字典包含列表

Question

我有一個返回以下字典

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"} 

我想恰克下列方式字典的功能：

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

的邏輯如下：如果有s_count在字典中，然後創建一個列表，其中包含以s開頭的所有值。例如，在我的情況下創建包含與每個字典不同的字典包含S_NAME和S_ID例如在我的情況下，有兩個字典在結果列表中AA列表：

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

，做同樣的用b，以及如果b_count計數存在。

有人能幫我解決嗎？

Answer 1

我會使用一個輔助函數：

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

使用該函數來創建新的詞典或就地改造現有字典：

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))] 

此功能通過前綴提取鍵：

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 

這會將您的abc樣品輸入到：

>>> pprint(xyz) 
{'b': [{'id': '9', 'name': 'henrik'}], 
'id': '1', 
'name': 'peter', 
's': [{'id': '2', 'name': 'melisa'}, {'id': '4', 'name': 'graham'}], 
'type': 'insecure'} 

Answer 2

我改進了一點，使它變得更加靈活。

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))] 

請考慮以下詞典。

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"} 

使用該函數來創建新的詞典或就地改造現有字典：

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 

，並轉化輸出爲：

print xyz 

{'b': [{'id': '9', 'name': 'henrik'}], 
'id': '1', 
'name': 'peter', 
's_a': [{'id': '2', 'name': 'melisa'}, {'id': '4', 'name': 'graham'}], 
'type': 'insecure'}