以 python 保留顺序合并两个或多个列表的最佳方法

Question

有没有更pythonic的方式来做到这一点？输入：list1 =  [['a',1],['b',2],['c',3],['d',4]]list2 =  [['e',5],['f',6],['g',7],['h',8]]期望的输出：out = [['a',1],['e',5],['b',2],['f',6],['c',3],['g',7],['d',4] ,['h',8]]我已经做好了：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  )

Answer 1

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]]

Answer 2

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]]

Answer 3

最通用的解决方案是模块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)))

产生你想要的输出。

chain与+不同zip，此解决方案适用于不均匀长度的输入，其中zip将静默丢弃超出最短输入长度的所有元素。