How do I multiply each element in a list by a number?
A blazingly faster approach is to do the multiplication in a vectorized manner instead of looping over the list. Numpy has already provided a very simply and handy way for this that you can use.
>>> import numpy as np
>>>
>>> my_list = np.array([1, 2, 3, 4, 5])
>>>
>>> my_list * 5
array([ 5, 10, 15, 20, 25])
Note that this doesn't work with Python's native lists. If you multiply a number with a list it will repeat the items of the as the size of that number.
In [15]: my_list *= 1000
In [16]: len(my_list)
Out[16]: 5000
If you want a pure Python-based approach using a list comprehension is basically the most Pythonic way to go.
In [6]: my_list = [1, 2, 3, 4, 5]
In [7]: [5 * i for i in my_list]
Out[7]: [5, 10, 15, 20, 25]
Beside list comprehension, as a pure functional approach, you can also use built-in map()
function as following:
In [10]: list(map((5).__mul__, my_list))
Out[10]: [5, 10, 15, 20, 25]
This code passes all the items within the my_list
to 5
's __mul__
method and returns an iterator-like object (in python-3.x). You can then convert the iterator to list using list()
built in function (in Python-2.x you don't need that because map
return a list by default).
benchmarks:
In [18]: %timeit [5 * i for i in my_list]
463 ns ± 10.6 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [19]: %timeit list(map((5).__mul__, my_list))
784 ns ± 10.7 ns per loop (mean ± std. dev. of 7 runs, 1000000 loops each)
In [20]: %timeit [5 * i for i in my_list * 100000]
20.8 ms ± 115 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [21]: %timeit list(map((5).__mul__, my_list * 100000))
30.6 ms ± 169 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)
In [24]: arr = np.array(my_list * 100000)
In [25]: %timeit arr * 5
899 µs ± 4.98 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
You can just use a list comprehension:
my_list = [1, 2, 3, 4, 5]
my_new_list = [i * 5 for i in my_list]
>>> print(my_new_list)
[5, 10, 15, 20, 25]
Note that a list comprehension is generally a more efficient way to do a for
loop:
my_new_list = []
for i in my_list:
my_new_list.append(i * 5)
>>> print(my_new_list)
[5, 10, 15, 20, 25]
As an alternative, here is a solution using the popular Pandas package:
import pandas as pd
s = pd.Series(my_list)
>>> s * 5
0 5
1 10
2 15
3 20
4 25
dtype: int64
Or, if you just want the list:
>>> (s * 5).tolist()
[5, 10, 15, 20, 25]
Finally, one could use map
, although this is generally frowned upon.
my_new_list = map(lambda x: x * 5, my_list)
Using map
, however, is generally less efficient. Per a comment from ShadowRanger on a deleted answer to this question:
The reason "no one" uses it is that, in general, it's a performance pessimization. The only time it's worth considering
map
in CPython is if you're using a built-in function implemented in C as the mapping function; otherwise,map
is going to run equal to or slower than the more Pythonic listcomp or genexpr (which are also more explicit about whether they're lazy generators or eagerlist
creators; on Py3, your code wouldn't work without wrapping themap
call inlist
). If you're usingmap
with alambda
function, stop, you're doing it wrong.
And another one of his comments posted to this reply:
Please don't teach people to use
map
withlambda
; the instant you need alambda
, you'd have been better off with a list comprehension or generator expression. If you're clever, you can makemap
work withoutlambda
s a lot, e.g. in this case,map((5).__mul__, my_list)
, although in this particular case, thanks to some optimizations in the byte code interpreter for simpleint
math,[x * 5 for x in my_list]
is faster, as well as being more Pythonic and simpler.