我的数据框如下：
Name_ID | URL | Count | Rating
------------------------------------------------
ABC | www.example.com/ABC | 10 | 5
123 | www.example.com/123 | 9 | 4
XYZ | www.example.com/XYZ | 5 | 2
ABC111 | www.example.com/ABC111 | 5 | 2
ABC121 | www.example.com/ABC121 | 5 | 2
222 | www.example.com/222 | 5 | 3
abc222 | www.example.com/abc222 | 4 | 2
ABCaaa | www.example.com/ABCaaa | 4 | 2
我正在尝试按如下方式创建JSON：
{
"name": "sampledata",
"children": [
{
"name": 9,
"children": [
{
"name": 4,
"children": [
{
"name": "123",
"size": 100
}
]
}
]
},
{
"name": 10,
"children": [
{
"name": 5,
"children": [
{
"name": "ABC",
"size": 100
}
]
}
]
},
{
"name": 4,
"children": [
{
"name": 2,
"children": [
{
"name": "abc222",
"size": 50
},
{
"name": "ABCaaa",
"size": 50
}
]
}
]
},
{
"name": 5,
"children": [
{
"name": 2,
"children": [
{
"name": "ABC",
"size": 16
},
{
"name": "ABC111",
"size": 16
},
{
"name": "ABC121",
"size": 16
}
]
},
{
"name": 3,
"children": [
{
"name": "222",
"size": 50
}
]
}
]
}
]
}
为了做到这一点：
"name"和"children"等标签。 我试过像
这样的东西results = [{"name": i, "children": j} for i,j in results.items()]
但我相信它不会标记它。
此外，添加标签为“＆＃34;尺寸＆＃34”的另一个字段;我计划根据公式计算该字段：
(Rating*Count*10000)/number_of_children_to_the_immediate_parent
这是我的脏代码：
import pandas as pd
from collections import defaultdict
import json
data =[('ABC', 'www.example.com/ABC', 10 , 5), ('123', 'www.example.com/123', 9, 4), ('XYZ', 'www.example.com/XYZ', 5, 2), ('ABC111', 'www.example.com/ABC111', 5, 2), ('ABC121', 'www.example.com/ABC121', 5, 2), ('222', 'www.example.com/222', 5, 3), ('abc222', 'www.example.com/abc222', 4, 2), ('ABCaaa', 'www.example.com/ABCaaa', 4, 2)]
df = pd.DataFrame(data, columns=['Name', 'URL', 'Count', 'Rating'])
gp = df.groupby(['Count'])
dict_json = {"name": "flare"}
children = []
for name, group in gp:
temp = {}
temp["name"] = name
temp["children"] = []
rgp = group.groupby(['Rating'])
for n, g in rgp:
temp2 = {}
temp2["name"] = n
temp2["children"] = g.reset_index().T.to_dict().values()
for t in temp2["children"]:
t["size"] = (t["Rating"] * t["Count"] * 10000) / len(temp2["children"])
t["name"] = t["Name"]
del t["Count"]
del t["Rating"]
del t["URL"]
del t["Name"]
del t["index"]
temp["children"].append(temp2)
children.append(temp)
dict_json["children"] = children
print json.dumps(dict_json, indent=4)
虽然上面的代码确实打印了我需要的东西，但我正在寻找更高效，更清晰的方法来做同样的事情，主要是因为实际的数据集可能更加嵌套和复杂。任何帮助/建议将不胜感激。
答案 0 :(得分：9)
相当有趣的问题和一个很好的问题！
您可以通过重新组织循环内的代码并使用list comprehensions来改进您的方法。无需删除内容并在循环中引入临时变量：
dict_json = {"name": "flare"}
children = []
for name, group in gp:
temp = {"name": name, "children": []}
rgp = group.groupby(['Rating'])
for n, g in rgp:
temp["children"].append({
"name": n,
"children": [
{"name": row["Name"],
"size": row["Rating"] * row["Count"] * 10000 / len(g)}
for _, row in g.iterrows()
]
})
children.append(temp)
dict_json["children"] = children
或者，＆＃34;包裹＆＃34;版本：
dict_json = {
"name": "flare",
"children": [
{
"name": name,
"children": [
{
"name": n,
"children": [
{
"name": row["Name"],
"size": row["Rating"] * row["Count"] * 10000 / len(g)
} for _, row in g.iterrows()
]
} for n, g in group.groupby(['Rating'])
]
} for name, group in gp
]
}
我为您打印了以下字典示例输入数据框：
{
"name": "flare",
"children": [
{
"name": 4,
"children": [
{
"name": 2,
"children": [
{
"name": "abc222",
"size": 40000
},
{
"name": "ABCaaa",
"size": 40000
}
]
}
]
},
{
"name": 5,
"children": [
{
"name": 2,
"children": [
{
"name": "XYZ",
