list.extend和list comprehension

Question

當我需要在列表中添加幾個相同的項目時，我使用list.extend：

a = ['a', 'b', 'c']
a.extend(['d']*3)

結果

['a', 'b', 'c', 'd', 'd', 'd']

但是，如何與列表理解類似？

a = [['a',2], ['b',2], ['c',1]]
[[x[0]]*x[1] for x in a]

結果

[['a', 'a'], ['b', 'b'], ['c']]

但我需要這個

['a', 'a', 'b', 'b', 'c']

有任何想法嗎？

Answer 1

堆疊式LC。

[y for x in a for y in [x[0]] * x[1]]

Answer 2

>>> a = [['a',2], ['b',2], ['c',1]]
>>> [i for i, n in a for k in range(n)]
['a', 'a', 'b', 'b', 'c']

Answer 3

一個itertools方法：

import itertools

def flatten(it):
    return itertools.chain.from_iterable(it)

pairs = [['a',2], ['b',2], ['c',1]]
flatten(itertools.repeat(item, times) for (item, times) in pairs)
# ['a', 'a', 'b', 'b', 'c']

Answer 4

如果您更喜歡擴展列表推導：

a = []
for x, y in l:
    a.extend([x]*y)

Answer 5

import operator
a = [['a',2], ['b',2], ['c',1]]
nums = [[x[0]]*x[1] for x in a]
nums = reduce(operator.add, nums)

Answer 6

>>> a = [['a',2], ['b',2], ['c',1]]
>>> sum([[item]*count for item,count in a],[])
['a', 'a', 'b', 'b', 'c']