tuples

问题 I have a simple question. I have list, or a tuple, and I want to split it into many lists (or tuples) that contain the same elements. I'll try to be more clear using an example: (1,1,2,2,3,3,4) --> (1,1),(2,2),(3,3),(4,) (1,2,3,3,3,3) --> (1,),(2,),(3,3,3,3) [2,2,3,3,2,3] --> [2,2],[3,3],[2],[3] How can I do? I know that tuples and lists do not have the attribute "split" so i thought that i could turn them into strings before. This is what i tried: def splitt(l) x=str(l) for i in range (len(x

问题 I have a simple question. I have list, or a tuple, and I want to split it into many lists (or tuples) that contain the same elements. I'll try to be more clear using an example: (1,1,2,2,3,3,4) --> (1,1),(2,2),(3,3),(4,) (1,2,3,3,3,3) --> (1,),(2,),(3,3,3,3) [2,2,3,3,2,3] --> [2,2],[3,3],[2],[3] How can I do? I know that tuples and lists do not have the attribute "split" so i thought that i could turn them into strings before. This is what i tried: def splitt(l) x=str(l) for i in range (len(x

问题 i have a array of tuple like var contactsname = [(String,String)]()//firstname,lastname example = [(alex,joe),(catty,drling),(alex,fox),(asta,alex)] i need to search elements which are matching the firstname or lastname and return those all elements matching the key func searchElementsForkey(key:String)->[(string,String)]{ //code here } searchElementsForKey("alex") = [(alex,joe),(alex,fox),(asta,alex)] 回答1: You can go like this. var contactsname = [(String,String)]()//firstname,lastname

问题 My purpose is to determine the maximum number of consecutive equal elements in a given string or list. I'll try to be more clear using an example: (1,1,1,1) -> {1:4} (1,1,'a','a',1) ->{1:2, 'a':2} (2,2,2,0,2,2,0) -> {2:3, 0:1} I have tired with something like that but it does not work: d={} for i in range(len(l)): while l[i]==l[i+1]: d[i]=l.count(i) return d 回答1: One possible approach, using itertools.groupby: from itertools import groupby t = 1,1,'a','a',1 out = {} for v, g in groupby(t): l

问题 Consider I have a custom type (which I can extend): struct Foo { int a; string b; }; How can I make an instance of this object assignable to a std::tie , i.e. std::tuple of references? Foo foo = ...; int a; string b; std::tie(a, b) = foo; Failed attempts: Overloading the assignment operator for tuple<int&,string&> = Foo is not possible, since assignment operator is one of the binary operators which have to be members of the left hand side object. So I tried to solve this by implementing a

问题 I have a list of tuples, some of which are nested: [(name,(6,9.0,2.4),link),(name,(7.8,9.0,5),link)...] I would like to un-nest the inner tuple for each item in the list, but preserve the outer tuple: [(name,6,9.0,2.4,link),(name,7.8,9.0,5,link)...] This is different from the solution to the question posed here in which the solution sought to preserve the pairs. 回答1: Given lst = [('xyz',(6,9.0,2.4),'link1'),('abc',(7.8,9.0,5),'link2')] Iterate over lst and unpack the inner tuples into the

问题 I have a list of tuples, some of which are nested: [(name,(6,9.0,2.4),link),(name,(7.8,9.0,5),link)...] I would like to un-nest the inner tuple for each item in the list, but preserve the outer tuple: [(name,6,9.0,2.4,link),(name,7.8,9.0,5,link)...] This is different from the solution to the question posed here in which the solution sought to preserve the pairs. 回答1: Given lst = [('xyz',(6,9.0,2.4),'link1'),('abc',(7.8,9.0,5),'link2')] Iterate over lst and unpack the inner tuples into the

问题 I have a list of tuples, some of which are nested: [(name,(6,9.0,2.4),link),(name,(7.8,9.0,5),link)...] I would like to un-nest the inner tuple for each item in the list, but preserve the outer tuple: [(name,6,9.0,2.4,link),(name,7.8,9.0,5,link)...] This is different from the solution to the question posed here in which the solution sought to preserve the pairs. 回答1: Given lst = [('xyz',(6,9.0,2.4),'link1'),('abc',(7.8,9.0,5),'link2')] Iterate over lst and unpack the inner tuples into the

问题 In Python and Ruby (and others as well, I'm sure). you can prefix an enumerable with * ("splat") to use it as an argument list. For instance, in Python: >>> def foo(a,b): return a + b >>> foo(1,2) 3 >>> tup = (1,2) >>> foo(*tup) 3 Is there something similar in Haskell? I assume it wouldn't work on lists due to their arbitrary length, but I feel that with tuples it ought to work. Here's an example of what I'd like: ghci> let f a b = a + b ghci> :t f f :: Num a => a -> a -> a ghci> f 1 2 3 ghci

问题 I play with C++0x for some time and now I want to use variadic templates and tuple to implement class "Task". I'm going to pass Task objects into newly created threads (using pthread). Task class will contain function pointer to function which should be called inside thread and arguments for this function, simplified code: class TaskCaller { // ... virtual bool dispatch (void); }; template<typename ...T_arguments> Task : public TaskCaller { public: // ... Task (bool (*function) (T_arguments&.