在单行中总结熊猫数据框答案

【问题标题】：Summarising pandas dataframe in single row在单行中总结熊猫数据框
【发布时间】：2021-01-22 20:05:21
【问题描述】：

我希望在将下面详述的数据框总结为一行摘要时得到一些帮助，如页面下方所需的输出所示。非常感谢。

employees = {'Name of Employee': ['Mark','Mark','Mark','Mark','Mark','Mark', 'Mark','Mark','Mark','Mark','Mark','Mark','Mark'],
                         'Department': ['21','21','21','21','21','21', '21','21','21','21','21','21','21'],
                         'Team': ['2','2','2','2','2','2','2','2','2','2','2','2','2'],
                         'Log': ['2020-02-19 09:01:17', '2020-02-19 09:54:02', '2020-04-10 11:00:31', '2020-04-11 12:39:08', '2020-04-18 09:45:22', '2020-05-05 09:01:17', '2020-05-23 09:54:02', '2020-07-03 11:00:31', '2020-07-03 12:39:08', '2020-07-04 09:45:22', '2020-07-05 09:01:17', '2020-07-06 09:54:02', '2020-07-06 11:00:31'],
                         'Call Duration' : ['0.01178', '0.01736','0.01923','0.00911','0.01007','0.01206','0.01256','0.01006','0.01162','0.00733','0.01250','0.01013','0.01308'],
                         'ITT': ['NO','YES', 'NO', 'Follow up', 'YES','YES', 'NO', 'Follow up','YES','YES', 'NO','YES','YES']
                        }
            
df = pd.DataFrame(employees)

期望的输出：

Name  Dept  Team     Start      End      Weeks Total Calls  Ave. Call time  Sold  Rejected  more info
Mark   21    2    2020-02-19 2020-07-06  19.71      13          0.01207       7       4        2

我试图应用的逻辑是（虽然我猜我在下面写的语法有错误，但我希望你仍然能够理解计算）：

开始 = df['Log'] 中的最小日期
End = df['Log'] 中的最大日期
周 =（df['log'] 中的最大日期 - df['Log'] 中的最小日期）/7
总调用次数 = df['Log'].count
大道。通话时间 = (df['Call Duration'].sum)/(df['Log'].count)
已售出 = (df['ITT']=='YES').count
拒绝 = (df['ITT']=='NO').count
更多信息 = (df['ITT']=='Follow up').count

【问题讨论】：

标签： python pandas dataframe row summarize

【解决方案1】：

使用pd.NamedAgg 和groupby 试试这个：

df['Log'] = pd.to_datetime(df['Log'])
df['Call Duration'] = df['Call Duration'].astype(float)

df.groupby(['Name of Employee', 'Team', 'Department'])\
  .agg(Start = ('Log','min'),
       End = ('Log', 'max'),
        Weeks = ('Log', lambda x: np.ptp(x) / np.timedelta64(7, 'D')),
        Total_Calls = ('Log', 'count'),
        Avg_Call_Time = ('Call Duration', 'mean'),
        Sold = ('ITT', lambda x: (x == 'YES').sum()),
        Rejected = ('ITT', lambda x: (x == 'NO').sum()),
        More_info = ('ITT', lambda x: (x=='Follow up').sum()))

输出：

                                               Start                 End      Weeks  Total_Calls  Avg_Call_Time  Sold  Rejected  More_info
Name of Employee Team Department                                                                                                          
Mark             2    21         2020-02-19 09:01:17 2020-07-06 11:00:31  19.726114           13       0.012068     7         4          2

【讨论】：

谢谢，真是太棒了。我对您的代码做了一个小改动，使其包含 'Team' df.groupby(['Name of Employee', 'Department', 'Team']
@windwalker 谢谢。是的，我错过了。 :) 我现在添加了它。快乐编码。保持安全并保持健康。
我注意到 groupby 标题不被视为列，因此在保存到数据库时会消失。如何将 groupby 数据转换为列并像其他数据一样显示？
我之前的问题有些为时过早。经过反复试验，我发现 df.reset_index() 完成了将 groupby 数据转换为 df 列的技巧。

【解决方案2】：

你有语法错误，你忘了在每个键的末尾加上逗号。现在你可以处理这个数据框了。

import pandas as pd
    employees = {'Name=': ['Mark','Mark','Mark','Mark','Mark','Mark', 'Mark','Mark','Mark','Mark','Mark','Mark','Mark'],
                             'Department': ['21','21','21','21','21','21', '21','21','21','21','21','21','21'],
                             'Team': ['2','2','2','2','2','2','2','2','2','2','2','2','2'],
                             'Log': ['2020-02-19 09:01:17', '2020-02-19 09:54:02', '2020-04-10 11:00:31', '2020-04-11 12:39:08', '2020-04-18 09:45:22', '2020-05-05 09:01:17', '2020-05-23 09:54:02', '2020-07-03 11:00:31', '2020-07-03 12:39:08', '2020-07-04 09:45:22', '2020-07-05 09:01:17', '2020-07-06 09:54:02', '2020-07-06 11:00:31'],
                             'Call Duration' : ['0.01178', '0.01736','0.01923','0.00911','0.01007','0.01206','0.01256','0.01006','0.01162','0.00733','0.01250','0.01013','0.01308'],
                             'ITT': ['NO','YES', 'NO', 'Follow up', 'YES','YES', 'NO', 'Follow up','YES','YES', 'NO','YES','YES']
                            }
                
    df = pd.DataFrame(employees)
    print(df)

输出：-

              Name  Department  ... Call Duration        ITT
              Mark         21  ...       0.01178         NO
              Mark         21  ...       0.01736        YES
              Mark         21  ...       0.01923         NO
              Mark         21  ...       0.00911  Follow up
              Mark         21  ...       0.01007        YES
              Mark         21  ...       0.01206        YES
              Mark         21  ...       0.01256         NO
              Mark         21  ...       0.01006  Follow up
              Mark         21  ...       0.01162        YES
              Mark         21  ...       0.00733        YES
              Mark         21  ...       0.01250         NO
              Mark         21  ...       0.01013        YES
              Mark         21  ...       0.01308        YES

[13 rows x 6 columns]

【讨论】：

感谢您发现我的疏忽。您能建议我如何制作单行摘要吗？