How do I multiply each element in a list by a number?

Question

I have a list:

my_list = [1, 2, 3, 4, 5]

How can I multiply each element in my_list by 5? The output should be:

[5, 10, 15, 20, 25]

Answer 1

var1 = [2,4,6,8,10,12]

#for Integer multiplier us   int(x).__mull__ with map    
var2 = list( map(int(2).__mul__,var1 ))
#for float multiplier
var2 = list( map(float(2.5).__mul__,var1 ))

Answer 2

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 eager list creators; on Py3, your code wouldn't work without wrapping the map call in list). If you're using map with a lambda 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 with lambda; the instant you need a lambda, you'd have been better off with a list comprehension or generator expression. If you're clever, you can make map work without lambdas 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 simple int math, [x * 5 for x in my_list] is faster, as well as being more Pythonic and simpler.

Answer 3

I found it interesting to use list comprehension or map with just one object name x. Note that whenever x is reassigned, its id(x) changes, i.e. points to a different object.

x = [1, 2, 3]
id(x)
2707834975552
x = [1.5 * x for x in x]
id(x)
2707834976576
x
[1.5, 3.0, 4.5]
list(map(lambda x : 2 * x / 3, x))
[1.0, 2.0, 3.0]
id(x) # not reassigned
2707834976576
x = list(map(lambda x : 2 * x / 3, x))
x
[1.0, 2.0, 3.0]
id(x)
2707834980928

Answer 4

With map (not as good, but another approach to the problem):

list(map(lambda x: x*5,[5, 10, 15, 20, 25]))

also, if you happen to be using numpy or numpy arrays, you could use this:

import numpy as np
list(np.array(x) * 5)

Answer 5

Multiplying each element in my_list by k:

k = 5
my_list = [1,2,3,4]
result = list(map(lambda x: x * k, my_list))

resulting in: [5, 10, 15, 20]

Answer 6

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)

Answer 7

Best way is to use list comprehension:

def map_to_list(my_list, n):
# multiply every value in my_list by n
# Use list comprehension!
    my_new_list = [i * n for i in my_list]
    return my_new_list
# To test:
print(map_to_list([1,2,3], -1))

Returns: [-1, -2, -3]

Answer 8

Since I think you are new with Python, lets do the long way, iterate thru your list using for loop and multiply and append each element to a new list.

using for loop

lst = [5, 20 ,15]
product = []
for i in lst:
    product.append(i*5)
print product

using list comprehension, this is also same as using for-loop but more 'pythonic'

lst = [5, 20 ,15]

prod = [i * 5 for i in lst]
print prod

Answer 9

You can do it in-place like so:

 l = [1, 2, 3, 4, 5]
 l[:] = [x * 5 for x in l]

This requires no additional imports and is very pythonic.

Answer 10

from functools import partial as p
from operator import mul
map(p(mul,5),my_list)

is one way you could do it ... your teacher probably knows a much less complicated way that was probably covered in class