您可以使用zip
和列表理解這裏:
>>> a = ['x','y','z']
>>> b = [1,2,3]
>>> [[x]*y for x,y in zip(a,b)]
[['x'], ['y', 'y'], ['z', 'z', 'z']]
或:
>>> [[x for _ in xrange(y)] for x,y in zip(a,b)]
[['x'], ['y', 'y'], ['z', 'z', 'z']]
zip
將首先在內存中創建整個列表,以獲得一個迭代器使用itertools.izip
如果a
包含列表或列表列表等可變對象,則可能需要使用copy.deepcopy
這裏是因爲改變一個副本會改變其他副本以及:
>>> from copy import deepcopy as dc
>>> a = [[1 ,4],[2, 5],[3, 6, 9]]
>>> f = [[dc(x) for _ in xrange(y)] for x,y in zip(a,b)]
#now all objects are unique
>>> [[id(z) for z in x] for x in f]
[[172880236], [172880268, 172880364], [172880332, 172880492, 172880428]]
timeit
比較(忽略進口):
>>> a = ['x','y','z']*10**4
>>> b = [100,200,300]*10**4
>>> %timeit [[x]*y for x,y in zip(a,b)]
1 loops, best of 3: 104 ms per loop
>>> %timeit [[x]*y for x,y in izip(a,b)]
1 loops, best of 3: 98.8 ms per loop
>>> %timeit map(lambda v: [v[0]]*v[1], zip(a,b))
1 loops, best of 3: 114 ms per loop
>>> %timeit map(list, map(repeat, a, b))
1 loops, best of 3: 192 ms per loop
>>> %timeit map(list, imap(repeat, a, b))
1 loops, best of 3: 211 ms per loop
>>> %timeit map(mul, [[x] for x in a], b)
1 loops, best of 3: 107 ms per loop
>>> %timeit [[x for _ in xrange(y)] for x,y in zip(a,b)]
1 loops, best of 3: 645 ms per loop
>>> %timeit [[x for _ in xrange(y)] for x,y in izip(a,b)]
1 loops, best of 3: 680 ms per loop
我建議提一提'itertools.izip'。 – kirelagin
謝謝!雖然'for'循環仍然會減慢代碼的速度嗎? – user2444731
@ user2444731你是什麼意思「慢」?你爲什麼認爲''''循環慢?比較慢什麼? – kirelagin