list

问题 I have a base interface public interface IBase { ... } and a interface that derives from this base public interface IChild : IBase { ... } Within my code, I call a method which will return me a List<IBase> (legacy code). With this List I am trying to fill a ObservableCollection<IChild> : List<IBase> baseList= GetListofBase(); ChildList = new ObservableCollection<IChild>(); // how to fill ChildList with the contents of baseList here? I know it is not possible to cast from a base to a derived

问题 I have a base interface public interface IBase { ... } and a interface that derives from this base public interface IChild : IBase { ... } Within my code, I call a method which will return me a List<IBase> (legacy code). With this List I am trying to fill a ObservableCollection<IChild> : List<IBase> baseList= GetListofBase(); ChildList = new ObservableCollection<IChild>(); // how to fill ChildList with the contents of baseList here? I know it is not possible to cast from a base to a derived

问题 I'm having trouble finding the length of a list. I know how to deal with lists such as say [a,b,c,d] or [a,b,[],d] or even [a,[[[]]],c,[]] each of which have a length of 4. The problem I'm having is trying to figure out the length of the list in a case like this [a,[b,c],d] . There are 4 elements, but when I run my code, it'll print 3. It considers the inner list [b,c] as a single element and I'm not sure how to count those separately. Here's what I have: % this is the base case mylen([],0).

问题 I am trying to append values to a list. But, as it's multiprocessing, the list is ending up with just one value finally. Is there a way where I can append sizes of all the images to the list rather than just one? import cv2 import concurrent.futures import os length = [] def multi_proc(image): name = image[0:-4] im = cv2.imread(image) final_im = cv2.resize(im, (100,100)) cv2.imwrite(name+"rs"+".png", final_im) l = im.shape print(l) length.append(l) with concurrent.futures.ProcessPoolExecutor(

问题 this question is already asked in here (convert JSON String to the list object in flutter) by someone else.. but right now.. I don't get the answer yet.. so is there a way to call List<LoginRespon> ... I have converted from json respond using https://app.quicktype.io/ and here is the code import 'dart:convert'; LoginRespon loginResponFromJson(String str) => LoginRespon.fromJson(json.decode(str)); String loginResponToJson(LoginRespon data) => json.encode(data.toJson()); class LoginRespon {

问题 I'm having trouble finding the length of a list. I know how to deal with lists such as say [a,b,c,d] or [a,b,[],d] or even [a,[[[]]],c,[]] each of which have a length of 4. The problem I'm having is trying to figure out the length of the list in a case like this [a,[b,c],d] . There are 4 elements, but when I run my code, it'll print 3. It considers the inner list [b,c] as a single element and I'm not sure how to count those separately. Here's what I have: % this is the base case mylen([],0).

问题 This question already has answers here : How to combine two lists in R (3 answers) Closed 12 months ago . I have list 1: l1 <- list(letters[1:26]) I have list 2: l2 <- list(c(1:26)) How would I combine l1 and l2 into a new list l4 with 52 elements: the first 26 elements are the elements of l1 and the next 26 are the elements of l2 ? 回答1: You could use c to combine them l4 <- c(l1[[1]], l2[[1]]) Or with unlist l4 <- c(unlist(l1), unlist(l2)) Maybe you want to wrap it in list l4 <- list(c(l1[[1

问题 This question already has answers here : How to combine two lists in R (3 answers) Closed 12 months ago . I have list 1: l1 <- list(letters[1:26]) I have list 2: l2 <- list(c(1:26)) How would I combine l1 and l2 into a new list l4 with 52 elements: the first 26 elements are the elements of l1 and the next 26 are the elements of l2 ? 回答1: You could use c to combine them l4 <- c(l1[[1]], l2[[1]]) Or with unlist l4 <- c(unlist(l1), unlist(l2)) Maybe you want to wrap it in list l4 <- list(c(l1[[1

问题 This question already has answers here : How to combine two lists in R (3 answers) Closed 12 months ago . I have list 1: l1 <- list(letters[1:26]) I have list 2: l2 <- list(c(1:26)) How would I combine l1 and l2 into a new list l4 with 52 elements: the first 26 elements are the elements of l1 and the next 26 are the elements of l2 ? 回答1: You could use c to combine them l4 <- c(l1[[1]], l2[[1]]) Or with unlist l4 <- c(unlist(l1), unlist(l2)) Maybe you want to wrap it in list l4 <- list(c(l1[[1

问题 I need to compare two columns together: "EMAIL" and "LOCATION". I'm using Email because it's more accurate than name for this issue. My objective is to find total number of locations each person worked at, sum up the total of locations to select which sheet the data will been written to and copy the original data over to the new sheet(tab). I need the original data copied over with all the duplicate locations, which is where this problem stumps me. Full Excel Sheet Had to use images because