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I am attempting to create list of lists using for loops in python. My plan is to use two separate loops; one loop to create big lists to put the small lists in, and another loop to create the small lists and append them to the big lists.

This is being used to make a 'battleship' type game where there is a 10x10 grid.

This is the chunk of code I am having difficult with:

for i in range(0,10):
    (newlist,(i))=[]

The only task this specific code is meant to complete is to create 10 new lists, each with a different name. For example, the first circulation of the loop would create list0, the second circulation list1 and so on up to list9.

Theoretically, I see nothing wrong with this code. It also does work when instead of creating a new list, you are putting a string into the new variable.

Whenever I run the program, I get the error 

ValueError: not enough values to unpack (expected 2, got 0)

I have no idea why this occurs, and would be very grateful if anyone here could help me out.

furas
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Link
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    Don't use `list` as an identifier -- it is a built-in. Also, it isn't very clear what you expect `(list,(i))=[]` to accomplish. – John Coleman Nov 20 '16 at 17:10
  • Thanks John. Unfortunately, changing list as the identifier doesn't do anything to help the problem. The purpose of the line of code you mentioned was to create 9 different lists, each with a different name. – Link Nov 20 '16 at 17:13
  • don't create lists with different names - create list with sublists or dictionary. – furas Nov 20 '16 at 17:13
  • @Link It wasn't meant to fix the problem -- it was meant to fix a naming blunder. – John Coleman Nov 20 '16 at 17:14
  • Possible duplicate of [How to define two-dimensional array in python](http://stackoverflow.com/questions/6667201/how-to-define-two-dimensional-array-in-python) – OneCricketeer Nov 20 '16 at 17:14

3 Answers3

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You seem to have a variable called list, which you want to populate with some lists. Your code doesn't work as this syntax is used to extract two pieces of data from the iterable on the right hand side. Clearly, [] is empty, so you can extract nothing out of it.

You could do it like this:

your_list = [list() for x in range(9)]

Note that you shouldn't call the variable list as there exists a built-in function with the same name that constructs an empty list. Right now the variable makes the built-in unaccessible.

Edit: If you need to have 10 lists of lists:

your_list = [[[] for x in range(9)] for y in range(10)]

Then, your_list will be a list containing 10 lists of lists.

ForceBru
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  • Cheers lad, much appreciated. However, that doesn't completely answer my full question. You've helped me out with creating one list of a list using a minimal amount of code, which is still part of the problem. However I need 10 lists of lists, each list having a different name identifier. For example, your_list1, your_list2 etc. I need 10 of these lists to be created by a loop to create a 10x10 grid. This is all so I can reference a position in the grid, such as list 4 item 7. – Link Nov 20 '16 at 17:23
  • You should absolutely never do that. – Mous May 18 '22 at 22:30
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Better create list with sublists because it is easier to use later,

all_lists = []
for i in range(10):
    all_lists.append([])

or shorter

all_lists = [ [] for i in range(10) ]

And now you can access lists using index, ie.

all_lists[0]

Eventually create dictionary - then you can use number or string as index 

all_lists = {}
for i in range(10):
    all_lists[i] = []  # first version
    # or 
    all_lists[str(i)] = [] # second version

or shorter

all_lists = { i:[] for i in range(10) } # first version
all_lists = { str(i):[] for i in range(10) } # first version

And now you can access lists using number or string, ie.

all_lists[0]   # first version
# or 
all_lists["0"] # second version

EDIT: to create 10x10 grid you can use

grid = [ [None]*10 for _ in range(10) ]

You can use different value instead of None as default.

And to get position (x,y) you can use 

grid[y][x]

First is y (row), second is x (column)

furas
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  • What's the reason for converting to `str` for use in the dictionary exactly? (and not showing the equivalent of `{i:[] for i in range(9)}`) ? – Jon Clements Nov 20 '16 at 17:16
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sometimes it happen because we declare and set values for two object in one line like this line: a,e=[]

not this one which is correct line a,e=[],[]

Spring98
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