list

问题 My question is: how to write a procedure that utilises tailcall, and that constructs a list not in the reverse order. To show what I mean, here is an example of a very simple procedure that is iterative, and that creates a copy of a list: (define (copy-list ls) (define (iter cp-ls rest-ls) (if (null? rest-ls) cp-ls (iter (cons (car rest-ls) cp-ls) (cdr rest-ls)))) (iter '() ls)) The problem is that, due to the iterative order in which the elements are cons ed together, the returned list ends

问题 I am trying to code a function where it gives me the indexes of where either list_2 or list_3 crosses list_ . So it would give me the points of intersection if there are any in numpy code. I want to get the crosses in order, so the list of indexes have to be formatted so that it would give me a cross in order of list_2 cross, list_3 cross , list_2 cross or list_3 cross, list_2 cross , list_3 cross etc. So if a cross has happened it has to wait for the other array values to cross the list

问题 I am trying to code a function where it gives me the indexes of where either list_2 or list_3 crosses list_ . So it would give me the points of intersection if there are any in numpy code. I want to get the crosses in order, so the list of indexes have to be formatted so that it would give me a cross in order of list_2 cross, list_3 cross , list_2 cross or list_3 cross, list_2 cross , list_3 cross etc. So if a cross has happened it has to wait for the other array values to cross the list

问题 I am trying to code a function where it gives me the indexes of where either list_2 or list_3 crosses list_ . So it would give me the points of intersection if there are any in numpy code. I want to get the crosses in order, so the list of indexes have to be formatted so that it would give me a cross in order of list_2 cross, list_3 cross , list_2 cross or list_3 cross, list_2 cross , list_3 cross etc. So if a cross has happened it has to wait for the other array values to cross the list

问题 I have a list that looks like this: [[[u'example', u'example2'], [u'example', u'example2'], [u'example', u'example2'], [u'example', u'example2'], [u'example', u'example2']], [[5.926582278481011, 10.012500000000001, 7.133823529411763, 8.257352941176471, 7.4767647058823545]]] I want to save this list to an Excel file in the following way: Column 1: [example, example, ..., example] Column 2: [example2, example2, ..., example2] Column 3: [5.926582278481011, 10.012500000000001, ..., 7

问题 I would like to remove non-alpha characters from a string and convert each word into a list component such that: "All, the above." -> ["all", "the", "above"] It would seem that the following function works: re.split('\W+', str) but it does not account for corner cases. For example: "The U.S. is where it's nice." -> ["the", "U", "S", "is", "where", "it", "s", "nice"] I want the period removed but neither the apostrophe or the periods in "U.S." My idea is to create a regex where spaces are

问题 I would like to remove non-alpha characters from a string and convert each word into a list component such that: "All, the above." -> ["all", "the", "above"] It would seem that the following function works: re.split('\W+', str) but it does not account for corner cases. For example: "The U.S. is where it's nice." -> ["the", "U", "S", "is", "where", "it", "s", "nice"] I want the period removed but neither the apostrophe or the periods in "U.S." My idea is to create a regex where spaces are

问题 I'd like to be able to change the number of loops and store all possibilities in a (list of lists). Let me explain by an example: You have the following list: initial_list=[1,2] Then you have a parameter n, which is the number of loops For example if n=2, I'd like my programm to return this list: final_list=[[x,x,1,2],[x,1,x,2],[x,1,2,x],[1,x,x,2],[1,x,2,x][1,2,x,x]] So for n=3, the program would do the same with 3 'x', etc Also, i'd like my initial_list could have different length, like [1,2

问题 I thought that [] and list() were two equal ways to create a list. But if you want a list with dictionnary keys, var = [a_dict.keys()] doesn't work since type(var) is [dict_keys] , correct syntax is : var = list(a_dict.keys()) I couldn't find an good explanation on this behaviour. Do you have one ? 回答1: TL;DR: list() is the same as [] list(obj) is not the same as [obj] a_dict.keys() is a dictionary view object, it returns an object which can be iterated to yield the keys of a_dict . So this

问题 I thought that [] and list() were two equal ways to create a list. But if you want a list with dictionnary keys, var = [a_dict.keys()] doesn't work since type(var) is [dict_keys] , correct syntax is : var = list(a_dict.keys()) I couldn't find an good explanation on this behaviour. Do you have one ? 回答1: TL;DR: list() is the same as [] list(obj) is not the same as [obj] a_dict.keys() is a dictionary view object, it returns an object which can be iterated to yield the keys of a_dict . So this