The following code tries to create an integer array filled with n times number 1.
import sys
def foo(n):
if n == 0:
return []
else:
return foo(n-1).append(1)
if __name__ == '__main__':
foo(5)
Executing this program yields in an error:
AttributeError: 'NoneType' object has no attribute 'append'
What am I doing wrong when creating the array?
Just look at the following code to understand why you are getting the error,
>>> x = [].append(1)
>>> x is None
True
When you append to a list, the return value is None! So you must do something like this,
def foo(n):
if n == 0:
return []
else:
return foo(n-1) + [1]
Using + operator is really like calling extend on a list for which the return value is the new list, unlike append.
>>> x = [1] + [1]
>>> x
[1, 1]
NOTE: Obviously for this simple example you should just use,
>>> [1] * 6
[1, 1, 1, 1, 1, 1]
Which is fine for immutable ints but if you are dealing with objects where you don't want references to the same one,
>>> [1 for _ in range(6)]
[1, 1, 1, 1, 1, 1]
But I'm assuming you are writing this to practice recursive solutions and such.
list(itertools.repeat(1, 6))The problem is in your else-clause. append does not return a new list, but rather adds an element to the list in-place, and then returns None (hence your error). Try this instead,
return foo(n-1) + [1] # creates a *new* list
result = foo(n-1) result.append(1) return result
it might be worth noting that python has some nice syntax to cover your usecase:
>>> [1]*5
[1, 1, 1, 1, 1]
Your program, slightly changed
import sys
def foo(n):
if n == 0:
return []
else:
return foo(n-1) + [1]
if __name__ == '__main__':
print(foo(5))
Prints
[1, 1, 1, 1, 1]
list.append() modifies the list, but doesn't return anything. So recursions of your function that reach that branch actually return nothing or None.
The method I listed appends an element to the list, and then returns the list, so your recursion works just as you had wanted.
append returns None. This is the problem.
