折叠列表列表以生成列表

Question

So I've got the following list:

cd = [[0.1,0.2,0.1,0.9], [0.2, 0.3, 0.4, 0.1],[0.2,0.3,0.1,0.5]

and I'd like to get the product of the individual elements of the nested lists in a list format...

so I'm going for:

p = [(0.1*0.2*0.2),(0.2*0.3*0.3),(0.1*0.4*0.1),(0.9*0.1*0.5)]

note that this isn't a 1 to 1 relationship between cd and p.

I'd like to do it simply...

In F#, for example, I would just do list.fold, and use a list as my accumulator. Is there a python equivalent, or do I have to do:

p = [cd[0]]
if len(cd) > 1:
    for i in range(len(cd) - 1):
        for j in range(len(p)):
            p[j] = p[j]*cd[i+1][j]
return p

Answer 1

You can do it using a list comprehension.

cd = [[0.1,0.2,0.1,0.9], [0.2, 0.3, 0.4, 0.1]]
print [i*j for i,j in zip(cd[0],cd[1])]

If you want only 2 decimal places.

print [round(i*j,2) for i,j in zip(cd[0],cd[1])]

If you have multiple strings use

cd = [[0.1,0.2,0.1,0.9], [0.2, 0.3, 0.4, 0.1],[1.0,2.0,3.0,4.0]]
from operator import mul
print [reduce(mul, i, 1) for i in zip(*cd)]

Answer 2

You can do this:

[reduce(lambda a, b: a*b, x)  for x in zip(*cd)]

This should work for multiple lists, and doesn't require any imports.

As @DSM mentions, you would also have to "import functools" and use "functools.reduce" for python3.

Answer 3

You can combine reduce with zip (and we could use a lambda here for multiplication, but since we're importing anyway, we might as well use mul ):

>>> from operator import mul
>>> from functools import reduce
>>> cd = [[0.1,0.2,0.1,0.9], [0.2, 0.3, 0.4, 0.1],[0.2,0.3,0.1,0.5]]
>>> [reduce(mul, ops) for ops in zip(*cd)]
[0.004000000000000001, 0.018, 0.004000000000000001, 0.045000000000000005]

(Python 3; you don't need to import reduce if you're using an outdated Python.)

Answer 4

You have a general solution with Numpy, working with more than two sublists:

import numpy as np
cd = [[0.1,0.2,0.1,0.9], [0.2, 0.3, 0.4, 0.1]]
out = np.apply_along_axis(np.prod, 0, cd) 

Answer 5

尝试：

out = [ reduce(lambda x, y: x*y, z) for z in zip(*cd) ]