Python-按前两个词分组行

我想按文件中的前两个单词对文件进行分组(然后重新排列并打印)

我想做

   lines=file.readlines() 
   i=0 
   for line in lines: 
    word1=line.split()[0] 
    word2=line.split()[1] 
    if word1==lines[i+1].split()[0] and word1==lines[i-1].split()[0] : 
        if word2=lines[i-1].split()[1] and word2==lines[i--1].split()[0]: 
              print line 
    else: 
       print "***new block of lines \n***" 

这是一个非常糟糕的解决方案，因为它不适用于第一行或最后一行，并且总体上效果不佳。非常感谢更好的解决方案

请您参考如下方法:

如果您尝试对共享文件中前两个单词的连续行进行分组，这就是 itertools.groupby 的用例，例如:

from itertools import groupby 
 
with open('somefile') as fin: 
    lines = ((line.split(None, 2)[:2], line) for line in fin if line.strip()) 
    for k, g in groupby(lines, lambda L: L[0]): 
        lines = [el[1] for el in g] 

此处 k 是分组键(最多前两个单词)，lines 是文件中共享该键的行。

示例 somefile 输入:

one two three four five 
one two five six seven 
three four something 
three four something else 
one two start of new one two block 

打印 k, lines 的结果:

['one', 'two'] ['one two three four five\n', 'one two five six seven\n'] 
['three', 'four'] ['three four something\n', 'three four something else\n'] 
['one', 'two'] ['one two start of new one two block\n'] 

要从 line 中排除前两个词，请使用:

with open('somefile') as fin: 
    lines = (line.split(None, 2) for line in fin if line.strip()) 
    for k, g in groupby(lines, lambda L: L[:2]): 
        lines = [el[2] for el in g]