1

我有一个返回以下字典的函数

abc= {"type":"insecure","id":"1",
       "name":"peter","s_count":"2",
      "b_count":"1", "s_1_name":"melisa",
        "s_1_id":"2","s_2_name":"graham",
      "s_2_id":"4", "b_1_name":"henrik", 
        "b_1_id": "9"}

我想通过以下方式查字典：

xyz={"type":"insecure","id":"1",
     "name":"peter",
      "s" : [{"id" : "2", "name": "melisa"},
             {"id" : "4", "name": "graham"}],
      "b" : [{"id" : "9", "name": "henrik"}]}

逻辑如下：如果字典中有 s_count，则创建一个列表，其中包含所有以 s 开头的值。例如，在我的情况下，创建一个包含不同字典的列表，每个字典包含 s_name 和 s_id 例如，在我的情况下，结果列表中有两个字典：

"s" : [{"id" : "2", "name": "melisa"},{"id" : "4", "name": "graham"}]

如果 b_count count 存在，也对 b 做同样的事情。

有人可以帮我吗？

4

2 回答 2

4

我会使用一个辅助函数：

from itertools import groupby
from operator import itemgetter

def extract_keys(mapping, prefix):
    prefix = '{}_'.format(prefix)

    # test for the `.._count` key first, if it's not there, bail out early
    if prefix + 'count' not in mapping:
        return None

    # find all matching keys, split out the counter for sorting and grouping
    keys = [(k, int(k.split('_', 2)[1]))
        for k in mapping if k.startswith(prefix) and k != prefix + 'count']
    keys.sort(key=itemgetter(1))

    # group keys on the counter, then generate a dictionary per counter value
    return [{k[0].split('_', 2)[-1]: mapping[k[0]] for k in group}
        for c, group in groupby(keys, itemgetter(1))]

此函数按前缀提取键：

>>> extract_keys(abc, 's')
[{'id': '2', 'name': 'melisa'}, {'name': 'graham', 'id': '4'}]
>>> extract_keys(abc, 'b')
[{'name': 'henrik', 'id': '9'}]

使用该功能创建新字典或就地转换现有字典：

xyz = {k: v for k, v in abc.iteritems() if not k.startswith('s_') and not k.startswith('b_')}
s_values = extract_keys(abc, 's')
if s_values is not None:
    xyz['s'] = s_values
b_values = extract_keys(abc, 'b')
if b_values is not None:
    xyz['b'] = b_values

这会将您的abc样本输入转换为：

>>> pprint(xyz)
{'b': [{'id': '9', 'name': 'henrik'}],
 'id': '1',
 'name': 'peter',
 's': [{'id': '2', 'name': 'melisa'}, {'id': '4', 'name': 'graham'}],
 'type': 'insecure'}
于 2013-05-10T15:16:32.350 回答
0

我对其进行了更多改进，使其变得更加灵活。

def extract_keys(mapping, prefix):
prefix = "{}_".format(prefix)

# test for the `.._count` key first, if it's not there, bail out early
if prefix + "count" not in mapping:
    return None

# find all matching keys, split out the counter for sorting and grouping
keys = [(k, int(k.split("_", -1)[-2])) for k in mapping if k.startswith(prefix) and k != prefix + "count"]
keys.sort(key=itemgetter(1))

# group keys on the counter, then generate a dictionary per counter value
return [{k[0].split("_", -1)[-1]: mapping[k[0]] for k in group} for c, group in groupby(keys, itemgetter(1))]

考虑以下字典。

abc= {"type":"insecure","id":"1",
   "name":"peter","s_a_count":"2",
  "b_count":"1", "s_a_1_name":"melisa",
    "s_a_1_id":"2","s_a_2_name":"graham",
  "s_a_2_id":"4", "b_1_name":"henrik", 
    "b_1_id": "9"}

使用该功能创建新字典或就地转换现有字典：

xyz = {k: v for k, v in abc.iteritems() if not k.startswith('s_a') and not k.startswith('b_')}
s_values = extract_keys(abc, 's_a')
if s_values is not None:
    xyz['s_a'] = s_values
b_values = extract_keys(abc, 'b')
if b_values is not None:
    xyz['b'] = b_values

转换后的输出是：

print xyz

{'b': [{'id': '9', 'name': 'henrik'}],
'id': '1',
'name': 'peter',
's_a': [{'id': '2', 'name': 'melisa'}, {'id': '4', 'name': 'graham'}],
'type': 'insecure'}
于 2013-05-11T14:04:23.610 回答