Pandas - Groupby Company 并根据基于乱序值日期的标准删除行
Pandas - Groupby Company and drop rows according to criteria based off the Dates of values being out of order
我有一个历史数据日志，想按公司计算进度之间的天数（前期的时间戳必须小于后期）。
Company Progress Time
AAA 3. Contract 07/10/2020
AAA 2. Discuss 03/09/2020
AAA 1. Start 02/02/2020
BBB 3. Contract 11/13/2019
BBB 3. Contract 07/01/2019
BBB 1. Start 06/22/2019
BBB 2. Discuss 04/15/2019
CCC 3. Contract 05/19/2020
CCC 2. Discuss 04/08/2020
CCC 2. Discuss 03/12/2020
CCC 1. Start 01/01/2020
预期产出：
进展（1.开始 --> 2.讨论）
Company Progress Time
AAA 1. Start 02/02/2020
AAA 2. Discuss 03/09/2020
CCC 1. Start 01/01/2020
CCC 2. Discuss 03/12/2020
进展（2.讨论 --> 3.合同）
Company Progress Time
AAA 2. Discuss 03/09/2020
AAA 3. Contract 07/10/2020
CCC 2. Discuss 03/12/2020
CCC 3. Contract 05/19/2020
我确实尝试了一些愚蠢的方法来完成这项工作，但在 excel 中仍然需要 manualyl 过滤器，下面是我的编码：
df_stage1_stage2 = df[(df['Progress']=='1. Start')|(df['Progress']=='2. Discuss ')]
pd.pivot_table(df_stage1_stage2 ,index=['Company','Progress'],aggfunc={'Time':min})
谁能帮忙解决这个问题？谢谢
创建一些掩码以过滤掉相关行。 m1 和 m2 过滤掉 1. Start 不是“第一个”日期时间（如果以相反顺序查看）的组）因为您的日期按 Company 升序和日期 descending).如果您还需要检查 2. Discuss 和 3. Contract 是否有序，而不是仅检查以确保 1. 有序的当前逻辑，则可以创建更多掩码。但是，根据您提供的数据 returns 正确的输出：
m1 = df.groupby('Company')['Progress'].transform('last')
m2 = np.where((m1 == '1. Start'), 'drop', 'keep')
df = df[m2=='drop']
df
中间输出：
Company Progress Time
0 AAA 3. Contract 07/10/2020
1 AAA 2. Discuss 03/09/2020
2 AAA 1. Start 02/02/2020
7 CCC 3. Contract 05/19/2020
8 CCC 2. Discuss 04/08/2020
9 CCC 2. Discuss 03/12/2020
10 CCC 1. Start 01/01/2020
从那里开始，按照您的指示进行过滤，方法是根据前两列的子集对重复项进行排序和删除，并保留 'first' 重复项：
最终的 df1 和 df2 输出：
df1
df1 = df[df['Progress'] != '3. Contract'] \
.sort_values(['Company', 'Time'], ascending=[True,True]) \
.drop_duplicates(subset=['Company', 'Progress'], keep='first')
df1 输出：
Company Progress Time
2 AAA 1. Start 02/02/2020
1 AAA 2. Discuss 03/09/2020
10 CCC 1. Start 01/01/2020
9 CCC 2. Discuss 03/12/2020
df2
df2 = df[df['Progress'] != '1. Start'] \
.sort_values(['Company', 'Time'], ascending=[True,True]) \
.drop_duplicates(subset=['Company', 'Progress'], keep='first')
df2 输出：
Company Progress Time
1 AAA 2. Discuss 03/09/2020
0 AAA 3. Contract 07/10/2020
9 CCC 2. Discuss 03/12/2020
7 CCC 3. Contract 05/19/2020
假设一个已经排序的 df:
（完整示例）
data = {
'Company':['AAA', 'AAA', 'AAA', 'BBB','BBB','BBB','BBB','CCC','CCC','CCC','CCC',],
'Progress':['3. Contract', '2. Discuss', '1. Start', '3. Contract', '3. Contract', '2. Discuss', '1. Start', '3. Contract', '2. Discuss', '2. Discuss', '1. Start', ],
'Time':['07-10-2020','03-09-2020','02-02-2020','11-13-2019','07-01-2019','06-22-2019','04-15-2019','05-19-2020','04-08-2020','03-12-2020','01-01-2020',],
}
df = pd.DataFrame(data)
df['Time'] = pd.to_datetime(df['Time'])
# We want to measure from the first occurrence (last date) if duplicated:
df.drop_duplicates(subset=['Company', 'Progress'], keep='first', inplace=True)
# Except for the rows of 'start', calculate the difference in days
df['days_delta'] = np.where((df['Progress'] != '1. Start'), df.Time.diff(-1), 0)
输出：
Company Progress Time days_delta
0 AAA 3. Contract 2020-07-10 123 days
1 AAA 2. Discuss 2020-03-09 36 days
2 AAA 1. Start 2020-02-02 0 days
3 BBB 3. Contract 2019-11-13 144 days
5 BBB 2. Discuss 2019-06-22 68 days
6 BBB 1. Start 2019-04-15 0 days
7 CCC 3. Contract 2020-05-19 41 days
8 CCC 2. Discuss 2020-04-08 98 days
10 CCC 1. Start 2020-01-01 0 days
如果您不想在输出中使用 'days' 单词：
df['days_delta'] = df['days_delta'].dt.days
第一个问题
#Coerce Time to Datetime
df['Time']=pd.to_datetime(df['Time'])
#`groupby().nth[]` `to slice the consecutive order`
df2=(df.merge(df.groupby(['Company'])['Time'].nth([-2,-1]))).sort_values(by=['Company','Time'], ascending=[True, True])
