How can I use a list comprehension to extend a lis

问题:

This question already has an answer here:

• list.extend and list comprehension 6 answers

I'm not experienced in Python, and I often write code that (simplified) looks like this:

accumulationList = []
for x in originalList:
    y = doSomething(x)
    accumulationList.append(y)
return accumulationList

Then after my test passes, I refactor to

return [doSomething(x) for x in originalList]

But suppose it turns out a little different, and my loop looks like this:

accumulationList = []
for x in originalList:
    y = doSomething(x)
    accumulationList.extend(y)
return accumulationList

where the doSomething list returns a list. What is the most Pythonic way to accomplish this? Obviously, the previous list comprehension would give a list of lists.

回答1:

Much simpler and cleaner with list comprehension:

[y for x in originalList for y in doSomething(x)]

回答2:

Do you mean something like this?

accumulationList = []
for x in originalList:
    accumulationList.extend(doSomething(x))
return accumulationList

or shorter code (but not optimal):

return sum((doSomething(x) for x in originalList), [])

or the same:

return sum(map(doSomething, originalList), [])

Thanks to @eyquem for the hint (if using Python 2.x):

import itertools as it

return sum(it.imap(doSomething, originalList), [])

回答3:

Python's in-place add operator (+=, available as iadd in operator module) is equivalent of .extend for list. Pair it with reduce to get what you want.

import operator

reduce(operator.iadd, (doSomething(x) for x in originalList)
, accumulation_list)

回答4:

I think the answers involving add or iadd run in quadratic time, which probably isn't good. I'd try:

from itertools import chain
accumulation_list = list(chain.from_iterable(doSomething(x) for x in originalList))

回答5:

I don't think there is special syntax for this case. But you could make the for loop shorter:

accumulationList += doSomething(x)

If you insist, you could use functional programming to flatten the list:

result = reduce(lambda a,b: a+b, [[i,i*2] for i in range(3)])

But I wouldn't call this pythonic, I think it's harder to read than a for loop.

回答6:

Functionally, you can use itertools.chain with map. For an input list L:

res = list(chain.from_iterable(map(doSomething, L)))

If you need an iterator, simply remove the list call. Here's a demo:

def doSomething(value):
    return [value * i for i in range(1, 4)]

def original(originalList):
    accumulationList = []
    for x in originalList:
        y = doSomething(x)
        accumulationList.extend(y)
    return accumulationList

def new(L):
    return list(chain.from_iterable(map(doSomething, L)))

x = [1, 2, 3]

assert original(x) == new(x)

print(new(x))

[1, 2, 3, 2, 4, 6, 3, 6, 9]