each item and nlapiInsertLineItem method to insert each item on the correct order. However, for this approach I would need to store all columns values before removing and

问题 How do I create a function sublist() that takes two lists, list1 and list2 , and returns True if list1 is a sublist of list2 , and False otherwise. list1 is a sublist of list2 if the numbers in list1 appear in list2 in the same order as they appear in list1 , but not necessarily consecutively. For example, >>> sublist([1, 12, 3],[25, 1, 30, 12, 3, 40]) True >>> sublist([5, 90, 2],[90, 20, 5, 2, 17]) False 回答1: Here's one way to do it in linear time (and constant space) with an iterator: def

问题 I want to write a code that contains a sublist , that only stops once the element of the list is a certain number , for example 9. I´ve already tried using different operators , if statements . def sublist (list): return [x for x in list if x <9] [7,8,3,2,4,9,51] the output for the list above should be : [7,8,3,2,4] 回答1: List comprehensions really are for making mapping/filtering combinations. If the length depends on some previous state in the iteration, you're better off with a for-loop, it

问题 I am trying to get Python to search my list for a sublist which contains both of my search terms, but instead I get any sublist which contains one or more of the searches. Here is my code: search1 = 4 search2 = 3 data = [[4,3],[4,7], [6,3], [9,2]] found = False for sublist in data: if search1 in sublist: print("there", sublist) if search2 in sublist: print("there", sublist) found = True if(found==False): print("not there") print(data) if(found==False): print("not there") print(data) The

问题 Is there a way to remove duplicate sublists from a list of lists, even if they are not the same ordering? So could I do something like make: x = [[1,2],[3,4],[5,6],[2,1],[7,8]] into x = [[1,2],[3,4],[5,6],[7,8]] Is there an itertools function or something with a for loop? Thanks! 回答1: this will preserve the order of list and sublists, with possible duplicates in sublists: y, s = [], set() for t in x: w = tuple(sorted(t)) if not w in s: y.append(t) s.add(w) if x = [[1,2],[3,4],[5,6],[2,1,1],[2