How to fill a list

Question

I have to make a function that takes an empty list as first argument and n as secound argument, so that:

L=[]
function(L,5)
print L
returns:
[1,2,3,4,5]

I was thinking:

def fillList(listToFill,n):
    listToFill=range(1,n+1)

but it is returning an empty list.

score 25 · Accepted Answer · answered Dec 08 '11 at 11:28

Consider the usage of extend:

>>> l = []
>>> l.extend(range(1, 6))
>>> print l
[1, 2, 3, 4, 5]
>>> l.extend(range(1, 6))
>>> print l
[1, 2, 3, 4, 5, 1, 2, 3, 4, 5]

If you want to make a function (doing the same):

def fillmylist(l, n):
    l.extend(range(1, n + 1))
l = []
fillmylist(l, 5)

score 4 · Answer 2 · answered Dec 08 '11 at 11:02

4

A function without an explicit return or yield returns None. What you want is

def fill_list(l, n):
    for i in xrange(1, n+1):
        l.append(i)
    return l

but that's very unpythonic. You'd better just call range(1, n+1) which also returns the list [1,2,3,4,5] for n=5:

def fill_list(n):
    return range(1, n+1)

answered Dec 08 '11 at 11:02

Fred Foo

355,277
75
744
836

The function should not return anything but fill the list passed to it with integers from 1 to n. – Linus Svendsson Dec 08 '11 at 11:04
@LinusSvendsson: then remove the `return` from my suggestion. – Fred Foo Dec 08 '11 at 11:05
the function should take two arguments, the list to fill and n! – Linus Svendsson Dec 08 '11 at 11:15
@LinusSvendsson: my first suggestion does that. – Fred Foo Dec 08 '11 at 11:20

score 2 · Answer 3 · answered Sep 24 '19 at 10:36

2

And a bit shorter example of what you want to do:

l = []
l.extend(range(1, 5))
l.extend([0]*3)

print(l)

answered Sep 24 '19 at 10:36

user5214530

467
5
11

Antoine Gallix · Answer 4 · 2014-03-05T17:36:02.030

If you do :

def fillList(listToFill,n): listToFill=range(1,n+1)

a new list is created inside the function scope and disappears when the function ends. useless.

With :

def fillList(listToFill,n): listToFill=range(1,n+1) return listToFill()

you return the list and you must use it like this:

newList=fillList(oldList,1000)

And finally without returning arguments:

def fillList(listToFill,n): listToFill.extend(range(1,n+1))

and call it like this:

fillList(oldList,1000)

Conclusion:

Inside a function , if you want to modify an argument you can reassign it and return it, or you can call the object's methods and return nothing. You cannot just reassign it like if you were outside of the function and return nothing, because it's not going to have effect outside the function.

score 0 · Answer 5 · answered Dec 08 '11 at 11:34

In

def fillList(listToFill,n):
    listToFill=range(1,n+1)

you change only the pointer of listToFill, if you don't return the new pointer; the new pointer isn't available out of the function and you have still the pointer of your empty list (in outer scope).

How to fill a list

5 Answers5

Linked