[ad_1] There are a decent amount of ways to accomplish what you are asking, but here is one way to read a file into a program and split each line by specific delimiters into a list, while still keeping the delimiters in the sentence. All of the functionality for turning a file to a list … Read more
[ad_1] You need for-loop with if/else to run different code for different lists. For list with “PRODUCT NAME PACK” you have to keep list with names in variable – so you may use it in next loops when you get list with numbers For other list you can keep first number as index and rest … Read more
[ad_1] First of all, you always must include the description of what you have tried so far and what error you encountered while doing so. It is not nice to ask a question directly without showing your efforts in it. Now coming back to your question, it is actually quite very easy, what you can … Read more
[ad_1] We have some list of lists a. We want to get counts for each of the sublists, and the total of those counts. Note that the total counts will just be the sum of the counts of the sublists. from collections import Counter a = [[‘das’,’sadad’,’asdas’,’das’],[‘das’,’sadad’,’da’],[‘aaa’,’8.9′]] def count(list_of_lists): counts = [Counter(sublist) for sublist in … Read more
[ad_1] Try this : In case the if(listOne.contains(object)) return listOne; The list.contains() uses equals. so, if list.contains() does not give you the expected result, you should be overriding the equals method. [ad_2] solved how to search in a list in java?
[ad_1] This should help you: lst = [‘apple’,’banana’,’cherry’] strings = input().split(‘,’) for x in strings: lst.remove(x) print(lst) Output: >>> apple,banana [‘cherry’] [ad_2] solved How to remove multiple input items from a list [closed]
[ad_1] add this into your code here is a fiddle ul#trainers, ul#keepers, ul#verdedigers, ul#middenvelders, ul#aanvallers {display:block!important; max-width:1000px; overflow-x:auto;} 6 [ad_2] solved Different don’t stand under eachother
[ad_1] Are you looking for the product of [5, 6, 7] with itself, 57 times? itertools.product() does that for you: from itertools import product for combination in product([5, 6, 7], repeat=57): print combination This will take a while to print all output, however. It’ll eventually print all 3 ** 57 == 1570042899082081611640534563 possible combinations. Don’t … Read more
[ad_1] All objects must have keys. You define an object using curly bracers {}. Basically what you are saying is, add an array with one object that has no keys defined. If you want an array with the values a,b,c,d you can remove the bracers: myObjectList[0] = [“a”, “b”, “c”, “d”]; You always define objects … Read more
[ad_1] A good approach might be to use count(*) and GROUP BY. So you get a count of the number of answer rows per user per list in the completed table then join to a subquery which gets you a total count of items in the list. Something like this: select user_id, c.list_id, count(user_id) as … Read more
[ad_1] Not beautiful solution, but working one values = [{‘id’: 1, ‘x’: 2}, {‘id’: 1, ‘y’: 4}, {‘id’: 1, ‘z’: 6}, {‘id’: 2, ‘j’: 5}, {‘id’: 2, ‘k’: 10}, {‘id’: 3, ‘w’: 1}, {‘id’: 3, ‘x’: 3}, {‘id’: 3, ‘y’: 5}, {‘id’: 3, ‘z’: 7}] # get all unique possible keys unique_keys = set((x[‘id’] for … Read more
[ad_1] The exception originates from my_data.get(position) in your onProgressChanged() listener. This listener is called asynchronously, when progress changes, but it refers to the original position provided, when you perform the onBindViewHolder(). So when at time X you do the onBindViewHolder(), position with value 2 is valid (if there are at least 3 entries in the … Read more
[ad_1] Based on my understanding you want to check if the type is a list and if the list’s elements is type X. You can do the following: var type = (new List<int>()).GetType(); if (type.GetInterface(“IList”) != null && type.IsGenericType && type.GenericTypeArguments.Length == 1 && type.GenericTypeArguments[0] == typeof(int)) Console.WriteLine(true); //Outputs True 3 [ad_2] solved Item type … Read more
[ad_1] You can use a nested list comprehension to generate a new list keeping the nested structure, using if/else statements to populate it with * or a space accordingly: L = [[0, 0, 1, 0, 0], [0, 1, 1, 1, 0], [1, 1, 1, 1, 1]] [[‘*’ if i else ‘ ‘ for i in … Read more
[ad_1] You can just use the str.index(char, [beg, end]) method to retrieve the position of a character inside a string. The method allows you to add a beginning and end position that you can use to check for the different occurrences. Here the code: s=”CNCCN” dist = s.index(‘N’, s.index(‘N’) + 1) – s.index(‘N’) And its … Read more