我希望使用功能和迭代工具将长数据集转换为宽数据集,我的理解是这是groupby的任务.我以前曾经问了几个关于这个的问题,并且认为我有这个问题,但在这种情况下并不完全,这应该更简单：

> Python functional transformation of JSON list of dictionaries from long to wide
> Correct use of a fold or reduce function to long-to-wide data in python or javascript?

这是我的数据：

from itertools import groupby
from operator import itemgetter
from pprint import pprint

>>> longdat=[
{"id":"cat", "name" : "best meower", "value": 10},
{"id":"cat", "name" : "cleanest paws", "value": 8},
{"id":"cat", "name" : "fanciest", "value": 9},
{"id":"dog", "name" : "smelly", "value": 9},
{"id":"dog", "name" : "dumb", "value": 9},
]

这是我想要的格式：

>>> widedat=[
{"id":"cat", "best meower": 10, "cleanest paws": 8, "fanciest": 9},
{"id":"dog", "smelly": 9, "dumb": 9},
]

以下是我失败的尝试：

# WRONG
>>> gh = groupby(sorted(longdat,key=id),itemgetter('id'))
>>> list(gh)
[('cat', <itertools._grouper object at 0x5d0b550>), ('dog', <itertools._grouper object at 0x5d0b210>)]

好的,需要从迭代器中获取第二个项目,这是公平的.

#WRONG
>>> gh = groupby(sorted(longdat,key=id),itemgetter('id'))
>>> for g,v in gh:
...     {"id":i["id"], i["name"]:i["value"] for i in v}
                                      ^
SyntaxError: invalid syntax

很奇怪,它看起来很有效.让我们解开那些循环以确保.

#WRONG
gb = groupby(sorted(longdat,key=id),itemgetter('id'))
data = {}
for g,v in gb:
    data[g] = {}
    for i in v:
        data[g] = i

#WRONG
gb = groupby(sorted(longdat,key=id),itemgetter('id'))
data = []
for g,v in gb:
    for i in v:
        data[g] = i

啊!好的,让我们回到单行表格

#WRONG
>>> gb = groupby(sorted(longdat,key=id),itemgetter('id'))
>>> [{"id":g, i["name"]:i["value"]} for i in k for g,k in gb]
[]

什么？为什么空？！让我们再次基本上解开这个：

#WRONG
gb = groupby(sorted(longdat,key=id),itemgetter('id'))
for g,k in gb:
    for i in k:
       print(g, i["name"],i["value"])
cat best meower 10
cat fanciest 9
cat cleanest paws 8
dog smelly 9
dog dumb 9

现在,最后一个显然是最糟糕的 – 很明显我的数据基本上就在它开始的地方,好像我甚至没有组合.

为什么这不起作用,我怎么能以我正在寻找的格式得到它？

此外,是否可能完全迭代地对此进行说明,以便我可以这样做

>>> result[0]
{"id":"cat", "best meower": 10, "cleanest paws": 8, "fanciest": 9}

并且只获得第一个结果而不处理整个列表(除了必须查看/ all / where id ==’cat’？)

解决方法:

传递给排序函数的键函数是id.它将返回所有列表项的所有不同值.

它应该是itemgetter(‘id’)或lambda x：x.id.

>>> id(longdat[0])
41859624L
>>> id(longdat[1])
41860488L
>>> id(longdat[2])
41860200L
>>> itemgetter('id')(longdat[1])
'cat'
>>> itemgetter('id')(longdat[2])
'cat'
>>> itemgetter('id')(longdat[3])
'cat'

from itertools import groupby
from operator import itemgetter

longdat = [
    {"id":"cat", "name" : "best meower", "value": 10},
    {"id":"cat", "name" : "cleanest paws", "value": 8},
    {"id":"cat", "name" : "fanciest", "value": 9},
    {"id":"dog", "name" : "smelly", "value": 9},
    {"id":"dog", "name" : "dumb", "value": 9},
]

getid = itemgetter('id')
result = [
    dict([['id', key]] + [[d['name'], d['value']] for d in grp])
    for key, grp in groupby(sorted(longdat, key=getid), key=getid)
]
print(result)

输出：

[{'best meower': 10, 'fanciest': 9, 'id': 'cat', 'cleanest paws': 8},
 {'dumb': 9, 'smelly': 9, 'id': 'dog'}]