我有一个数据集，有8个属性(按第一个属性排序)，格式如下(仅举个例子，是制表符分隔的)
AX 0123 December 20, 2010 1 2 8.0 hello this
AX 2313 April 19, 2009 2 3 4.0 hi there
AX 4532 December 19, 2010 6 2 8.0 nice tie
AX 1244 January 10, 2011 3 4 8.0 king tale
BX 0214 September 10, 2009 2 3 9.0 this king
BX 0114 February 9, 2003 4 9 4.0 his brought
BX 3214 September 1, 2006 1 3 3.0 is great
MG 980 April 20, 2007 2 4 7.1 not available
MG 246 May 8, 2005 5 1 2.1 make goat
现在，文件已根据第一个属性排序，现在我需要根据第一个属性在内部根据日期排序，输出应该是这样的(我不想使用数据库，这是一个巨大的文件(2 GB) 所以我认为可能需要一个特殊的 python 代码(不确定是否可以用一个简单的代码来做到这一点)
AX 2313 April 19, 2009 2 3 4.0 hi there
AX 4532 December 19, 2010 6 2 8.0 nice tie
AX 0123 December 20, 2010 1 2 8.0 hello this
AX 1244 January 10, 2011 3 4 8.0 king tale
BX 0114 February 9, 2003 4 9 4.0 his brought
BX 3214 September 1, 2006 1 3 3.0 is great
BX 0214 September 10, 2009 2 3 9.0 this king
MG 246 May 8, 2005 5 1 2.1 make goat
MG 980 April 20, 2007 2 4 7.1 not available
非常感谢任何回复。如果您有任何其他问题，请告诉我。
请您参考如下方法:
好的，又快又脏。你可以改进它:
from datetime import datetime as dt
from collections import defaultdict
dd = defaultdict(list)
with open('test.txt') as f:
for line in f:
lines = line.split('\t')
dd[lines[0]].append(lines)
def mydate(line):
return dt.strptime(line[2], "%B %d, %Y")
keys = sorted(dd.keys())
my_list = []
for key in keys:
dd[key].sort(key=mydate)
my_list.extend(dd[key])
for item in my_list:
print item
这会产生:
['AX', '2313', 'April 19, 2009', '2', '3', '4.0', 'hi there\n']
['AX', '4532', 'December 19, 2010', '6', '2', '8.0', 'nice tie\n']
['AX', '0123', 'December 20, 2010', '1', '2', '8.0', 'hello this\n']
['AX', '1244', 'January 10, 2011', '3', '4', '8.0', 'king tale\n']
['BX', '0114', 'February 9, 2003', '4', '9', '4.0', 'his brought\n']
['BX', '3214', 'September 1, 2006', '1', '3', '3.0 is great\n']
['BX', '0214', 'September 10, 2009', '2', '3', '9.0 this king\n']
['MG', '246', 'May 8, 2005', '5', '1', '2.1', 'make goat']
['MG', '980', 'April 20, 2007', '2', '4', '7.1', 'not available\n']
那么你只需要string.join()所有的列表
text_lines = []
for item in my_list:
text_lines.append('\t'.join(item))
full_text = ''.join(text_lines)
