blocks|key|2619422|text|您可以将文件读入内存并使用|type|unstyled|depth|inlineStyleRanges|entityRanges|data|2619423|r'(?ms)%5E\d%2B\.\s*(.*?)(?=%5E\d%2B\.%7C\Z)'|code-block|syntax|javascript|2619424|请参阅regex+demo|offset|length|2619425|详细信息|style|BOLD|2619426|2619427|(?ms)+-启用组和组modes|unordered-list-item|CODE|2619428|%5E+-+line|2619429|\d%2B+-+1%2B+_+digits|2619430|\.+-a+dot|2619431|\s*+-1%2B_+0%2B+-组1(这是re.findall在这里返回的内容)：任何字符，直到(但不包括)第一个出现的0%2B+-1%2B-0%2B行的开始，结束数字和.|2619432|%7C+-+or|2619433|\Z+-+string.|2619434|结束|2619435|​|2619436|Python：|2619437|with+open('lines.txt')+as+fh:
++++print(re.findall(r'(?ms)%5E\d%2B\.\s*(.*?)(?=%5E\d%2B\.%7C\Z)',+fh.read()))|2619438|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/r/humIzT/1^0|0|0|3|A|0|0|0|4|0|0|0|5|0|0|1|0|0|3|0|0|2|0|0|3|I|A|24|1|0|0|1|0|0|2|0|0|0|0|0^^$0|@$1|2|3|4|5|6|7|1L|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|1M|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|1N|8|@]|9|@$I|1O|J|1P|1|1Q]]|A|$]]|$1|K|3|L|5|6|7|1R|8|@$I|1S|J|1T|M|N]]|9|@]|A|$]]|$1|O|3|-4|5|6|7|1U|8|@]|9|@]|A|$]]|$1|P|3|Q|5|R|7|1V|8|@$I|1W|J|1X|M|S]]|9|@]|A|$]]|$1|T|3|U|5|R|7|1Y|8|@$I|1Z|J|20|M|S]]|9|@]|A|$]]|$1|V|3|W|5|R|7|21|8|@$I|22|J|23|M|S]]|9|@]|A|$]]|$1|X|3|Y|5|R|7|24|8|@$I|25|J|26|M|S]]|9|@]|A|$]]|$1|Z|3|10|5|R|7|27|8|@$I|28|J|29|M|S]|$I|2A|J|2B|M|S]|$I|2C|J|2D|M|S]]|9|@]|A|$]]|$1|11|3|12|5|R|7|2E|8|@$I|2F|J|2G|M|S]]|9|@]|A|$]]|$1|13|3|14|5|R|7|2H|8|@$I|2I|J|2J|M|S]]|9|@]|A|$]]|$1|15|3|16|5|6|7|2K|8|@]|9|@]|A|$]]|$1|17|3|18|5|6|7|2L|8|@]|9|@]|A|$]]|$1|19|3|1A|5|6|7|2M|8|@]|9|@]|A|$]]|$1|1B|3|1C|5|D|7|2N|8|@]|9|@]|A|$E|F]]|$1|1D|3|-4|5|6|7|2O|8|@]|9|@]|A|$]]]|1E|$1F|$5|1G|1H|1I|A|$1J|1K]]]]

You may read in the file into memory and use

<pre><code>r'(?ms)^\d+\.\s*(.*?)(?=^\d+\.|\Z)'
</code></pre>

See the <a href="https://regex101.com/r/humIzT/1" rel="nofollow noreferrer">regex demo</a>

Details

<ul>
<li><code>(?ms)</code> - enable <code>re.MULTILINE</code> and <code>re.DOTALL</code> modes</li>
<li><code>^</code> - start of a line</li>
<li><code>\d+</code> - 1+ digits</li>
<li><code>\.</code> - a dot</li>
<li><code>\s*</code> - 0+ whitespaces</li>
<li><code>(.*?)</code> - Group 1 (this is what <code>re.findall</code> returns here): any 0+ chars, as few as possible</li>
<li><code>(?=^\d+\.|\Z)</code> - up to (but not inlcuding) the first occurrence of

<ul>
<li><code>^\d+\.</code> - start of a line, 1+ digits and <code>.</code></li>
<li><code>|</code> - or</li>
<li><code>\Z</code> - end of string.</li>
</ul></li>
</ul>

Python:

<pre><code>with open('lines.txt') as fh:
 print(re.findall(r'(?ms)^\d+\.\s*(.*?)(?=^\d+\.|\Z)', fh.read()))
</code></pre>

blocks|key|2619461|text|使用re.findall查找从\d\.\s%2B模式到下一个'\n\d‘模式或直到结束的所有内容|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2619462|>>>+import+re
>>>+re.findall(r'\d%2B\.\s%2B(.*?(?=\n\d%7C$))',+text,+flags=re.DOTALL)
['robert\n+++smith',+'harry',+'john']|code-block|syntax|javascript|2619463|entityMap^0|2|A|F|7|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]|$9|P|A|Q|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|R|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|S|8|@]|D|@]|E|$]]]|L|$]]

