python - 在 Python 中包装多行字符串(保留现有的换行符)？

Question

考虑这个例子:

import textwrap
import pprint

mystr=r"""
First line.
Second line.
The third line is a very long line, which I would like to somehow wrap; wrap at 80 characters - or less, or more! ... can it really be done ??"""

pprint.pprint(textwrap.wrap(mystr,80))

字符串 mystr 已经是一个多行字符串，因为它包含换行符；但是，如果我运行这个脚本，我会得到输出:

[' First line. Second line. The third line is a very long line, which I would like',
 'to somehow wrap; wrap at 80 characters - or less, or more! ... can it really be',
 'done ??']

... 这意味着 textwrap.wrap 首先“加入”多行字符串(即删除其中现有的换行符)，然后才将其包装(即在给定的位置拆分它字符数)。

如何包装多行字符串，以便保留换行符？也就是说，在这种情况下，预期的输出将是:

['First line.', 
 'Second line.', 
 'The third line is a very long line, which I would like to somehow wrap; wrap at',
 '80 characters - or less, or more! ... can it really be done ??']

---

编辑；感谢@u_mulder 的评论，我试过了:

textwrap.wrap(mystr,80,replace_whitespace=False)

然后我得到:

['\nFirst line.\nSecond line.\nThe third line is a very long line, which I would like',
 'to somehow wrap; wrap at 80 characters - or less, or more! ... can it really be',
 'done ??']

换行符似乎被保留，但作为“内联”字符；所以这里的第一个元素本身就是一个多行字符串——所以它不是我所要求的，每一行都是一个数组元素。

Answer 1

只需在拆分后添加换行符:

import textwrap
import pprint
import itertools

mystr=r"""
First line.
Second line.
The third line is a very long line, which I would like to somehow wrap; wrap at 80 characters - or less, or more! ... can it really be done ??"""

wrapper = textwrap.TextWrapper(width = 80)
mylist = [wrapper.wrap(i) for i in mystr.split('\n') if i != '']
mylist = list(itertools.chain.from_iterable(mylist))

pprint.pprint(mylist)

输出:

['First line.',
 'Second line.',
 'The third line is a very long line, which I would like to somehow wrap; wrap at',
 '80 characters - or less, or more! ... can it really be done ??']