Question

我想在python中使用相同长度的交错4列表。

我搜索这个网站，只看到如何在python中交错2： Interleaving two lists in Python

可以为4个清单提供建议吗？

我有这样的名单

l1 = ["a","b","c","d"]
l2 = [1,2,3,4]
l3 = ["w","x","y","z"]
l4 = [5,6,7,8]

我想要列表

l5 = ["a",1,"w",5,"b",2,"x",6,"c",3,"y",7,"d",4,"z",8]

Answer 1

如果列表的长度相同，zip()可用于交换四个列表，就像它在您链接的问题中用于交错两个列表一样：

>>> l1 = ["a", "b", "c", "d"]
>>> l2 = [1, 2, 3, 4]
>>> l3 = ["w", "x", "y", "z"]
>>> l4 = [5, 6, 7, 8]
>>> l5 = [x for y in zip(l1, l2, l3, l4) for x in y]
>>> l5
['a', 1, 'w', 5, 'b', 2, 'x', 6, 'c', 3, 'y', 7, 'd', 4, 'z', 8]

Answer 2

itertools.chain和zip：

from itertools import chain

l1 = ["a", "b", "c", "d"]
l2 = [1, 2, 3, 4]
l3 = ["w", "x", "y", "z"]
l4 = [5, 6, 7, 8]

print(list(chain(*zip(l1, l2, l3, l4))))

或者@PatrickHaugh建议使用chain.from_iterable：

list(chain.from_iterable(zip(l1, l2, l3, l4)))

Answer 3

来自itertools食谱

itertool recipes建议使用名为roundrobin的解决方案，该解决方案允许使用不同长度的列表。

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))

print(*roundrobin(*lists)) # a 1 w 5 b 2 x 6 c 3 y 7 d 4 z 8

切片

或者，这里的解决方案仅依赖于切片，但要求所有列表具有相同的长度。

l1 = ["a","b","c","d"]
l2 = [1,2,3,4]
l3 = ["w","x","y","z"]
l4 = [5,6,7,8]

lists = [l1, l2, l3, l4]

lst = [None for _ in range(sum(len(l) for l in lists))]

for i, l in enumerate(lists):
    lst[i:len(lists)*len(l):len(lists)] = l

print(lst) # ['a', 1, 'w', 5, 'b', 2, 'x', 6, 'c', 3, 'y', 7, 'd', 4, 'z', 8]

Answer 4

有关其他多样性（或者如果您需要使用Pandas进行此操作）

import pandas as pd
l1 = ["a","b","c","d"]
l2 = [1,2,3,4]
l3 = ["w","x","y","z"]
l4 = [5,6,7,8]

df = pd.DataFrame([l1 ,l2, l3, l4])
result = list(df.values.flatten('A'))

['a'，1，'w'，5，'b'，2，'x'，6，'c'，3，'y'，7，'d'，4，'z'， 8]

Answer 5

只是为了多样性，numpy.dstack然后flatten可以做同样的伎俩。

>>> import numpy as np
>>> l1 = ["a","b","c","d"]
>>> l2 = [1,2,3,4]
>>> l3 = ["w","x","y","z"]
>>> l4 = [5,6,7,8]
>>> np.dstack((np.array(l1),np.array(l2),np.array(l3),np.array(l4))).flatten()                                                                                                       
array(['a', '1', 'w', '5', 'b', '2', 'x', '6', 'c', '3', 'y', '7', 'd',                                                                                                              
       '4', 'z', '8'],                                                                                                                                                               
      dtype='|S21')

顺便说一下，你实际上不需要制作一个阵列，短版本也可以使用

>>> np.dstack((l1,l2,l3,l4)).flatten()                                                                                                       
array(['a', '1', 'w', '5', 'b', '2', 'x', '6', 'c', '3', 'y', '7', 'd',                                                                                                              
       '4', 'z', '8'],                                                                                                                                                               
      dtype='|S21')

Answer 6

另一种方法可能是zip使用np.concatenate：

import numpy as np
l5 = np.concatenate(list(zip(l1, l2, l3, l4)))
print(l5)

结果：

['a' '1' 'w' '5' 'b' '2' 'x' '6' 'c' '3' 'y' '7' 'd' '4' 'z' '8']

注意：l5是numpy.ndarray类型，您可以使用list或list(l5)将其转换为l5.tolist()

Answer 7

使用zip和reduce：

import functools, operator
>>> functools.reduce(operator.add, zip(l1,l2,l3,l4))
('a', 1, 'w', 5, 'b', 2, 'x', 6, 'c', 3, 'y', 7, 'd', 4, 'z', 8)

交错4列出相同长度的python

7 个答案:

来自itertools食谱

切片