Python 为每个换行符将一个列表拆分成几个列表

Question

我有以下列表，当列表中的元素为“\n”时，我想将其拆分为多个列表。

输入：

['chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290\n', '27\n','\n','chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064\n','27\n','\n']

预计 output：

[
 ['chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290', '27'],
 ['chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064', '27']
]

我尝试在它们的末尾剥离带有“\n”的元素，并使用并修改了这篇文章中接受的答案：

for i, n in enumerate(lst):
    if n != "\n":
        lst[i] = lst[i].rstrip("\n")

[item.split(",") for item in ','.join(lst).split('\n') if item]

但是由于我使用逗号而不是单个空格来连接和拆分，所以在拆分成几个列表后我得到了“”。 我怎样才能防止这种情况发生？

[
 ['chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290','27',''],
 ['','chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064','27','']
]

Answer 1

这对你有用吗？

list1 = ['chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290\n', '27\n','\n','chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064\n','27\n','\n']

list2 = []
tmp = []
for item in list1:
    if item != '\n':
        tmp.append(item.rstrip('\n'))
    else:
        #Note we aren't actually processing this item of the input list, as '\n' by itself is unwanted
        list2.append(tmp)
        tmp = []

Answer 2

我建议使用more_itertools.split_at拆分您的列表。

因为您的原始列表以分隔符'\n'结尾，所以拆分它会导致列表的最后一项成为空子列表。 if检查排除了这一点。

from more_itertools import split_at

original = [
    'chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290\n',
    '27\n',
    '\n',
    'chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064\n',
    '27\n',
    '\n'
]

processed = [
    [item.rstrip() for item in sublist]
    for sublist in split_at(original, lambda i: i == '\n')
    if sublist
]

print(processed)

Output（为清楚起见添加了换行符）：

[['chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290', '27'],
 ['chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064', '27']]

Answer 3

使用列表理解从列表中删除\n项，并使用.strip()删除要保留的行末尾的\n 。 然后遍历临时列表 ( blist ) 并创建最终列表。

alist = ['chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290\n', '27\n','\n','chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064\n','27\n','\n']

# use line comprehension to remove the \n from the list
# and use the .strip() to remove the trailing \n in the strings
blist = [i.strip() for i in alist if i != '\n']

final_list = []
for i in range(0,len(blist),2):
    final_list.append( [blist[i], blist[i+1]] )

Answer 4

你可以使用groupby ：

from itertools import groupby

seq = ['chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290\n', '27\n','\n','chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064\n','27\n','\n']

result = [
    [string.rstrip() for string in group]
    for key, group in groupby(seq, lambda s: s != "\n")
    if key
]

结果：

[
    ['chain 2109 chrY 59373566 + 1266734 1266761 chrX 156040895 + 1198245 1198272 20769290', '27'],
    ['chain 2032 chrY 59373566 + 1136192 1136219 chrX 156040895 + 1086629 1086656 4047064', '27']
]