交错4个相同长度的python列表[重复]

Question

我想在 python 中交错 4 个相同长度的列表。

我搜索了这个站点，只看到如何在 python 中交错 2： Interleaving two lists in Python

可以为 4 个列表提供建议吗？

我有这样的列表

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

我想要这样的列表

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

Answer 1

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))))

或者正如@PatrickHaugh 建议的那样使用chain.from_iterable：

list(<b>chain.from_iterable(zip</b>(l1, l2, l3, l4)))

Answer 2

如果列表长度相同，zip() 可用于交错四个列表，就像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 3

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)

Answer 4

来自 itertools 食谱

itertool recipes 提出了一个名为 roundrobin 的解决方案，它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

带切片

或者，这里有一个完全依赖切片的解决方案，但要求所有列表的长度相同。

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 5

另一种方法是使用zip 和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(l5) 或l5.tolist() 将其转换为list

Answer 6

为了额外的多样性（或者如果你需要用 Pandas 来做这件事）

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 7

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') 

顺便说一句，您实际上不需要制作数组，短版也可以使用

>>> 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')