Please try below code. It might work for you. Use predicate to find element from List. public static class A { public static readonly List<string> alist = new List<string> { //some big data about 70 rows }; public struct astruct { public const string adata = “a data”; } } public class B { string … Read more
The first one is a valid array declaration. The second one has an incorrect reference type. In the third one, you cannot create an object of type List as it is an interface For the last one, it’s ArrayList, not Arraylist. List[] myList2 = new List[5]; List myList5 = new ArrayList(); solved Which Lists will/won’t … Read more
Collections.rotate will do that you want String[] daysArr = { “Mon”, “Tue”, “Wed”, “Thu”, “Fri”, “Sat” }; List<String> daysList = Arrays.asList(daysArr); String input = “Fri”; int index = daysList.indexOf(input); if (index > 0) { Collections.rotate(daysList, -index); } System.out.println(daysList); Hope it helps! solved How to arrange the datas to display in array by given string of … Read more
dictionaries are unordered (well pre 3.6 at least), instead sort the items d = {3: ‘__init__’, 5: ‘other’, 7: ‘hey ‘, 11: ‘hey’} print(sorted(d.items(),key=lambda item:(item[0].strip(),item[1]))) # output => [(3, ‘__init__’), (7, ‘hey ‘), (11, ‘hey’), (5, ‘other’)] if you really want it as a dict (for reasons i cant fathom) and you are using 3.6+ … Read more
list = [1,2,3,4,5,6] for x in range(len(list)): if list[x] % 3 == 0: list[x] = list[x]**2 print(list) solved How to divide specific elements in a list and then square the values? [closed]
I think this is what you mean. a = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5] b = [1, 2, 3, 4, 5] for x in a: for y in b: if a[x] == b[y]: #Do stuff here print “found a match ” + str(a[x]) elif y == len(b) – 1: break … Read more
You can create a recursive function, public void loopThrough(List<Region> regions) { for(Region region : regions) { if(region.hasSubRegion()) { loopThrough(region.getSubRegions()); } } } 1 solved Iterate through a possibly infinite amount of nested ArrayLists
Let’s start from the middle of this expression. ‘3’ * 2 evaluates to ’33’. This is because multiplication of a str object by a int object results in concatenating string with itself n times (providing you are multiplying by n). int(’33’) is just 33 as an integer. When you feed int class a string it … Read more
As the error states, it cannot convert from ‘System.Collections.Generic.List’ to ‘string’. However I never knew the function SingleOrDefult() existed. All credit to @maccettura even though he didn’t know what I was trying to do! xD Code change below for the answer: List<string> listFrom = new List<string>(); //Contains a list of strings List<string> listTo = new … Read more
map does a delayed execution, so when you cast the result to a list, the map is executed, which amounts in computing sum(1), sum(2), sum(-3) and sum(4) Any of those will fail because sum takes something like a list (an iterable object) in input. You pass integers which are not iterable, hence the error message. … Read more
Your ENUM values on the map seem like another map. Therefore I put it like this Map<Enum, Map<String, Object>> However, question is not clear! Here is one possible solution. import java.util.*; public class Main { public static void main(String[] args) { Map<Enum, Map<String, Object>> map = new HashMap<>(); Map<String, Object> value = new HashMap<>(); value.put(“String1”, … Read more
Are you looking for this? var tupleList = new List<Tuple<string, string>>(); for (int i = 0; i < data.Count; i++) { for (int j = i + 1; j < data.Count; j++) { tupleList.Add(new Tuple<string, string>(data[i], data[j])); } } 1 solved Combine elements of a list [closed]
Here is a function that works fine for all cases and return the list of all first candidates encountered if no choice can be made to separate them. def find_best(list_of_lists): i = 0 while len(list_of_lists[i]) == 0: i+=1 list_containing_candidates = list_of_lists[i][:] if len(list_containing_candidates) == 1 : return list_containing_candidates[0] else: if i+1 < len(list_of_lists): for next_list … Read more
Try this: nest = [[0,0,0,0,0],[0,0,0,0,0],[0,0,0,0,0]] x = 1 y = 3 for array in nest: for i in range(x, y+1): array[i] = 1 print(nest) I think you would benefit from a basic Python tutorial too, for example try codecademy.com. 3 solved Python – Replacing Indice Values in Nested Lists
you can use something like that: sentence = input().split() ans = [] for i in range(len(sentence)): if i%2: ans.append(sentence[i]+’…’) else: ans.append(sentence[i]) print(‘ ‘.join(ans)) solved How to iterate odd indexes and add them back to original list