在dthon中将dict转换为已排序的dict

时间:2012-10-25 05:45:03

标签: python pandas sorteddictionary

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}

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}}

(按值的降序排列,结果应为dict)

1 个答案:

答案 0 :(得分:17)

您无法对dict进行排序,因为词典没有排序。

相反,请使用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

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)。这样的事情就是你想要的:

{
  "count": 24,
  "top 5": [
    {"Gears of war 3": 6},
    {"Batman": 5},
    {"Rocksmith": 5},
    {"gears of war 3": 4},
    {"Madden": 3}
  ]
}

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)]

现在您只需将l传递给m['top5']并转储它:

m["Top 5"]= l
k = json.dumps(m)