Python: Append item to list N times

Question

This seems like something Python would have a shortcut for. I want to append an item to a list N times, effectively doing this:

l = []
x = 0
for i in range(100):
    l.append(x)

It would seem to me that there should be an "optimized" method for that, something like:

l.append_multiple(x, 100)

Is there?

_{If you want to append elements from another sequence (instead of the same value repeatedly, you can just .extend with that sequence directly. See How to append to the end of an empty list?.}

Ended up going with `l.extend([x] * 100)` since it fit my situation best. Amber gets the answer credit for being the most comprehensive. Thanks! — Toji, Jan 11 '11 at 05:27
Make sure you understand Python's reference semantics before using this for anything non-trivial. The x's appended to l are all the same x, so mutating one of them affects them all (all one of them, referred to 100 times). — Karl Knechtel, Jan 11 '11 at 09:35
For clarification: I am using it for integers, so my usage is safe. — Toji, Jan 11 '11 at 14:36

Amber · Accepted Answer · 2011-01-11T05:20:06.760

For immutable data types:

l = [0] * 100
# [0, 0, 0, 0, 0, ...]

l = ['foo'] * 100
# ['foo', 'foo', 'foo', 'foo', ...]

For values that are stored by reference and you may wish to modify later (like sub-lists, or dicts):

l = [{} for x in range(100)]

(The reason why the first method is only a good idea for constant values, like ints or strings, is because only a shallow copy is does when using the <list>*<number> syntax, and thus if you did something like [{}]*100, you'd end up with 100 references to the same dictionary - so changing one of them would change them all. Since ints and strings are immutable, this isn't a problem for them.)

If you want to add to an existing list, you can use the extend() method of that list (in conjunction with the generation of a list of things to add via the above techniques):

a = [1,2,3]
b = [4,5,6]
a.extend(b)
# a is now [1,2,3,4,5,6]

Or combine both suggestions into the form of `a.extend(b*n)` where `n` is the times you want to repeat elements in b — DarkCygnus, Feb 08 '18 at 19:07

score 41 · Answer 2 · edited Jun 21 '13 at 22:53

41

Use extend to add a list comprehension to the end.

l.extend([x for i in range(100)])

See the Python docs for more information.

edited Jun 21 '13 at 22:53

Zero Piraeus

56,143
27
150
160

answered Jan 11 '11 at 05:15

Jake

12,713
18
66
96

2

Or get rid of the loop: `l.extend([x] * 100)` – EzPizza Jul 16 '23 at 17:35

score 13 · Answer 3 · answered Jan 11 '11 at 13:28

13

Itertools repeat combined with list extend.

from itertools import repeat
l = []
l.extend(repeat(x, 100))

answered Jan 11 '11 at 13:28

kevpie

25,206
2
24
28

score 9 · Answer 4 · answered Oct 09 '17 at 10:20

9

I had to go another route for an assignment but this is what I ended up with.

my_array += ([x] * repeated_times)

answered Oct 09 '17 at 10:20

CTS_AE

12,987
8
62
63

1

this is so ridiculously simple compared to the other ones, i love it – M1n3c4rt May 16 '23 at 09:07

score 7 · Answer 5 · answered Dec 07 '19 at 14:22

7

l = []
x = 0
l.extend([x]*100)

answered Dec 07 '19 at 14:22

Rajan saha Raju

794
7
13

score 3 · Answer 6 · answered Jan 11 '11 at 05:14

3

You could do this with a list comprehension

l = [x for i in range(10)];

answered Jan 11 '11 at 05:14

Mark Elliot

75,278
22
140
160

score 0 · Answer 7 · answered May 24 '21 at 20:40

you can add any value like this for several times:

a = [1,2,3]
b = []
#if you want to add on item 3 times for example:
for i in range(len(a)):
    j = 3
    while j != 0:
        b.append(a[i])
        j-=1
#now b = [1,1,1,2,2,2,3,3,3]

Python: Append item to list N times

7 Answers7

Linked

Related