List append() in for loop
The list.append
function does not return any value(but None
), it just adds the value to the list you are using to call that method.
In the first loop round you will assign None
(because the no-return of append
) to a
, then in the second round it will try to call a.append
, as a is None
it will raise the Exception you are seeing
You just need to change it to:
a=[]
for i in range(5):
a.append(i)
print(a)
# [0, 1, 2, 3, 4]
list.append
is what is called a mutating or destructive method, i.e. it will destroy or mutate the previous object into a new one(or a new state).
If you would like to create a new list based in one list without destroying or mutating it you can do something like this:
a=['a', 'b', 'c']
result = a + ['d']
print result
# ['a', 'b', 'c', 'd']
print a
# ['a', 'b', 'c']
As a corollary only, you can mimic the append
method by doing the following:
a=['a', 'b', 'c']
a = a + ['d']
print a
# ['a', 'b', 'c', 'd']
You don't need the assignment, list.append(x)
will always append x
to a
and therefore there's no need te redefine a
.
a = []
for i in range(5):
a.append(i)
print(a)
is all you need. This works because list
s are mutable.
Also see the docs on data structures.
No need to re-assign.
a=[]
for i in range(5):
a.append(i)
a