I want to create a function that takes a 2 dimensional list and outputs a 1 dimensional list with the same contents. Here is what I have:
twoDlist= [[23, 34, 67],[44,5,3],[7,8,9]]
def twoone (list1):
for x in range(len(list1)):
for y in range(len(list1)):
list2=[]
list2.append(list1[x][y])
print(twoone(twoDlist))
Only problem it returns 'None'. What is wrong, and how can I fix it?
Two issues here, the first is that you are not returning a value and the second is that you are resetting list2 to an empty list inside of your nested loop, so at the end you would only have a single element in it.
Here is how you could fix your code:
def twoone(list1):
list2 = []
for x in range(len(list1)):
for y in range(len(list1)):
list2.append(list1[x][y])
return list2
>>> twoone(twoDlist)
[23, 34, 67, 44, 5, 3, 7, 8, 9]
However, there are much better ways to do this, see the link in jtbandes' comment.
The best way is itertools.chain():
>>> from itertools import chain
>>> list(chain.from_iterable(twoDlist))
[23, 34, 67, 44, 5, 3, 7, 8, 9]
The reason your method returns None is that it does not have a return statement. Even with a return statement, however, the result is incorrect. As other answers point out, initializing list2 outside the loops (and returning list2) solves the issue.
You could also use a nested list comprehension to accomplish the task more compactly.
def twoone(list1):
return [val for lst in list1 for val in lst]
The reason that its happening is that you are reinitialise list2 as [] in every iteration. Try initialising the list2 part outside the for loop in twoone.