"size": 33333
},
{
"name": "ABC111",
"size": 33333
},
{
"name": "ABC121",
"size": 33333
}
]
},
{
"name": 3,
"children": [
{
"name": "222",
"size": 150000
}
]
}
]
},
{
"name": 9,
"children": [
{
"name": 4,
"children": [
{
"name": "123",
"size": 360000
}
]
}
]
},
{
"name": 10,
"children": [
{
"name": 5,
"children": [
{
"name": "ABC",
"size": 500000
}
]
}
]
}
]
}
答案 1 :(得分：3)
如果我理解正确你要做的是将一个groupby放入嵌套的json中，如果是这种情况，那么你可以使用pandas groupby并将其转换为嵌套的列表列表，如下所示：
lol = pd.DataFrame(df.groupby(['Count','Rating'])\
.apply(lambda x: list(x['Name_ID']))).reset_index().values.tolist()
[['10', '5', ['ABC']],
['4', '2', ['abc222', 'ABCaaa']],
['5', '2', ['XYZ ', 'ABC111', 'ABC121']],
['5', '3', ['222 ']],
['9', '4', ['123 ']]]
之后你可以循环lol把它放到dict中，但是既然你想设置嵌套项，你就必须使用autovivification（检查出来）：
class autovividict(dict):
def __missing__(self, key):
value = self[key] = type(self)()
return value
d = autovividict()
for l in lol:
d[l[0]][l[1]] = l[2]
现在您可以使用json包进行打印和导出：
print json.dumps(d,indent=2)
如果你需要多个groupby，你可以用pandas连接你的组，转换为lol，删除任何nans，然后循环，让我知道一个完整的例子是否有帮助。
答案 2 :(得分：1)
设置
from io import StringIO
import pandas as pd
txt = """Name_ID,URL,Count,Rating
ABC,www.example.com/ABC,10,5
123,www.example.com/123,9,4
XYZ,www.example.com/XYZ,5,2
ABC111,www.example.com/ABC111,5,2
ABC121,www.example.com/ABC121,5,2
222,www.example.com/222,5,3
abc222,www.example.com/abc222,4,2
ABCaaa,www.example.com/ABCaaa,4,2"""
df = pd.read_csv(StringIO(txt))
<强> size
预先计算它
df['size'] = df.Count.mul(df.Rating) \
.mul(10000) \
.div(df.groupby(
['Count', 'Rating']).Name_ID.transform('count')
).astype(int)
<强> 溶液
创建递归函数
def h(d):
if isinstance(d, pd.Series): d = d.to_frame().T
rec_cond = d.index.nlevels > 1 or d.index.nunique() > 1
return {'name': str(d.index[0]), 'size': str(d['size'].iloc[0])} if not rec_cond else \
[dict(name=str(n), children=h(g.xs(n))) for n, g in d.groupby(level=0)]
演示
import json
my_dict = dict(name='flare', children=h(df.set_index(['Count', 'Rating', 'Name_ID'])))
json.dumps(my_dict)
'{“name”：“flare”，“children”：[{“name”：“4”，“children”：[{“name”：“2”，“children”：[{“name” ：“ABCaaa”，“children”：{“name”：“ABCaaa”，“size”：“40000”}}，{“name”：“abc222”，“children”：{“name”：“abc222”， “size”：“40000”}}]}]}，{“name”：“5”，“children”：[{“name”：“2”，“children”：[{“name”：“ABC111” ，“children”：{“name”：“ABC111”，“size”：“33333”}}，{“name”：“ABC121”，“children”：{“name”：“ABC121”，“size”： “33333”}}，{“name”：“XYZ”，“children”：{“name”：“XYZ”，“size”：“33333”}}]}，{“name”：“3”，“儿童“：{”name“：”222“，”size“：”150000“}}]}，{”name“：”9“，”children“：[{”name“：”4“，”children“ ：{“name”：“123”，“size”：“360000”}}]}，{“name”：“10”，“children”：[{“name”：“5”，“children”：{ “name”：“ABC”，“size”：“500000”}}]}]}'
my_dict
{'children': [{'children': [{'children': [{'children': {'name': 'ABCaaa',
'size': '40000'},
'name': 'ABCaaa'},
{'children': {'name': 'abc222', 'size': '40000'}, 'name': 'abc222'}],
'name': '2'}],
'name': '4'},
{'children': [{'children': [{'children': {'name': 'ABC111', 'size': '33333'},
'name': 'ABC111'},
{'children': {'name': 'ABC121', 'size': '33333'}, 'name': 'ABC121'},
{'children': {'name': 'XYZ', 'size': '33333'}, 'name': 'XYZ'}],
'name': '2'},
{'children': {'name': '222', 'size': '150000'}, 'name': '3'}],
'name': '5'},
{'children': [{'children': {'name': '123', 'size': '360000'}, 'name': '4'}],
'name': '9'},
{'children': [{'children': {'name': 'ABC', 'size': '500000'}, 'name': '5'}],
'name': '10'}],
'name': 'flare'}