Use <code>re.findall</code> to find all from <code>\d\.\s+</code> pattern to next '\n\d' pattern or upto end

<pre><code>&gt;&gt;&gt; import re
&gt;&gt;&gt; re.findall(r'\d+\.\s+(.*?(?=\n\d|$))', text, flags=re.DOTALL)
['robert\n smith', 'harry', 'john']
</code></pre>

blocks|key|2408782|text|您可以使用re.split。|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|2408783|Regex：\n?\d%2B\.\s*|BOLD|2408784|详细信息：|2408785|2408786|\n+-+Newline|unordered-list-item|2408787|?+-匹配0到1次，match+if+'new+line‘exists|2408788|\d%2B+-匹配1到无限次之间的数字(%2B)+times|2408789|\.+-+Dot|2408790|\s*+-匹配0到无限次之间的任何空白字符(等于[\r\n\t\f\v+])+(*)+|2408791|2408792|Python代码|2408793|re.split(r'\n?\d%2B\.\s*',+lines)[1:]|code-block|syntax|javascript|2408794|[1:]删除第一项，因为它是空字符串|2408795|输出：|2408796|['robert\n+++smith',+'harry',+'john']|2408797|entityMap|0|LINK|mutability|MUTABLE|url|https://regex101.com/r/ZZeVWH/2^0|5|8|0|0|5|6|B|6|B|0|0|0|0|0|2|0|0|1|0|0|3|0|0|2|0|0|3|O|D|0|0|0|8|0|0|0|4|0|0|0^^$0|@$1|2|3|4|5|6|7|1I|8|@$9|1J|A|1K|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|1L|8|@$9|1M|A|1N|B|H]|$9|1O|A|1P|B|C]]|D|@$9|1Q|A|1R|1|1S]]|E|$]]|$1|I|3|J|5|6|7|1T|8|@]|D|@]|E|$]]|$1|K|3|-4|5|6|7|1U|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|1V|8|@$9|1W|A|1X|B|C]]|D|@]|E|$]]|$1|O|3|P|5|N|7|1Y|8|@$9|1Z|A|20|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|N|7|21|8|@$9|22|A|23|B|C]]|D|@]|E|$]]|$1|S|3|T|5|N|7|24|8|@$9|25|A|26|B|C]]|D|@]|E|$]]|$1|U|3|V|5|N|7|27|8|@$9|28|A|29|B|C]|$9|2A|A|2B|B|C]]|D|@]|E|$]]|$1|W|3|-4|5|6|7|2C|8|@]|D|@]|E|$]]|$1|X|3|Y|5|6|7|2D|8|@$9|2E|A|2F|B|H]]|D|@]|E|$]]|$1|Z|3|10|5|11|7|2G|8|@]|D|@]|E|$12|13]]|$1|14|3|15|5|6|7|2H|8|@$9|2I|A|2J|B|C]]|D|@]|E|$]]|$1|16|3|17|5|6|7|2K|8|@]|D|@]|E|$]]|$1|18|3|19|5|11|7|2L|8|@]|D|@]|E|$12|13]]|$1|1A|3|-4|5|6|7|2M|8|@]|D|@]|E|$]]]|1B|$1C|$5|1D|1E|1F|E|$1G|1H]]]]

You can use <code>re.split</code>.

Regex: <a href="https://regex101.com/r/ZZeVWH/2" rel="nofollow noreferrer"><code>\n?\d+\.\s*</code></a>

Details:

<ul>
<li><code>\n</code> - Newline</li>
<li><code>?</code> - Matches between zero and one times, match if 'new line' exists</li>
<li><code>\d+</code> - Matches a digit (+) between one and unlimited times</li>
<li><code>\.</code> - Dot</li>
<li><code>\s*</code> - Matches any whitespace character (equal to <code>[\r\n\t\f\v ]</code>) (*) between zero and unlimited times</li>
</ul>

Python code:

<pre><code>re.split(r'\n?\d+\.\s*', lines)[1:]
</code></pre>

<code>[1:]</code> removes the first item because its empty string

Output:

