问题
Hello so I want to multiply the integers inside a list.
For example;
l = [1, 2, 3]
l = [1*2, 2*2, 3*2]
output:
l = [2, 4, 6]
So I was searching online and most of the answers were regarding multiply all the integers with each other such as:
[1*2*3]
回答1:
Try a list comprehension:
l = [x * 2 for x in l]
This goes through l, multiplying each element by two.
Of course, there's more than one way to do it. If you're into lambda functions and map, you can even do
l = map(lambda x: x * 2, l)
to apply the function lambda x: x * 2 to each element in l. This is equivalent to:
def timesTwo(x):
return x * 2
l = map(timesTwo, l)
回答2:
The most pythonic way would be to use a list comprehension:
l = [2*x for x in l]
If you need to do this for a large number of integers, use numpy arrays:
l = numpy.array(l, dtype=int)*2
A final alternative is to use map
l = list(map(lambda x:2*x, l))
回答3:
Another functional approach which is maybe a little easier to look at than an anonymous function if you go that route is using functools.partial to utilize the two-parameter operator.mul with a fixed multiple
>>> from functools import partial
>>> from operator import mul
>>> double = partial(mul, 2)
>>> list(map(double, [1, 2, 3]))
[2, 4, 6]
回答4:
The simplest way to me is:
map((2).__mul__, [1, 2, 3])
回答5:
using numpy :
In [1]: import numpy as np
In [2]: nums = np.array([1,2,3])*2
In [3]: nums.tolist()
Out[4]: [2, 4, 6]
回答6:
#multiplying each element in the list and adding it into an empty list
original = [1, 2, 3]
results = []
for num in original:
results.append(num*2)# multiply each iterative number by 2 and add it to the empty list.
print(results)
来源：https://stackoverflow.com/questions/26446338/how-to-multiply-all-integers-inside-list