blocks|key|5763790|text|您可以使用带有适当键函数的itertools.groupby来分块列表，并使用str.join将分块转换为字符串，从而获得所需字符串的列表：|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|5763791|from+itertools+import+groupby

l+=+['2019','Melbourne','region1','suburb1','\n',+'2018','region1','Melbourne','\n']

[',+'.join(g)+for+k,+g+in+groupby(l,+key=lambda+s:+s+!=+'\n')+if+k]
#+['2019,+Melbourne,+region1,+suburb1',+'2018,+region1,+Melbourne']|code-block|syntax|javascript|5763792|请注意，groupby返回一个惰性迭代器，该迭代器产生同样的惰性组。因此，这不会构建任何不必要的中间内存列表。|5763793|entityMap|0|LINK|mutability|MUTABLE|url|https://docs.python.org/3/library/itertools.html#itertools.groupby|1|https://docs.python.org/3/library/stdtypes.html#str.join^0|D|H|13|8|D|H|0|13|8|1|0|0|4|7|0^^$0|@$1|2|3|4|5|6|7|W|8|@$9|X|A|Y|B|C]|$9|Z|A|10|B|C]]|D|@$9|11|A|12|1|13]|$9|14|A|15|1|16]]|E|$]]|$1|F|3|G|5|H|7|17|8|@]|D|@]|E|$I|J]]|$1|K|3|L|5|6|7|18|8|@$9|19|A|1A|B|C]]|D|@]|E|$]]|$1|M|3|-4|5|6|7|1B|8|@]|D|@]|E|$]]]|N|$O|$5|P|Q|R|E|$S|T]]|U|$5|P|Q|R|E|$S|V]]]]

You can get a list of the desired strings, using <a href="https://docs.python.org/3/library/itertools.html#itertools.groupby" rel="nofollow noreferrer"><code>itertools.groupby</code></a> with an appropriate key function to chunk the list and <a href="https://docs.python.org/3/library/stdtypes.html#str.join" rel="nofollow noreferrer"><code>str.join</code></a> to turn the chunks into strings:

<pre><code>from itertools import groupby

l = ['2019','Melbourne','region1','suburb1','\n', '2018','region1','Melbourne','\n']

[', '.join(g) for k, g in groupby(l, key=lambda s: s != '\n') if k]
# ['2019, Melbourne, region1, suburb1', '2018, region1, Melbourne']
</code></pre>

Note that <code>groupby</code> returns a lazy iterator producing equally lazy groups. So this does not build any unnecessary intermediate in-memory lists.

blocks|key|4295129|text|你可以做这样的事情！使用re|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|4295130|import+re
l+=+['2019','Melbourne','region1','suburb1','\n',+'2018','region1','Melbourne','\n']
res+=+re.split("[\r\n]%2B",'+'.join(l))[:-1]
string1+=+res[0].strip().replace('+',',+')
string2+=+res[1].strip().replace('+',',+')

#+string1+-->+'2019,+Melbourne,+region1,+suburb1'
#+string2+-->+'2018,+region1,+Melbourne'|code-block|syntax|javascript|4295131|entityMap^0|C|2|0|0^^$0|@$1|2|3|4|5|6|7|M|8|@$9|N|A|O|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|P|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Q|8|@]|D|@]|E|$]]]|L|$]]

You can do something like this! using <code>re</code>

<pre><code>import re
l = ['2019','Melbourne','region1','suburb1','\n', '2018','region1','Melbourne','\n']
res = re.split("[\r\n]+",' '.join(l))[:-1]
string1 = res[0].strip().replace(' ',', ')
string2 = res[1].strip().replace(' ',', ')

# string1 --&gt; '2019, Melbourne, region1, suburb1'
# string2 --&gt; '2018, region1, Melbourne'
</code></pre>

blocks|key|4295160|text|我提出这样的解决方案：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|4295161|listin=['2019','Melbourne','region1','suburb1','\n',+'2018','region1','Melbourne','\n'];
listsout=[];sublist=[];
for+val+in+listin:
++++if+val.__contains__('\n')==False:
++++++++sublist.append(val);
++++else:
++++++++listsout.append([i+for+i+in+sublist]);
++++++++sublist+=+[];
#+--------------------------------------------
strings+=+[',+'.join(sublist)+for+sublist+in+listsout];|code-block|syntax|javascript|4295162|entityMap^0|0|0^^$0|@$1|2|3|4|5|6|7|I|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|J|8|@]|9|@]|A|$E|F]]|$1|G|3|-4|5|6|7|K|8|@]|9|@]|A|$]]]|H|$]]

I propose solution like this:

