dict(groupby) 和 groupby 的区别 [英] difference between dict(groupby) and groupby
本文介绍了dict(groupby) 和 groupby 的区别的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我有一个这样的清单
[u'201003', u'200403', u'200803', u'200503', u'201303',
u'200903', u'200603', u'201203', u'200303', u'200703', u'201103']
我们将此列表称为years_list"
lets call this list as 'years_list'
当我按年分组时,
group_by_yrs_list = groupby(years_list, key = lambda year_month: year_month[:-2])
for k,v in group_by_yrs_list:
print k, list(v)
我得到了想要的输出:
2010 [u'201003']
2004 [u'200403']
2008 [u'200803']
2005 [u'200503']
2013 [u'201303']
2009 [u'200903']
2006 [u'200603']
2012 [u'201203']
2003 [u'200303']
2007 [u'200703']
2011 [u'201103']
然后,我像这样稍微改变了我的实现,
Then, I slightly changed my implementation like this,
group_by_yrs_list = dict(groupby(years_list, key = lambda year_month: year_month[:-2]))
for k,v in group_by_yrs_list.items():
print k, list(v)
我刚刚添加了一个字典,但输出不同,
I have just added a dict, but the output is different,
2003 []
2006 []
2007 []
2004 []
2005 []
2008 []
2009 []
2011 [u'201103']
2010 []
2013 []
2012 []
我不知道为什么.请帮我找出字典实际上在做什么.
I couldn't find out why. Please help me to find what the dict is doing actually.
(Python 2.7)
(Python 2.7)
推荐答案
groupby
产生一对 (key, iterator-of-group).如果您要迭代第二对,则第一对的迭代器组已被消耗,因此您会得到空列表.
groupby
yields pairs of (key, iterator-of-group). If you are iterating the second pair, the iterator-of-group of the first pair is already consumed, so you get empty list.
试试下面的代码:
group_by_yrs_list = {year:list(grp) for year, grp in groupby(years_list, key=lambda year_month: year_month[:-2])}
for k, v in group_by_yrs_list.items():
print k, v
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