<pre><code>['robert\n smith', 'harry', 'john']
</code></pre>

blocks|key|2619619|text|我提出了一个解决方案，与其他一些解决方案相反，它只收集的名称，名称中间没有不必要的空格。|type|unstyled|depth|inlineStyleRanges|offset|length|style|BOLD|entityRanges|data|2619620|这个想法是：|2619621|2619622|保存一个元组列表(+number，name_segment)，如果当前行中缺少组号，则“复制”上一行的组号。要保存的对是使用\n作为结果列表中的这些连接名称的separator.|unordered-list-item|2619623|Save，按数字(第一个元素)连接这些元组准备的。getPair+|2619624|连接来自每个组的名称段。|2619625|2619626|使用列表理解允许以一种非常简洁的方式编写程序。如下所示：|2619627|import+re,+itertools

def+getPair(line):
++global+grp
++nr,+nameSegm+=+re.match(r'%5E(\d%2B\.)?\s%2B(\w%2B)$',+line).groups()
++if+nr:++#+Number+present
++++grp+=+nr
++return+grp,+nameSegm+++

grp+=+''++++#+Group+label+(number)
with+open('lines.txt')+as+fh:
++lst+=+[getPair(line)+for+line+in+fh.readlines()]
res+=+['\n'.join([t[1]+for+t+in+g])
++for+_,+g+in+itertools.groupby(lst,+lambda+x:+x[0])]
print(f"Result:+{res}")|code-block|syntax|javascript|2619628|总而言之，该程序比其他程序要长一点，但它只给提供名称，没有额外的空格。|2619629|entityMap^0|R|1|0|0|0|0|0|0|0|0|0|M|2|0^^$0|@$1|2|3|4|5|6|7|11|8|@$9|12|A|13|B|C]]|D|@]|E|$]]|$1|F|3|G|5|6|7|14|8|@]|D|@]|E|$]]|$1|H|3|-4|5|6|7|15|8|@]|D|@]|E|$]]|$1|I|3|J|5|K|7|16|8|@]|D|@]|E|$]]|$1|L|3|M|5|K|7|17|8|@]|D|@]|E|$]]|$1|N|3|O|5|K|7|18|8|@]|D|@]|E|$]]|$1|P|3|-4|5|6|7|19|8|@]|D|@]|E|$]]|$1|Q|3|R|5|6|7|1A|8|@]|D|@]|E|$]]|$1|S|3|T|5|U|7|1B|8|@]|D|@]|E|$V|W]]|$1|X|3|Y|5|6|7|1C|8|@$9|1D|A|1E|B|C]]|D|@]|E|$]]|$1|Z|3|-4|5|6|7|1F|8|@]|D|@]|E|$]]]|10|$]]

I propose a solution, which gathers only names,
without unnecessary spaces in the middle of names,
contrary to some other solutions.

The idea is:

<ul>
<li>Save a list of tuples (number, name_segment), "copying"
the group number from previous line if absent in the current
line. The pair to be saved is prepared by getPair function.</li>
<li>Group these tuples on the number (first element).</li>
<li>Join name segments from each group, using \n as separator.</li>
<li>Save these joined names in a result list.</li>
</ul>

Using list comprehensions allows to write the program in
a quite concise way. See below:

<pre><code>import re, itertools

def getPair(line):
 global grp
 nr, nameSegm = re.match(r'^(\d+\.)?\s+(\w+)$', line).groups()
 if nr: # Number present
 grp = nr
 return grp, nameSegm 

grp = '' # Group label (number)
with open('lines.txt') as fh:
 lst = [getPair(line) for line in fh.readlines()]
res = ['\n'.join([t[1] for t in g])
 for _, g in itertools.groupby(lst, lambda x: x[0])]
print(f"Result: {res}")
</code></pre>

To sum up, the program is a bit longer than other, but is gives
only names, without additional spaces.

I have file with following lines:

lines.txt

<pre><code>1. robert
 smith
2. harry
3. john
</code></pre>

I want to get array as follows:

<pre><code>["robert\nsmith","harry","john"]
</code></pre>

I tried something like this:

<pre><code>with open('lines.txt') as fh:
 m = [re.match(r"^\d+\.(.*)",line) for line in fh.readlines()]
 print(m)
 for i in m:
 print(i.groups())
</code></pre>

It outputs following:

<pre><code>[&lt;_sre.SRE_Match object; span=(0, 9), match='1. robert'&gt;, None, &lt;_sre.SRE_Match object; span=(0, 8), match='2. harry'&gt;, &lt;_sre.SRE_Match object; span=(0, 7), match='3. john'&gt;]
(' robert',)
Traceback (most recent call last):
 File "D:\workspaces\workspace6\PdfGenerator\PdfGenerator.py", line 5, in &lt;module&gt;
 print(i.groups())
AttributeError: 'NoneType' object has no attribute 'groups'
</code></pre>

It seems that I am approaching this problem in very wrong way. How you will solve this?

Fetching content from numbered list in file with python regular expressions

我有一个包含以下行的文件：lines.txt1. robert   smith2. harry3. john我想像下面这样获取数组：["robert\nsmith","harry","john"]我尝试了这样的东西：with open('lines.txt') as fh:    m = [re.match(r"^\d...

问使用python正则表达式从文件的编号列表中获取内容
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