<pre><code>listin=['2019','Melbourne','region1','suburb1','\n', '2018','region1','Melbourne','\n'];
listsout=[];sublist=[];
for val in listin:
 if val.__contains__('\n')==False:
 sublist.append(val);
 else:
 listsout.append([i for i in sublist]);
 sublist = [];
# --------------------------------------------
strings = [', '.join(sublist) for sublist in listsout];
</code></pre>

blocks|key|1585999|text|一种可能的解决方案是使用join()将给定的列表连接成一个字符串，然后对该字符串使用split来拆分到list。代码如下：|type|unstyled|depth|inlineStyleRanges|entityRanges|data|1586000|test=['2019','Melbourne','region1','suburb1','\n',+'2018','region1','Melbourne','\n']
test_string=",".join(test)

sep_list=+test_string.split("\n")|code-block|syntax|javascript|1586001|您将获得以"\n“分隔的字符串列表|1586002|entityMap^0|0|0|0^^$0|@$1|2|3|4|5|6|7|K|8|@]|9|@]|A|$]]|$1|B|3|C|5|D|7|L|8|@]|9|@]|A|$E|F]]|$1|G|3|H|5|6|7|M|8|@]|9|@]|A|$]]|$1|I|3|-4|5|6|7|N|8|@]|9|@]|A|$]]]|J|$]]

One possible solution would be to concatenate the given list into a string using join() and then use split on the string to split in to list. Following is the code:

<pre><code>test=['2019','Melbourne','region1','suburb1','\n', '2018','region1','Melbourne','\n']
test_string=",".join(test)

sep_list= test_string.split("\n")
</code></pre>

You will get a list of strings separated by "\n"

blocks|key|1978688|text|您可以使用split()，但我得到了意外的结果，所以我使用正则表达式|type|unstyled|depth|inlineStyleRanges|offset|length|style|CODE|entityRanges|data|1978689|import+re
l+=+['2019','Melbourne','region1','suburb1','\n',+'2018','region1','Melbourne','\n']
string1+,+string2+=+re.findall("(?!,)([%5E\n]%2B)(?=,)"+,",".join(l))
print(string1,string2)|code-block|syntax|javascript|1978690|1978691|(?!，)：用于匹配以,+->+match开头但不包含在result|ordered-list-item|1978692|(%5E\n%2B)中的字符串:匹配字符串直到,+)：匹配以,+->+match结尾但不包含在result|1978693|中的字符串|1978694|entityMap^0|5|7|0|0|0|B|1|0|J|1|Q|1|0|0^^$0|@$1|2|3|4|5|6|7|U|8|@$9|V|A|W|B|C]]|D|@]|E|$]]|$1|F|3|G|5|H|7|X|8|@]|D|@]|E|$I|J]]|$1|K|3|-4|5|6|7|Y|8|@]|D|@]|E|$]]|$1|L|3|M|5|N|7|Z|8|@$9|10|A|11|B|C]]|D|@]|E|$]]|$1|O|3|P|5|N|7|12|8|@$9|13|A|14|B|C]|$9|15|A|16|B|C]]|D|@]|E|$]]|$1|Q|3|R|5|6|7|17|8|@]|D|@]|E|$]]|$1|S|3|-4|5|6|7|18|8|@]|D|@]|E|$]]]|T|$]]

You can use <code>split()</code>,but i get an unexpected result so i use regex

<pre><code>import re
l = ['2019','Melbourne','region1','suburb1','\n', '2018','region1','Melbourne','\n']
string1 , string2 = re.findall("(?!,)([^\n]+)(?=,)" ,",".join(l))
print(string1,string2)
</code></pre>

<ol>
<li>(?!,) : use to match string began without <code>,</code> -> match but not contain in result</li>
<li>([^\n]+) : match str until <code>\n</code></li>
<li>(?=,) : match str which end with <code>,</code> -> match but not contain in result</li>
</ol>

I have a list similar to the following one:

<pre><code>['2019','Melbourne','region1','suburb1','\n', '2018','region1','Melbourne','\n']
</code></pre>

I want to break(split) and store the information of the list based on '\n'. It means I want to create following strings from the above list:

<pre><code>string1='2019, Melbourne, region1, suburb1'
string2='2018, region1, Melbourne'
</code></pre>

I know how to do that using a simple for loop and playing around indices but I wonder if someone could assist me with doing the mentioned task in a more professional way. Because playing around the indices makes my code a bit complex and unreadable.

How to break a list based on list item and create relevant strings

翻译质量差，导致语言生硬或混乱。

没有提供实际的解决方法或示例。

解答不清晰，无法理解或解决问题。

页面排版不美观，阅读体验差。

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我有一个类似于下面的列表：['2019','Melbourne','region1','suburb1','\n', '2018','region1','Melbourne','\n']我想要中断(拆分)并存储基于'\n‘的列表信息。这意味着我想从上面的列表中创建以下字符串：string1='2019, Melbour...

问如何根据列表项拆分列表并创建相关字符串
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问如何根据列表项拆分列表并创建相关字符串EN