从pandas数据帧列（python3）中提取所有模式

我正在使用jupyter笔记本（python 3）。我正试图从我的列表中提取熊猫数据框关键字。我将有大约50个关键字在列表中

例如：

import pandas as pd
import re

rgx_words1 = ['algaecid','algaecide','algaecides','anti-bakterien']

 

pattern = "\\b("+'|'.join(rgx_words1)+")\\b"

re_patt = re.compile(pattern)

 

pattern2 = "("+'|'.join(rgx_words1)+")"

re_patt2 = re.compile(pattern2)

 

data = [[1, 'I, will, find, algaecide, dd, algaecid, algaecides'], [2, 'fff, algaecid, dd, algaecide'], [3, 'ssssalgaecidllll, algaecides']]

  

# Create the pandas DataFrame

mydf = pd.DataFrame(data, columns = ['id', 'text'])

 

mydf['matches'] = mydf.apply(lambda x : re.findall(re_patt,x['text']),axis=1)

mydf['matches2'] = mydf.apply(lambda x : re.findall(re_patt2,x['text']),axis=1)

通过Reu patt，我提取了精确的单词，得到了正确的结果。在id 1中，我的输出是algaecide，algaecid，algaecides。有了re_patt2，我希望所有模式都像“SSSSALGAECIDLLL”一样，输出为“algaecid”。id 1中re_patt2的输出为algaecid、algaecid、algaecid，我想要的输出为algaecid、algaecid、algaecides。

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如蒙指教，我将不胜感激。提前感谢。

您可以更改

模式2

以选择匹配非空白字符，但左侧和右侧的逗号

[^\s，]*

除外

pattern2 = "[^\s,]*(?:"+'|'.join(rgx_words1)+")[^\s,]*"

代码可能看起来像

import pandas as pd
import re

rgx_words1 = ['algaecid','algaecide','algaecides','anti-bakterien']

pattern = "\\b("+'|'.join(rgx_words1)+")\\b"
re_patt = re.compile(pattern)

pattern2 = "[^\s,]*(?:"+'|'.join(rgx_words1)+")[^\s,]*"
re_patt2 = re.compile(pattern2)

data = [[1, 'I, will, find, algaecide, dd, algaecid, algaecides'], [2, 'fff, algaecid, dd, algaecide'], [3, 'ssssalgaecidllll, algaecides']]
mydf = pd.DataFrame(data, columns = ['id', 'text'])

mydf['matches'] = mydf.apply(lambda x : re.findall(re_patt, x['text']), axis=1)
mydf['matches2'] = mydf.apply(lambda x : re.findall(re_patt2, x['text']), axis=1)

print(mydf)

输出

   id                                               text                            matches                           matches2
0   1  I, will, find, algaecide, dd, algaecid, algaec...  [algaecide, algaecid, algaecides]  [algaecide, algaecid, algaecides]
1   2                       fff, algaecid, dd, algaecide              [algaecid, algaecide]              [algaecid, algaecide]
2   3                       ssssalgaecidllll, algaecides                       [algaecides]     [ssssalgaecidllll, algaecides]

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例如，有没有一种方法可以在id 3“SSSSALGAECIDLL”中获得列匹配中的输出2“algaecid”而不是“SSSSSSALGAECIDLL”？@DinkoJantoš您可以通过重新定义单词，如

rgx_words1=['algaecides'，'algaecide'，'algaecid'，'anti-bakerien]

然后用管道状的

pattern2='124;连接这些单词。连接（rgx_words1）

我这样做了：rgx_words1.sort（key=len，reverse=True）['anti-bakterien'，'algaecides'，'algaecide'，'algaecid']mypatt='124;'。join（rgx_words1）re_mypatt=re.compile（mypatt）mydf['matches3']=mydf.apply（lambda x:re.findall（re_mypatt，x['text']），axis=1）在我的新列匹配3中，我有输出algaecid，algaecid，在列匹配2中，输出是ssssssalgaecidlll，algaecides。对于单词ssssssalgaecidlll，我的输出algaecide是正确的，但是对于单词algaecides，我的输出是algaecides，我需要algaecides。@DinkoJantoš如果您仅使用

连接单词，则替换将是正确的tch第一个。排序后，备选方案看起来像这样

anti-bakterien | algaecides | algaecides | algaecides | algaecides

，因此它将首先匹配

algaecides

，如果可以的话。那么有什么解决方案吗？或者这是错误的->pattern2='|。join（rgx|words1）？我可以从单词'sssalgaecidesllll'-->algaecides而不是algaecid获得输出吗？