有沒有更pythonic的方式來做到這一點？輸入：list1 = [['a',1],['b',2],['c',3],['d',4]]list2 = [['e',5],['f',6],['g',7],['h',8]]期望的輸出：out = [['a',1],['e',5],['b',2],['f',6],['c',3],['g',7],['d',4] ,['h',8]]我已經做好了：def mergePreserveOrder(*argv): for arg in argv: for arg2 in argv: if(len(arg) != len(arg2)) : print("arrays size do not match" + str(arg) + str(arg2)) return output = [] for index in range (len(argv[0])): for arg in argv: output.append(arg[index]) return outputmergePreserveOrder (list1 ,list2 )
3 回答
倚天杖
TA貢獻1828條經驗 獲得超3個贊
您可以zip一起使用chain.from_iterable：
list(chain.from_iterable(zip(list1, list2)))
示例：
from itertools import chain
list1 = [['a',1],['b',2],['c',3],['d',4]]
list2 = [['e',5],['f',6],['g',7],['h',8]]
print(list(chain.from_iterable(zip(list1, list2))))
# [['a', 1], ['e', 5], ['b', 2], ['f', 6], ['c', 3], ['g', 7], ['d', 4], ['h', 8]]
ABOUTYOU
TA貢獻1812條經驗 獲得超5個贊
只需itertools.chain使用zip：
>>> import itertools
>>> list(itertools.chain(*zip(list1, list2)))
[['a', 1], ['e', 5], ['b', 2], ['f', 6], ['c', 3], ['g', 7], ['d', 4], ['h', 8]]
qq_遁去的一_1
TA貢獻1725條經驗 獲得超8個贊
最通用的解決方案是模塊roundrobin中的配方：itertools
from itertools import cycle, islice
def roundrobin(*iterables):
? ? "roundrobin('ABC', 'D', 'EF') --> A D E B F C"
? ? # Recipe credited to George Sakkis
? ? num_active = len(iterables)
? ? nexts = cycle(iter(it).__next__ for it in iterables)
? ? while num_active:
? ? ? ? try:
? ? ? ? ? ? for next in nexts:
? ? ? ? ? ? ? ? yield next()
? ? ? ? except StopIteration:
? ? ? ? ? ? # Remove the iterator we just exhausted from the cycle.
? ? ? ? ? ? num_active -= 1
? ? ? ? ? ? nexts = cycle(islice(nexts, num_active))
list1 =? [['a',1],['b',2],['c',3],['d',4]]
list2 =? [['e',5],['f',6],['g',7],['h',8]]
print(list(roundrobin(list1, list2)))
產生你想要的輸出。
chain與+不同zip，此解決方案適用于不均勻長度的輸入，其中zip將靜默丟棄超出最短輸入長度的所有元素。
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