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Enter the user's input in the n-inner lists

I am trying to figure out, how would it be possible to insert in an input number of inner lists of the list different users inputs? n = int (input ("Please enter a number:")) L = [[] for i in range (n)] for i in range (n): L[0:n].append (input ("Please enter 3 words or letters or numbers:")) For example, user gives n = 3. Then the user is supposed to receive three times the question to enter a string. All three strings should enter accordingly each one into a separate inner list. Three days of battle with this thing... Probably still dont understand how the Loop iteration works. :(

12/28/2018 11:10:17 AM

Daniella

5 Answers

New Answer

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Did you mean something like this: some_list=[ [] for i in range(n) ] for i in range(n) : some_list[i].append(input('Enter a string here: ')

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I'd do it like this: inner_list = [] for i in range(n): inner_list.append([input()]) L.append(inner_list) (Or the same model in list comp shape.) One question: Why don't you just append the user inputs? Is it necessary to put them into an extra list? Without it, an inner list might look like this: ['first', 'second', 'and third input'] And one opinion: Usually it would be more convenient to let the user decide on the fly, how many inputs they want to do. That could look like this: inner_list = [] while True: x = input() if not x: break inner_list.append(x) L.append(inner_list) In this pattern, the user keeps appending as long as they like, but when they just hit enter, the list gets appended to the larger list.

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Thank you! The second option is understanble for me and it works. Another question has popped up: the user enters a string like "abc", can i slice it directly into the inner list as three strings "a", "b", "c"? And the second string "def" into "d", "e", "f" etc?

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You can split any string by just using list(string).

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Since 'Strings' are iterable in Python so you can use extend method of list to add each character of a string into a list or you can convert it directly into a list as above said by HonFu.