Wan从数据中按日期和状态获取分组答案

【问题标题】：Wan to get group by date and status from the dataWan从数据中按日期和状态获取分组
【发布时间】：2022-01-17 18:14:21
【问题描述】：

我有以下数据

data=[
{"name":"aa"..."created_at":"2022-01-17 07:38:26.403Z","status":"success"},
{"name":"bb"..."created_at":"2021-12-1 07:38:26.403Z","status":"failed"},
{"name":"kk"..."created_at":"2022-01-13 07:38:26.403Z","status":"success"},

{"name":"ll"..."created_at":"2021-12-17 07:38:26.403Z","status":"success"},

]

我想要一份类似的报告

created_at,success

December,1
Jan,2

我在下面尝试过

 d = defaultdict(int)
 for i in kk:
     if(i["status"] == 'success')
     d[i["status"]]+=1

它只给出成功计数，但不能按月计算。

【问题讨论】：

标签： python

【解决方案1】：

您可以使用pd.to_datetime 将您的“created_at”列转换为日期时间，过滤您的“成功”行，然后每月使用groupby 聚合count：

df['created_at'] = pd.to_datetime(df['created_at'])
out = df.loc[df.status.eq('success')].groupby(df.created_at.dt.month).status.count()

打印：

   created_at  success
0           1        2
1          12        1

# as a dict

>>> out.set_index('created_at').to_dict()
{'success': {1: 2, 12: 1}}

然后您可以将月份数字转换为月份名称，如图所示here

【讨论】：

@Rajarshi DAs 我相信接受的答案使用pandas 库进行分析。因此，请在将来适当地标记您的问题以避免任何混淆。目前它仅在python 下标记。

【解决方案2】：

您必须先使用datetime.strptime 解析日期，然后使用itetools.groupby 根据月份进行分组。

>>> from datetime import datetime
>>> import calendar
>>>
>>> from itertools import groupby
>>>
>>> data = [
...     {"name": "aa", "created_at": "2022-01-17 07:38:26.403Z", "status": "success"},
...     {"name": "bb", "created_at": "2021-12-1 07:38:26.403Z", "status": "failed"},
...     {"name": "kk", "created_at": "2022-01-13 07:38:26.403Z", "status": "success"},
...     {"name": "ll", "created_at": "2021-12-17 07:38:26.403Z", "status": "success"},
... ]
>>>
>>>
>>> def custom_key_fun(row):
...     created_at_as_datetime = datetime.strptime(
...         row["created_at"], "%Y-%m-%d %H:%M:%S.%fZ"
...     )
...     month = calendar.month_abbr[created_at_as_datetime.month]
...     return month
...
>>>
>>> result = {
...     key: sum(1 for row in group if row["status"] == "success")
...     for key, group in groupby(sorted(data, key=custom_key_fun), key=custom_key_fun)
... }
>>>
>>> print(result)
{'Dec': 1, 'Jan': 2}

【讨论】：

【解决方案3】：

加载熊猫

import pandas as pd
from datetime import datetime

data = [{"name": "aa", "created_at": "2022-01-17 07:38:26.403Z", "status": "success"},
        {"name": "bb", "created_at": "2021-12-1 07:38:26.403Z", "status": "failed"},
        {"name": "kk", "created_at": "2022-01-13 07:38:26.403Z", "status": "success"},
        {"name": "ll", "created_at": "2021-12-17 07:38:26.403Z", "status": "success"},
        ]
df = pd.DataFrame(data)
df
    name    created_at  status
0   aa  2022-01-17 07:38:26.403Z    success
1   bb  2021-12-1 07:38:26.403Z     failed
2   kk  2022-01-13 07:38:26.403Z    success
3   ll  2021-12-17 07:38:26.403Z    success

构建新列

def date_convert(x):
    x = x.split('.')[0]
    return datetime.strptime(x, "%Y-%m-%d %H:%M:%S").strftime("%B")


df['month_name'] = df.created_at.apply(date_convert)
df
    name    created_at  status  month_name
0   aa  2022-01-17 07:38:26.403Z    success     January
1   bb  2021-12-1 07:38:26.403Z     failed  December
2   kk  2022-01-13 07:38:26.403Z    success     January
3   ll  2021-12-17 07:38:26.403Z    success     December

计算并构建新的 DataFrame

group_month = df[df.status == "success"].groupby('month_name').groups
data = [{"created_at": i, "success": len(j)} for i, j in group_month.items()]
new_df = pd.DataFrame(data)
new_df
    created_at  success
0   December    1
1   January     2

【讨论】：