Question

我有一个对象列表，其中对象可以是列表或标量。我想要一个只有标量的扁平列表。例如：

L = [35,53,[525,6743],64,63,[743,754,757]]
outputList = [35,53,525,6743,64,63,743,754,757]

P.S。此问题的答案不适用于异构列表。 Flattening a shallow list in Python

Answer 1

这是一个相对简单的递归版本，它将展平列表的任何深度

l = [35,53,[525,6743],64,63,[743,754,757]]

def flatten(xs):
    result = []
    if isinstance(xs, (list, tuple)):
        for x in xs:
            result.extend(flatten(x))
    else:
        result.append(xs)
    return result

print flatten(l)

Answer 2

可以使用numpy

在一行中整齐地完成

import numpy as np
np.hstack(l)

你最终得到了一个ndarray

array([  35,   53,  525, 6743,   64,   63,  743,  754,  757])

Answer 3

>>> data = [35,53,[525,6743],64,63,[743,754,757]]
>>> def flatten(L):
        for item in L:
            if isinstance(item,list):
                for subitem in item:
                    yield subitem
            else:
                yield item


>>> list(flatten(data))
[35, 53, 525, 6743, 64, 63, 743, 754, 757]

以下是用于代码 - 高尔夫目的的单行版本（看起来不太好：D）

>>> [y for x in data for y in (x if isinstance(x,list) else [x])]
[35, 53, 525, 6743, 64, 63, 743, 754, 757]

Answer 4

l = [35,53,[525,6743],64,63,[743,754,757]]
outputList = []

for i in l:
    if isinstance(i, list):
        outputList.extend(i)
    else:
        outputList.append(i)

Answer 5

这是一个oneliner，基于the question you've mentioned：

list(itertools.chain(*((sl if isinstance(sl, list) else [sl]) for sl in l)))

更新：以及完全基于迭代器的版本：

from itertools import imap, chain
list(chain.from_iterable(imap(lambda x: x if isinstance(x, list) else [x], l)))

Answer 6

outputList = []
for e in l:
    if type(e) == list:
        outputList += e
    else:
        outputList.append(e)

>>> outputList
[35, 53, 525, 6743, 64, 63, 743, 754, 757]

Answer 7

def nchain(iterable):
    for elem in iterable:
        if type(elem) is list:
            for elem2 in elem:
                yield elem2
        else:
            yield elem

Answer 8

允许无限树深度的递归函数：

def flatten(l):
    if isinstance(l,(list,tuple)):
        if len(l):
            return flatten(l[0]) + flatten(l[1:])
        return []
    else:
        return [l]

>>>flatten([35,53,[525,[1,2],6743],64,63,[743,754,757]])
[35, 53, 525, 1, 2, 6743, 64, 63, 743, 754, 757]

我试图避免使用isinstance来允许泛型类型，但旧版本会对字符串进行无限循环。现在它正确地平整字符串（现在不是字符，但好像它假装一个字符串是一个标量）。

Answer 9

>>> L = [35,53,[525,6743],64,63,[743,754,757]]
>>> K = []
>>> [K.extend([i]) if type(i) == int else K.extend(i) for i in L ]
[None, None, None, None, None, None]
>>> K
[35, 53, 525, 6743, 64, 63, 743, 754, 757]

Answer 10

此解决方案仅适用于您的特定情况（列表中的标量）并假设标量为整数。这是一个可怕的解决方案，但它非常短暂。

outputlist = map(int,",".split(str(L).replace("[","").replace("]","")))

Answer 11

答案很简单。利用递归。

def flatten(nst_lst, final_list):

    for val in nst_lst:
        if isinstance(val, list):
            flatten(val, final_list)
        else:
            final_list.append(val)
    return final_list

#Sample usage
fl_list = []
lst_to_flatten = [["this",["a",["thing"],"a"],"is"],["a","easy"]]

print(flatten(lst_to_flatten, fl_list))

如何在python中将异构列表列表压缩成单个列表？

11 个答案: