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<p>我有一个字符串如下:</p>
<pre><code>'1:CH,AG,ME,GS,AP,CH,HE,AC,AC,AG,CA,HE,AT,AT,AC,AT,OG,NE,AG,AC,CS,OD\n&:TA,EB,PA,AC,BR,TH,PO,AC,2I,AC,TH,PE,TH,AZ,AZ,ZE,CS,OD,CH,EO,ZE,OG\n&:TH,ZE,ZE,HE,HE,HP,HP,OG,HP,ZE\n2:ZE,FD,FD,AG,EO,OG,AG,NE,RU,GS,HP,ZE,ZE,HM,HM,PC,PC,AS,AS,TY,TY,AG\n&:AG,GS,NO,EU,ZF,HE,AT,AT,OD,OD,EB,OD,GS,TR,OD,AC,TR,GS,OD,TR,OD,AT,GS\n&:CA,GS,NE,GS,AG,PS,HL,AG,NE,ID,AJ,AX,DI,OD,ME,AT,GS,MU,HO,PB,LT,9Z,PT,9Y\n&:9W,9X,AR,9V,9U,9T,AX,9S,9R,AT,AJ,DI,ST,EA,AG,ME,NE,MU,9Q,9P,9O,9N,9M,9L\n&:9K,ID,MG,OD,FY,AU,AU,HR,HR,9J,TL,9I,9H,9G,9F,AC,BR,AC,9E,9D,9C,9B,99\n'
</code></pre>
<p>如您所见,我希望将“\n(number or&;here):”替换为“,”
因为它们都以“\n”开头,以“:”结尾,所以我认为应该有一种方法可以同时替换它们。你知道吗</p>
<p>输出如下:</p>
<pre><code>'CH,AG,ME,GS,AP,CH,HE,AC,AC,AG,CA,HE,AT,AT,AC,AT,OG,NE,AG,AC,CS,OD,TA,EB,PA,AC,BR,TH,PO,AC,2I,AC,TH,PE,TH,AZ,AZ,ZE,CS,OD,CH,EO,ZE,OG,TH,ZE,ZE,HE,HE,HP,HP,OG,HP,ZE,ZE,FD,FD,AG,EO,OG,AG,NE,RU,GS,HP,ZE,ZE,HM,HM,PC,PC,AS,AS,TY,TY,AG,AG,GS,NO,EU,ZF,HE,AT,AT,OD,OD,EB,OD,GS,TR,OD,AC,TR,GS,OD,TR,OD,AT,GS,CA,GS,NE,GS,AG,PS,HL,AG,NE,ID,AJ,AX,DI,OD,ME,AT,GS,MU,HO,PB,LT,9Z,PT,9Y,9W,9X,AR,9V,9U,9T,AX,9S,9R,AT,AJ,DI,ST,EA,AG,ME,NE,MU,9Q,9P,9O,9N,9M,9L,9K,ID,MG,OD,FY,AU,AU,HR,HR,9J,TL,9I,9H,9G,9F,AC,BR,AC,9E,9D,9C,9B,99'
</code></pre>
<p>可行的办法是为数字和&;制作lop;。你知道吗</p>
<pre><code>string.replace('\n&:',',')
for i in range(1,20):
string.replace('\ni:',',')
</code></pre>
<p>但我相信一定有更好的办法。你知道吗</p>