Cod*_*nja 9 python sorteddictionary pandas
我想在dthon中将dict转换为已排序的dict
data = pandas.read_csv('D:\myfile.csv')
for colname, dtype in data.dtypes.to_dict().iteritems():
if dtype == 'object':
print colname
count = data[colname].value_counts()
d = dict((str(k), int(v)) for k, v in count.iteritems())
f = dict(sorted(d.iteritems(), key=lambda item: item[1], reverse = True)[:5])
print f
m ={}
m["count"]= int(sum(count))
m["Top 5"]= f
print m
k = json.dumps(m)
print k
f = {'Gears of war 3': 6, 'Batman': 5, 'gears of war 3': 4, 'Rocksmith': 5, 'Madden': 3}
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我想要的输出是:
f = {'Gears of war 3': 6, 'Batman': 5, 'Rocksmith': 5, 'gears of war 3': 4, 'Madden': 3}
k = {'count':24, 'top 5':{'Gears of war 3': 6, 'Batman': 5, 'Rocksmith': 5, 'gears of war 3': 4, 'Madden': 3}}
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(按值的降序排列,结果应为dict)
K Z*_*K Z 18
你无法排序,dict
因为字典没有排序.
相反,使用collections.OrderedDict
:
>>> from collections import OrderedDict
>>> d = {'Gears of war 3': 6, 'Batman': 5, 'gears of war 3': 4, 'Rocksmith': 5, 'Madden': 3}
>>> od = OrderedDict(sorted(d.items(), key=lambda x:x[1], reverse=True))
>>> od
OrderedDict([('Gears of war 3', 6), ('Batman', 5), ('gears of war 3', 4), ('Rocksmith', 5), ('Madden', 3)])
>>> od.keys()
['Gears of war 3', 'Batman', 'gears of war 3', 'Rocksmith', 'Madden']
>>> od.values()
[6, 5, 4, 5, 3]
>>> od['Batman']
5
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您在JSON对象中看到的"顺序"没有意义,因为JSON对象是无序的[ RFC4267 ].
如果您想在JSON中进行有意义的排序,则需要使用一个列表(按照您想要的方式排序).这样的东西就是你想要的东西:
{
"count": 24,
"top 5": [
{"Gears of war 3": 6},
{"Batman": 5},
{"Rocksmith": 5},
{"gears of war 3": 4},
{"Madden": 3}
]
}
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给定相同的dict d
,您可以通过以下方式生成排序列表(这是您想要的):
>>> l = sorted(d.items(), key=lambda x:x[1], reverse=True)
>>> l
[('Gears of war 3', 6), ('Batman', 5), ('Rocksmith', 5), ('gears of war 3', 4), ('Madden', 3)]
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现在你只是传递l
到m['top5']
和转储:
m["Top 5"]= l
k = json.dumps(m)
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