Question

我有一个数据框data（带有冗长且不一致的文本字符串注释）和匹配的ID。我的目标是使用list子字符串提取相关的子字符串，并为提取的子字符串创建一个新列。我被告知正则表达式是一个很好的起点，但我还没有想出一个可以产生匹配结果的好模式。我希望有人能够看到这一点并指导我以正确的方式解决这个问题。

list = ['sentara williamsburg regional medical',
       'shady grove adventist hospital',
       'sibley memorial hospital',
       'southern maryland hospital center',
       'st. mary`s hospital',
       'suburban hospital healthcare system',
       'the cancer center at lake manassas',
       'ucla medical center',
       'united medical center- greater southeast community',
       'univ of md charles regional medical ctr',
       'university of maryland medical center',
       'university of north carolina hospital',
       'university of virginia health system',
       'unknown facility',
       'va medical center',
       'virginia hospital center-arlington',
       'walter reed army medical center',
       'washington adventist hospital',
       'washington hospital center',
       'wellstar health system, inc',
       'winchester medical center']

 data:
     ID     Notes                             
     530.0  Cancer is best diag @Wwashington Adventist Hospital
     651.0  nan
     692.0  GMC-009 can be accessed at ST. Mary`s but not in UCLA Med. Center
     993.0  I'm not sure of Sibley; however, Shady Grove Adventist Hosp. is great hospital
     044.0  nan
     055.0  2015-01-20 was the day she visited WR Army Medical Center in WDC
     476.0  nan

预期输出 - 案例确实无关紧要！

 data_out: 
     ID     Notes                             
     530.0  Washington Adventist Hospital
     651.0  nan
     692.0  ST. Mary`s Hospital, UCLA Medical Center
     993.0  Sibley Memorial Hoapital, Shady Grove Adventist Hospital
     044.0  nan
     055.0  Walter Reed Army Medical Center
     476.0  nan

Answer 1

我会做的。像：

import re
reg = re.compile('|'.join(your_list))
results = reg.match(your_data)

Answer 2

更新：此代码遍历列表中的所有单词，并将它们与“Notes”列进行比较。如果在“列表”和“注释”中都有一个单词，则该单词将写入新列“输出”中。您必须使用正则表达式来获得所需的结果。注意：由于这个事实，“列表”中的单词可能看起来完全不同，但与“列”（缩写，拼写，错误，Csase敏感）中的单词具有相同的含义，它很难达到所有不同的情况。因此，使用“词汇方法”来解决这个问题可能是有用的......

#Create a new list
newlist=[]

#Split the sentences of the "Notes" column
[newlist.append(data.loc[i,"Notes"].split(" ")) for i in range(len(data["Notes"]))]

#Create the new column "output" and default the values to be the same as in the column "Notes"
data["output"]=data["Notes"]

#Run through all words
for i in range(len(list)):
    for j in range(len(newlist)):
        for element in range(len(newlist[j])):
            if re.search(newlist[j][element],list[i]):
                data.loc[j,"output"]= "' '{0}".format(newlist[j][element])

如果有更多的矢量化方法，我会感激评论

如何编写相关的REGEX模式以在python中提取较大文本字符串的子字符串

2 个答案: