在一行中找到一个子字符串并在python中打印该行并仅打印一行？

Question

我试图在我的文本文件中打印包含我列表中任何单词的所有行，并且显示“名称”。我遇到的问题是我的程序迭代太多，重复的行会因为多次迭代而打印出来。我怎样才能打印线路发生一次？另外我怎样才能将行打印到输出文件？

这是我到目前为止有：

names=[bob,carter,jim,mike] 
with open("base.txt") as openfile: 
     for line in openfile: 
      for part in line.split(): 
       for i in names: 
        if i in part: 

         print line 

Answer 1

无需分割线，只需检查线作为一个整体包含名称。另外，不需要检查每个名字，第一场比赛就会完成。 any将帮助你避免一些坎坷代码：

with open("base.txt") as openfile: 
    for line in openfile: 
     if any(name in line for name in names): 
      print line 

Answer 2

检查一次所有的名字，并使用any()如下。

names=['bob','carter','jim','mike'] 
with open("base.txt") as openfile: 
    for line in openfile: 
     if any([n in line for n in names]): 
      print line.strip() 

[n in line for n in names]所做的是检查行中的每个名称并返回一个布尔值列表。 any()检查列表中的任何元素是否为True。

Answer 3

您可以使用正则表达式匹配字符串：

import re 

names=["bob","carter","jim","mike"] 
match_string = "(" + ")|(".join(names)+")" #create a regex that can match all the words in the list names 
outfile = open("out.txt","w") #open output file 


with open("base.txt") as openfile: 
     for line in openfile: 
       if re.search(match_string,line): 
         outfile.write(line) #writes output to the file 
         print line 

outfile.close() 

Answer 4

正如其他已经发布，你可以使用any确认在该行的名称中至少一个的发生。使用列表理解把所有匹配的行成一个列表：

with open("base.txt") as openfile, open("output.txt", "w") as outputfile: 
    result = [line if any(n in line for n in names) for line in openfile] 
    outputfile.writelines(result) # wwii's comment: the lines already contain a separator 

要写入result到outputfile，你应该使用writelines方法，采取result序列参数（@二战的评论）。