将列表项与前一项连接起来

Question

如果条目不是以数字开头，我想将一个条目与前一个条目连接起来

例如:

l = ["1. first paragraph", "2. second paragraph", "end of second paragraph", "3. third paragraph"]
result = []
curr_str = ""
for item in l:
curr_str += item
if not item[0].isdigit():
result.append(curr_str)
curr_str = ""

What I want

result = ["1. first paragraph", "2. second paragraphend of second paragraph", "3. third paragraph"]

What I have

result=["1. first paragraph2. second paragraphend of second paragraph"]

Answer 1

你可以使用负索引来获取你想要的

l = ["1. first paragraph", "2. second paragraph", "end of second paragraph", "3. third paragraph"]
res = []
for i in l:
if i[0].isdigit():
res.append(i)
else:
res[-1] = res[-1] + i
print(res)

输出

['1. first paragraph', '2. second paragraphend of second paragraph', '3. third paragraph']

注意:

如果你的第一个元素不是以数字开头，这将不起作用。

如果列表为空或第一个字符为数字，则需要更改if条件，如if not res or i[0].isdigit():。

Answer 2

一种方法可能是将列表作为单个字符串连接在一起，然后用空格分隔，后跟一个数字段落头:

import re
l = ["1. first paragraph", "2. second paragraph", "end of second paragraph", "3. third paragraph"]
inp = ' '.join(l)
paragraphs = re.split(r's+(?=d+.)', inp)
print(paragraphs)

这个打印:

['1. first paragraph',
'2. second paragraph end of second paragraph',
'3. third paragraph']

Answer 3

在将当前项连接到curr_str之前，需要检查当前项是否以数字开头。

在循环结束时，您需要检查curr_str是否包含任何内容，因此您可以将最后的项目附加到列表中。

l = ["1. first paragraph", "2. second paragraph", "end of second paragraph", "3. third paragraph"]
result = []
curr_str = ""
for item in l:
if item[0].isdigit():
if curr_str:
result.append(curr_str)
curr_str = ""
curr_str += item
if curr_str:
result.append(curr_str)
print(result)

Answer 4

CODE

连接列表中的所有元素，您必须使用"".join(list)，但在您的情况下，它必须更具选择性，因此我们可以使用列表推导:

l = ["1. first paragraph", "2. second paragraph", "end of second paragraph", "3. third paragraph"]
import re
result = [elem for elem in l if re.match('^d.*$', elem)]
result = "n".join(result)
print(result)

输出:

1. first paragraph
2. second paragraph
3. third paragraph

解释

第一步:

['1. first paragraph2. second paragraph3. third paragraph']

使用列表推导式和正则表达式过滤输入列表中的值，并得到一个仅包含以数字开头的字符串的列表

输出:

['1. first paragraph', '2. second paragraph', '3. third paragraph']

第二步:

result = ["n".join(result)]

使用join内置字符串方法连接列表项

Answer 5

好了

l = ["1. first paragraph", "2. second paragraph", "end of second paragraph", "3. third paragraph"]
result = []
curr_str = l[0]
for item in l[1:]:
if item[0].isdigit():
result.append(curr_str)
curr_str = item
else:
curr_str+=item
result.append(curr_str)

这将工作。我真的不能评论你的解决方案，因为我不理解它背后的思考过程，重新开始。如果您有任何问题，请随时提问