Python-按前两个单词对行进行分组

我想按文件中的前两个单词对文件进行分组（然后重新排列和打印）

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***"

但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

如果您尝试对共享文件中前两个单词的连续行进行分组，那么这是一个用例itertools.groupby, eg:

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]

Here k是分组键（最多前两个单词）并且lines将是文件中共享该密钥的行。

Example somefile input:

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

的结果print k, lines:

['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']

从列表中排除前两个单词line, use:

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]