从python中的平面列表创建嵌套字典的通用方法

Question

我正在寻找最简单的通用方法来转换这个python列表：

x = [
        {"foo":"A", "bar":"R", "baz":"X"},
        {"foo":"A", "bar":"R", "baz":"Y"},
        {"foo":"B", "bar":"S", "baz":"X"},
        {"foo":"A", "bar":"S", "baz":"Y"},
        {"foo":"C", "bar":"R", "baz":"Y"},
    ]

成：

foos = [ 
         {"foo":"A", "bars":[
                               {"bar":"R", "bazs":[ {"baz":"X"},{"baz":"Y"} ] },
                               {"bar":"S", "bazs":[ {"baz":"Y"} ] },
                            ]
         },
         {"foo":"B", "bars":[
                               {"bar":"S", "bazs":[ {"baz":"X"} ] },
                            ]
         },
         {"foo":"C", "bars":[
                               {"bar":"R", "bazs":[ {"baz":"Y"} ] },
                            ]
         },
      ]

组合“foo”，“bar”，“baz”是唯一的，正如您所看到的，列表不一定按此键排序。

Answer 1

#!/usr/bin/env python3
from itertools import groupby
from pprint import pprint

x = [
        {"foo":"A", "bar":"R", "baz":"X"},
        {"foo":"A", "bar":"R", "baz":"Y"},
        {"foo":"B", "bar":"S", "baz":"X"},
        {"foo":"A", "bar":"S", "baz":"Y"},
        {"foo":"C", "bar":"R", "baz":"Y"},
    ]


def fun(x, l):
    ks = ['foo', 'bar', 'baz']
    kn = ks[l]
    kk = lambda i:i[kn]
    for k,g in groupby(sorted(x, key=kk), key=kk):
        kg = [dict((k,v) for k,v in i.items() if k!=kn) for i in g]
        d = {}
        d[kn] = k
        if l<len(ks)-1:
            d[ks[l+1]+'s'] = list(fun(kg, l+1))
        yield d

pprint(list(fun(x, 0)))

---

[{'bars': [{'bar': 'R', 'bazs': [{'baz': 'X'}, {'baz': 'Y'}]},
           {'bar': 'S', 'bazs': [{'baz': 'Y'}]}],
  'foo': 'A'},
 {'bars': [{'bar': 'S', 'bazs': [{'baz': 'X'}]}], 'foo': 'B'},
 {'bars': [{'bar': 'R', 'bazs': [{'baz': 'Y'}]}], 'foo': 'C'}]

---

注意： dict是无序的！ 但它和你的一样。

Answer 2

我将定义一个执行单个分组步骤的函数，如下所示：

from itertools import groupby
def group(items, key, subs_name):
    return [{
        key: g,
        subs_name: [dict((k, v) for k, v in s.iteritems() if k != key)
            for s in sub]
    } for g, sub in groupby(sorted(items, key=lambda item: item[key]),
        lambda item: item[key])]

然后呢

[{'foo': g['foo'], 'bars': group(g['bars'], "bar", "bazs")} for g in group(x,
     "foo", "bars")]

这给了foos想要的结果。

Answer 3

这是一个简单的数据循环，没有递归。 值为字典键的辅助树在构建时用作结果树的索引。

def make_tree(diclist, keylist):
    indexroot = {}
    root = {}
    for d in diclist:
        walk = indexroot
        parent = root
        for k in keylist:
            walk = walk.setdefault(d[k], {})
            node = walk.setdefault('node', {})
            if not node:
                node[k] = d[k]
                parent.setdefault(k+'s',[]).append(node)
            walk = walk.setdefault('children', {})
            parent = node
    return root[keylist[0]+'s']

foos = make_tree(x, ["foo","bar","baz"])