tuples

问题 I was hoping that if I use the built-in map function on a tuple of length N in TypeScript then the return type would also be a tuple of length N (perhaps with different type for the elements depending on the function being passed to map ). Instead, the return type is just a standard, variable-length array of whatever type the callback function returns. The tuple's length is lost. I wrote a custom function that does what I want, but I'm wondering if there's a better way that is eluding me. I'm

问题 I have an assignment in which I must create a list of tuples from a string. I do this by splitting the string with .split(). I then iterate over the list and add the items and the number of repetitions of each item into a dict, like so: for word in s: if word in count_dict: count_dict[word] += 1 else: count_dict[word] = 1 After this I create a list of tuples using: pairs = count_dict.items() Followed by: count_list = [] for pair in pairs: count_list.append(pair) count_list.sort(key=lambda x:

问题 I have a list o tuples: list = [(element1, element2, element3), (elementelel4, element5, elementelement6), (el7, el8, elel9)] I want to print it as a table (so the distance between elements should be the same): HEADER1 HEADER2 HEADER3 element1 element2 element3 elementelel4 element5 elementelement6 el7 el8 elel9 I was trying to use some examples which I found, but it was used to print a list of lists, and I have a list of tuples. 回答1: Using Format Specification Mini-Language print "HEADER1

问题 Just curious why I can't do this: let myFn (data : obj) = match data with | :? (string * string) as (s1, s2) -> sprintf "(%s, %s)" s1 s2 |> Some | :? (string * string * int) as (s1, s2, i) -> sprintf "(%s, %s, %d)" s1 s2 i |> Some | _ -> None How come? 回答1: See F# spec, section 7.3 "As patterns" An as pattern is of the form pat as ident Which means you need to use an identifier after as : let myFn (data : obj) = match data with | :? (string * string) as s1s2 -> let (s1, s2) = s1s2 in sprintf

问题 I want to ask if something like this is possible in python: a,b = [i,i+1 for i in range(5)] I know this isn't possible because I have got an error, but I think you understand what I am trying to achieve. Let me clear it up, I can do : a,b = 3+2,3 Edit ---> Or even better: a,b = [0,1,2,3,4],[1,2,3,4,5] I wan't a similar thing in my first code example. I am trying to assign variables 'a' and 'b' as list, with list comprehension, but using tuple as assignment, the point is I don't want to use

问题 Given n and k , I need to create all tuples of length k whose entries are from range(n) (0 to n-1) such that the tuple's entries are in dictionary order and there are parentheses in a particular format. Specifically, the tuple has parentheses around each pair, from inside out. For example, if n=3 and k=4 , then I would like the output to include something like (((0,0),1),2) , but not something like (((0,0),2),1) . The code below works for this specific instance. The issue is that I don't know

问题 This question already has answers here : F# how to return have value a tuple or null (3 answers) Closed 24 days ago . I have this type-correct C# function: static System.Tuple<int,int> f(int n) { switch (n) { case 0: return null; default: return System.Tuple.Create(n,n+1); } } I try to re-implement it in F#: let f = function | 0 -> null | n -> System.Tuple.Create(n, n+1) Type-checker disagrees, though: error FS0001: The type '('a * 'b)' does not have 'null' as a proper value. How do I re

问题 I have the following list of tuples: [('a', 1), ('a', 1), ('b', 1), ('c',1), ('a', 1), ('c', 1)] I would like to know if I can utilize python's reduce function to aggregate them and produce the following output : [('a', 3), ('b', 1), ('c', 2)] Or if there are other ways, I would like to know as well (loop is fine) 回答1: It seems difficult to achieve using reduce , because if both tuples that you "reduce" don't bear the same letter, you cannot compute the result. How to reduce ('a',1) and ('b'

问题 I have the following list of tuples: [('a', 1), ('a', 1), ('b', 1), ('c',1), ('a', 1), ('c', 1)] I would like to know if I can utilize python's reduce function to aggregate them and produce the following output : [('a', 3), ('b', 1), ('c', 2)] Or if there are other ways, I would like to know as well (loop is fine) 回答1: It seems difficult to achieve using reduce , because if both tuples that you "reduce" don't bear the same letter, you cannot compute the result. How to reduce ('a',1) and ('b'

问题 I have the following list of tuples: [('a', 1), ('a', 1), ('b', 1), ('c',1), ('a', 1), ('c', 1)] I would like to know if I can utilize python's reduce function to aggregate them and produce the following output : [('a', 3), ('b', 1), ('c', 2)] Or if there are other ways, I would like to know as well (loop is fine) 回答1: It seems difficult to achieve using reduce , because if both tuples that you "reduce" don't bear the same letter, you cannot compute the result. How to reduce ('a',1) and ('b'