基于某个键/值合并/总结一系列字符

Question

我有一个如下所示的字典列表。

[ 
    {"value1": 53, "day": "Thu, 07 May 2015", "value2": 70, "type_of": "foo"}, 
    {"value1": 17, "day": "Thu, 07 May 2015", "value2": 12, "type_of": "foo"}, 
    {"value1": 21, "day": "Thu, 12 May 2013", "value2": 40, "type_of": "foo"} 
] 

现在，我想这对在求和"value1"/"value2"键和离开单独"type_of"的方式"day"键相同的值类型的字典结合起来。

下面是我想要的结果：

[ 
    {"value1": 70, "day": "Thu, 07 May 2015", "value2": 82, "type_of": "foo"}, 
    {"value1": 21, "day": "Thu, 12 May 2013", "value2": 40, "type_of": "foo"} 
] 

Answer 1

更多与defaultdict紧凑的解决方案：

from collections import defaultdict 

# data array in df 

res = defaultdict (lambda : {"value1": 0, "value2": 0, "day": "", "type_of": "foo"}) 

for x in df: 
    res[x["day"]]["value1"] += x["value1"] 
    res[x["day"]]["value2"] += x["value2"] 
    res[x["day"]]["day"] = x["day"] 
    res[x["day"]]["type_of"] = x["type_of"] 


print (res) 

Answer 2

我建立类似一个你需要，希望它会帮助你完成你的任务，一些代码。

list_dicts = [ 
    {"value1": 53, "day": "Thu, 07 May 2015", "value2": 70, "type_of": "foo",}, 
    {"value1": 17, "day": "Thu, 07 May 2015","value2": 12,"type_of": "foo"}, 
    {"value1": 21, "day": "Thu, 12 May 2013", "value2": 40, "type_of": "foo"} 
] 

dicts_by_day = {} 

for ldict in list_dicts: 
    if ldict["day"] in dicts_by_day: 
     dicts_by_day[ldict["day"]].append(ldict) 
    else: 
     dicts_by_day[ldict["day"]] = [ldict] 

for day, values in dicts_by_day.items(): 
    print(day,values) 

在最后一次迭代中，您可以对每天的组进行操作。

尝试编辑您的问题（你的“list_dicts”不正确写入）

Answer 3

任何理由不使用熊猫吗？ 这是非常方便，让您的生活变得更轻松：

import pandas as pd 

l1 = [{"value1": 53, "day": "Thu, 07 May 2015", "value2": 70, "type_of": "foo"}, 
    {"value1": 17, "day": "Thu, 07 May 2015","value2": 12,"type_of" : "foo"}, 
    {"value1": 21, "day": "Thu, 12 May 2013", "value2": 40, "type_of": "foo"}] 


# Create a "table" with keys as column names and set as index the pair (day, type_of) 
df1 = pd.DataFrame(l1).set_index(['day', 'type_of']) 

# Here you sum values with same day and type_of. This is already what you are 
# looking for  
out_df = df1.groupby(level=[0, 1]).sum() 
# This last bit is to convert back to dictionary, although I would suggest to 
# use pandas dataframes rather than lists of dictionaries 
out_dict = out_df.reset_index().to_dict(orient = 'record') 

# out = [{'day': 'Thu, 07 May 2015', 'value1': 70L, 'value2': 82L}, 
#  {'day': 'Thu, 12 May 2013', 'value1': 21L, 'value2': 40L}]