Question

我有一个dicts列表，如下所示。

[
    {"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，我正在尝试将"day"键具有相同值的词组与"value1" / "value2"键相加并离开{{1}单独。

以下是我的预期结果：

"type_of"

Answer 1

任何不使用熊猫的理由？它非常方便，让您的生活更轻松：

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}]

Answer 2

我构建了一个类似于你需要的代码，希望它能帮助你完成任务。

android:orientation="vertical"

在最后一次迭代中，您可以为每一天的小组进行操作。

尝试编辑您的问题（您的“list_dicts”未正确写入）

Answer 3

使用defaultdict的更紧凑的解决方案：

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)

根据特定键/值

3 个答案: