Question

我有一个list strings，其中包含sublist操作系strings：

ids = [u'spotify:track:3ftnDaaL02tMeOZBunIwls', u'spotify:track:4CKjTXDDWIrS0cwSA9scgk', [u'spotify:track:6oRbm1KOqskLTFc1rvGi5F',
u'spotify:track:045sp2JToyTaaKyXkGejPy']]

我试图用它来压扁它：

[item for item in ids for item in sublist]

和

chain = itertools.chain(ids)

但这些解决方案将字符串分开......

如何将原始list展平为

[u'spotify:track:3ftnDaaL02tMeOZBunIwls', u'spotify:track:4CKjTXDDWIrS0cwSA9scgk', u'spotify:track:6oRbm1KOqskLTFc1rvGi5F',u'spotify:track:045sp2JToyTaaKyXkGejPy']

Answer 1

您可以使用带isinstance检查的简单循环。

out = []
for i in ids:
    if isinstance(i, list):
        out.extend(i)
    else:
        out.append(i)

print(out)

输出：

['spotify:track:3ftnDaaL02tMeOZBunIwls',
 'spotify:track:4CKjTXDDWIrS0cwSA9scgk',
 'spotify:track:6oRbm1KOqskLTFc1rvGi5F',
 'spotify:track:045sp2JToyTaaKyXkGejPy']

您也可以使用itertools.chain，但需要额外的预处理层：

from itertools import chain

out = list(chain.from_iterable([i if isinstance(i, list) else [i] for i in ids]))
print(out)

输出相同。

Answer 2

您只需要将任何字符串转换为列表：

import itertools
ids = [u'spotify:track:3ftnDaaL02tMeOZBunIwls', u'spotify:track:4CKjTXDDWIrS0cwSA9scgk',   [u'spotify:track:6oRbm1KOqskLTFc1rvGi5F',
u'spotify:track:045sp2JToyTaaKyXkGejPy']]

new_data = list(itertools.chain.from_iterable([[i] if not isinstance(i, list) else i for i in b]))

输出：

[u'spotify:track:3ftnDaaL02tMeOZBunIwls', u'spotify:track:4CKjTXDDWIrS0cwSA9scgk', u'spotify:track:6oRbm1KOqskLTFc1rvGi5F', u'spotify:track:045sp2JToyTaaKyXkGejPy']

Answer 3

您可以创建一个功能来展平列表。

def flatten(lst):
  if isinstance(lst, (str, unicode)):
    return [lst]
  return [unit for item in lst for unit in flatten(item)]

print(flatten(ids))

Answer 4

你可以这样做：

>>> [x for l in ids for x in (l if isinstance(l, list) else [l])]
[u'spotify:track:3ftnDaaL02tMeOZBunIwls', u'spotify:track:4CKjTXDDWIrS0cwSA9scgk', u'spotify:track:6oRbm1KOqskLTFc1rvGi5F', u'spotify:track:045sp2JToyTaaKyXkGejPy']

这类似于您对[item for item in ids for item in sublist]的列表理解，但添加了一个测试，看它是否实际上是我们正在查看的列表。

展平包含子列表的字符串列表

4 个答案: