使用 findall python 从推文中提取 @mentions(给出错误的结果)

问题描述

我有一个类似这样的 csv 文件

I have a csv file something like this

text
RT @CritCareMed: New Article: Male-Predominant Plasma Transfusion Strategy for Preventing Transfusion-Related Acute Lung Injury... htp://…
#CRISPR Inversion of CTCF Sites Alters Genome Topology & Enhancer/Promoter Function in @CellCellPress htp://.co/HrjDwbm7NN
RT @gvwilson: Where's the theory for software engineering? Behind a paywall, that's where. htp://.co/1t3TymiF3M #semat #fail
RT @sciencemagazine: What’s killing off the sea stars? htp://.co/J19FnigwM9 #ecology
RT @MHendr1cks: Eve Marder describes a horror that is familiar to worm connectome gazers. htp://.co/AEqc7NOWoR via @nucAmbiguous htp://…

我想从推文文本中提取所有提及(以@"开头).到目前为止，我已经这样做了

I want to extract all the mentions (starting with '@') from the tweet text. So far I have done this

import pandas as pd
import re

mydata = pd.read_csv("C:/Users/file.csv")
X = mydata.ix[:,:]
X=X.iloc[:,:1] #I have multiple columns so I'm selecting the first column only that is 'text'

for i in range(X.shape[0]):
result = re.findall("(^|[^@\w])@(\w{1,25})", str(X.iloc[:i,:]))

print(result);

这里有两个问题:首先:在 str(X.iloc[:1,:]) 它给了我 ['CritCareMed'] 这不正确，因为它应该给我 ['CellCellPress']，在 str(X.iloc[:2,:]) 它再次给了我 ['CritCareMed'] 这当然是又不行了.我得到的最终结果是

There are two problems here:First: at str(X.iloc[:1,:]) it gives me ['CritCareMed'] which is not ok as it should give me ['CellCellPress'], and at str(X.iloc[:2,:]) it again gives me ['CritCareMed'] which is of course not fine again. The final result I'm getting is

[(' ', 'CritCareMed'), (' ', 'gvwilson'), (' ', 'sciencemagazine')]

它不包括第二行中的提及和最后一行中的两个提及.我想要的应该是这样的:

It doesn't include the mentions in 2nd row and both two mentions in last row.What I want should look something like this:

我怎样才能达到这些结果?这只是一个示例数据，我的原始数据有很多推文，所以方法可以吗?

How can I achieve these results? this is just a sample data my original data has lots of tweets so is the approach ok?

推荐答案

您可以使用 str.findall 方法来避免for循环，使用负向后面替换(^|[^@\w]) 形成另一个您在正则表达式中不需要的捕获组:

You can use str.findall method to avoid the for loop, use negative look behind to replace (^|[^@\w]) which forms another capture group you don't need in your regex:

df['mention'] = df.text.str.findall(r'(?<![@\w])@(\w{1,25})').apply(','.join)
df
#                                                text   mention
#0  RT @CritCareMed: New Article: Male-Predominant...   CritCareMed
#1  #CRISPR Inversion of CTCF Sites Alters Genome ...   CellCellPress
#2  RT @gvwilson: Where's the theory for software ...   gvwilson
#3  RT @sciencemagazine: What’s killing off the se...   sciencemagazine
#4  RT @MHendr1cks: Eve Marder describes a horror ...   MHendr1cks,nucAmbiguous

还有 X.iloc[:i,:] 返回一个数据框，所以 str(X.iloc[:i,:]) 给你字符串数据框的表示，与单元格中的元素非常不同，要从text列中提取实际字符串，可以使用X.text.iloc[0]，或者更好的遍历列的方法，使用 迭代项:

---

Also X.iloc[:i,:] gives back a data frame, so str(X.iloc[:i,:]) gives you the string representation of a data frame, which is very different from the element in the cell, to extract the actual string from the text column, you can use X.text.iloc[0], or a better way to iterate through a column, use iteritems:

import re
for index, s in df.text.iteritems():
    result = re.findall("(?<![@\w])@(\w{1,25})", s)
    print(','.join(result))

#CritCareMed
#CellCellPress
#gvwilson
#sciencemagazine
#MHendr1cks,nucAmbiguous

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