Question

假设我在Pandas数据框中保存了以下数据集 - 请注意最后一列[Status]是我想要创建的列：

Department  Employee    Issue Date  Submission Date ***Status***
A   Joe 18/05/2014  25/06/2014  0
A   Joe 1/06/2014   28/06/2014  1
A   Joe 23/06/2014  30/06/2014  2
A   Mark    1/03/2015   13/03/2015  0
A   Mark    23/04/2015  15/04/2015  0
A   William 15/07/2016  30/07/2016  0
A   William 1/08/2016   23/08/2016  0
A   William 20/08/2016  19/08/2016  1
B   Liz 18/05/2014  7/06/2014   0
B   Liz 1/06/2014   15/06/2014  1
B   Liz 23/06/2014  16/06/2014  0
B   John    1/03/2015   13/03/2015  0
B   John    23/04/2015  15/04/2015  0
B   Alex    15/07/2016  30/07/2016  0
B   Alex    1/08/2016   23/08/2016  0
B   Alex    20/08/2016  19/08/2016  1

我想根据以下条件创建一个额外的列[状态]：

对于每个独特的[部门]＆amp; [员工]组合（例如，在部门A中有三行对应于Joe），将[发布日期]从最旧到最新排序
如果当前行[发布日期]大于所有前一行[提交日期]，则将[状态]标记为0;否[状态] =没有[发布日期]＆lt; [提交日期]

例如：对于部门A中的员工Joe。当[发布日期] =＆＃39; 1/06 / 2014＆＃39;时，前一行的[提交日期]在[发布日期]之后因此，[状态] =第2行为第1行。类似地，当[发布日期] =＆＃39; 23/06 / 2014＆＃39;时，第1行＆amp; 2 [提交日期]都在[发布日期]之后，因此第3行的[状态] = 2.我们需要对部门和员工的每个唯一组合执行此计算。

注意：真实数据集未按显示的示例排序。

Answer 1

这个问题是在6个月前发布的，但希望我的回答仍能提供一些帮助。

首先，导入库并创建数据框：

# import libraries
import numpy as np
import pandas as pd

# Make DataFrame
df = pd.DataFrame({'Department' : ['A']*8 + ['B']*8,
                   'Employee' : ['Joe']*3 +\
                                ['Mark']*2 +\
                                ['William']*3 +\
                                ['Liz']*3 +\
                                ['John']*2 +\
                                ['Alex']*3,
                   'Issue Date' : ['18/05/2014', '1/06/2014', '23/06/2014',
                                   '1/03/2015', '23/04/2015',
                                   '15/07/2016', '1/08/2016', '20/08/2016',
                                   '18/05/2014', '1/06/2014', '23/06/2014',
                                   '1/03/2015', '23/04/2015',
                                   '15/07/2016', '1/08/2016', '20/08/2016'],
                   'Submission Date' : ['25/06/2014', '28/06/2014', '30/06/2014',
                                        '13/03/2015', '15/04/2015',
                                        '30/07/2016', '23/08/2016', '19/08/2016',
                                        '7/06/2014', '15/06/2014', '16/06/2014',
                                        '13/03/2015', '15/04/2015',
                                        '30/07/2016', '23/08/2016', '19/08/2016']})

其次，将发行日期和提交日期转换为datetime：

    # Convert 'Issue Date', 'Submission Date' to pd.datetime
df.loc[:, 'Issue Date'] = pd.to_datetime(df.loc[:, 'Issue Date'],
                                         dayfirst = True)
df.loc[:, 'Submission Date'] = pd.to_datetime(df.loc[:, 'Submission Date'],
                                              dayfirst = True)

第三步，重置索引并按部门，员工和发布日期对值进行排序：

# Reset index and sort_values by 'Department', 'Employee', 'Issue Date'
df.reset_index(drop = True).sort_values(by = ['Department',
                                              'Employee',
                                              'Issue Date'],
                                        inplace = True)

第四，按部门分组，员工;累计计数行数;插入原始df：

# Group by 'Department', 'Employee'; cumulative count rows; insert into original df
df.insert(df.shape[1],
          'grouped count',
          df.groupby(['Department',
                      'Employee']).cumcount())

第五，创建一个no_issue和no_submission数据帧，并在Department和Employee上将它们合并在一起：

# Create df without 'Issue Date'
no_issue = df.drop('Issue Date', axis = 1)

# Create df without 'Submission Date'
no_submission = df.drop('Submission Date', axis = 1)

# Outer merge no_issue with no_submission on 'Department', 'Employee'
merged = no_issue.merge(no_submission,
                        how = 'outer',
                        on = ['Department',
                              'Employee'])

这会将提交日期与每个部门，员工组的发行日期数量

重复

以下是Joe的样子：

第六，创建一个数据框，只保留分组count_x小于分组count_y的行，然后按部门，员工和发布日期排序：

# Create merged1 df that keeps only rows where 'grouped count_x' < 'grouped count_y';
# sort by 'Department', 'Employee', 'Issue Date
merged1 = merged[merged.loc[:, 'grouped count_x'] <
                 merged.loc[:, 'grouped count_y']].sort_values(by = ['Department',
                                                                     'Employee',
                                                                     'Issue Date'])

第七，将状态列插入布尔值，其中发布日期小于提交日期：

# Insert 'Status' as a boolean when 'Issue Date' < 'Submission Date'
merged1.insert(merged.shape[1],
               'Status',
               merged1.loc[:, 'Issue Date'] < merged1.loc[:, 'Submission Date'])

第八，按部门，员工和发行日期分组，对状态求和，并重置索引：

# Group by 'Department', 'Employee', 'Issue Date' and sum 'Status'; reset index
merged1 = merged1.groupby(['Department',
                           'Employee',
                           'Issue Date']).agg({'Status' : np.sum}).reset_index()

这将返回一个数据框，其中包含所有正确的状态，减去每个部门，员工组的最短发布日期

第九，按部门和员工对原始合并数据框进行分组，找到最小发行日期，并重置索引：

# Group merged by 'Department', 'Employee' and find min 'Issue Date'; reset index
merged = merged.groupby(['Department',
                         'Employee']).agg({'Issue Date' : 'min'}).reset_index()

第十，将merged1与merged连接，将na填充为0（因为最小发行日期的状态始终为0）并按部门，员工和发布日期排序：

# Concatenate merged with merged1; fill na with 0; sort by 'Department', 'Employee', 'Issue Date'
concatenated = pd.concat([merged1, merged]).fillna(0).sort_values(by = ['Department',
                                                                        'Employee',
                                                                        'Issue Date'])

第十一，内部将合并后的数据框与Department，Employee和Issue Date上的连接数据框合并，然后删除分组计数：

# Merge concatenated with df; drop grouped count
final = df.merge(concatenated,
                 how = 'inner',
                 on = ['Department',
                       'Employee',
                       'Issue Date']).drop('grouped count',
                                           axis = 1)

瞧！这是您的最终数据框：

# Final df
final

Pandas数据框：根据日期列比较创建其他列

1 个答案: