python - 在 Python 中加入 2 个不同大小的列表

Question

尝试使用 2 个不同大小的列表时出现“索引列表超出范围”错误。

例子：

ListA = [None, None, None, None, None]
ListB = ['A', None, 'B']

for x, y in enumerate(ListA):
    if ListB[x]:
        ListA[x]=ListB[x]

这样做会得到“list of index out of range”错误，因为 ListB[3] 和 ListB[4] 不存在：
我希望加入 ListA 和 ListB 来让 ListA 看起来像这样：

ListA = ['A', None, 'B', None, None]

我怎样才能做到这一点？

Answer 1

使用itertools.izip_longest

from itertools import izip_longest
ListA = [b or a for a, b in izip_longest(ListA,ListB)]

Answer 2

最快的解决方案是使用切片分配

>>> ListA = [None, None, None, None, None]
>>> ListB = ['A', None, 'B']
>>> ListA[:len(ListB)] = ListB
>>> ListA
['A', None, 'B', None, None]

---

定时

>>> def merge_AO(ListA, ListB):
    return [ i[1] for i in map(None,ListA,ListB)]

>>> def merge_ke(ListA, ListB):
    for x in range(len(ListB)): #till end of b
        ListA[x]=ListB[x]
    return ListA

>>> def merge_JK(ListA, ListB):
    ListA = [b or a for a, b in izip_longest(ListA,ListB)]
    return ListA

>>> def merge_AB(ListA, ListB):
    ListA[:len(ListB)] = ListB
    return ListA

>>> funcs = ["merge_{}".format(e) for e in ["AO","ke","JK","AB"]]
>>> _setup = "from __main__ import izip_longest, ListA, ListB, {}"
>>> tit = [(timeit.Timer(stmt=f + "(ListA, ListB)", setup = _setup.format(f)), f) for f in funcs]
>>> for t, foo in tit:
    "{} took {} secs".format(t.timeit(100000), foo)


'0.259869612113 took merge_AO secs'
'0.115819095634 took merge_ke secs'
'0.204675467452 took merge_JK secs'
'0.0318886645255 took merge_AB secs'

Answer 3

试试这个：

>>> [i[1] for i in map(None,ListA,ListB)]
['A', None, 'B', None, None]

Answer 4

试试这个：

ListA = [None, None, None, None, None]
ListB = ['A', None, 'B']

for x in range(len(ListB)): #till end of b
    ListA[x]=ListB[x]

Answer 5

使用 MAP 避免列表索引超出范围错误

for iterator,tup in enumerate(map(None,ListA,ListB)):
    if tup[1]:
        ListA[iterator] = tup[1]

这将解决问题。