#Apply the universal rule for this problem which is, after groupby nth, drop any agroup with duplicates
df2[~df2.Company.isin(df2[df2.groupby('Company').Progress.transform('nunique')==1].Company.values)]
#Calculate the diff() in Time in each group
df2['diff'] = df2.sort_values(by='Progress').groupby('Company')['Time'].diff().dt.days.fillna(0)#.groupby('Company')['Time'].diff() / np.timedelta64(1, 'D')
#Filter out the groups where start and Discuss Time are in conflict
df2[~df2.Company.isin(df2.loc[df2['diff']<0, 'Company'].unique())]
Company Progress Time diff
1 AAA 1.Start 2020-02-02 0.0
0 AAA 2.Discuss 2020-03-09 36.0
5 CCC 1.Start 2020-01-01 0.0
4 CCC 2.Discuss 2020-03-12 71.0
第二题
#Groupbynth to slice right consecutive groups
df2=(df.merge(df.groupby(['Company'])['Time'].nth([0,1]))).sort_values(by=['Company','Time'], ascending=[True, True])
#Drop any groups after grouping that have duplicates
df2[~df2.Company.isin(df2[df2.groupby('Company').Progress.transform('nunique')==1].Company.values)]
Company Progress Time
1 AAA 2.Discuss 2020-03-09
0 AAA 3.Contract 2020-07-10
5 CCC 2.Discuss 2020-04-08
4 CCC 3.Contract 2020-05-19
我有一个历史数据日志，想按公司计算进度之间的天数（前期的时间戳必须小于后期）。
Company Progress Time
AAA 3. Contract 07/10/2020
AAA 2. Discuss 03/09/2020
AAA 1. Start 02/02/2020
BBB 3. Contract 11/13/2019
BBB 3. Contract 07/01/2019
BBB 1. Start 06/22/2019
BBB 2. Discuss 04/15/2019
CCC 3. Contract 05/19/2020
CCC 2. Discuss 04/08/2020
CCC 2. Discuss 03/12/2020
CCC 1. Start 01/01/2020
预期产出：
进展（1.开始 --> 2.讨论）
Company Progress Time
AAA 1. Start 02/02/2020
AAA 2. Discuss 03/09/2020
CCC 1. Start 01/01/2020
CCC 2. Discuss 03/12/2020
进展（2.讨论 --> 3.合同）
Company Progress Time
AAA 2. Discuss 03/09/2020
AAA 3. Contract 07/10/2020
CCC 2. Discuss 03/12/2020
CCC 3. Contract 05/19/2020
我确实尝试了一些愚蠢的方法来完成这项工作，但在 excel 中仍然需要 manualyl 过滤器，下面是我的编码：
df_stage1_stage2 = df[(df['Progress']=='1. Start')|(df['Progress']=='2. Discuss ')]
pd.pivot_table(df_stage1_stage2 ,index=['Company','Progress'],aggfunc={'Time':min})
谁能帮忙解决这个问题？谢谢
创建一些掩码以过滤掉相关行。 m1 和 m2 过滤掉 1. Start 不是“第一个”日期时间（如果以相反顺序查看）的组）因为您的日期按 Company 升序和日期 descending).如果您还需要检查 2. Discuss 和 3. Contract 是否有序，而不是仅检查以确保 1. 有序的当前逻辑，则可以创建更多掩码。但是，根据您提供的数据 returns 正确的输出：
m1 = df.groupby('Company')['Progress'].transform('last')
m2 = np.where((m1 == '1. Start'), 'drop', 'keep')
df = df[m2=='drop']
df
中间输出：
Company Progress Time
0 AAA 3. Contract 07/10/2020
1 AAA 2. Discuss 03/09/2020
2 AAA 1. Start 02/02/2020
7 CCC 3. Contract 05/19/2020
8 CCC 2. Discuss 04/08/2020
9 CCC 2. Discuss 03/12/2020
10 CCC 1. Start 01/01/2020
从那里开始，按照您的指示进行过滤，方法是根据前两列的子集对重复项进行排序和删除，并保留 'first' 重复项：
最终的 df1 和 df2 输出：
df1
df1 = df[df['Progress'] != '3. Contract'] \
.sort_values(['Company', 'Time'], ascending=[True,True]) \
.drop_duplicates(subset=['Company', 'Progress'], keep='first')
df1 输出：
Company Progress Time
2 AAA 1. Start 02/02/2020
1 AAA 2. Discuss 03/09/2020
10 CCC 1. Start 01/01/2020
9 CCC 2. Discuss 03/12/2020
df2
df2 = df[df['Progress'] != '1. Start'] \
.sort_values(['Company', 'Time'], ascending=[True,True]) \
.drop_duplicates(subset=['Company', 'Progress'], keep='first')
df2 输出：
Company Progress Time
1 AAA 2. Discuss 03/09/2020
0 AAA 3. Contract 07/10/2020
9 CCC 2. Discuss 03/12/2020
7 CCC 3. Contract 05/19/2020
假设一个已经排序的 df:
（完整示例）
data = {
'Company':['AAA', 'AAA', 'AAA', 'BBB','BBB','BBB','BBB','CCC','CCC','CCC','CCC',],
'Progress':['3. Contract', '2. Discuss', '1. Start', '3. Contract', '3. Contract', '2. Discuss', '1. Start', '3. Contract', '2. Discuss', '2. Discuss', '1. Start', ],
'Time':['07-10-2020','03-09-2020','02-02-2020','11-13-2019','07-01-2019','06-22-2019','04-15-2019','05-19-2020','04-08-2020','03-12-2020','01-01-2020',],
}
df = pd.DataFrame(data)
df['Time'] = pd.to_datetime(df['Time'])
# We want to measure from the first occurrence (last date) if duplicated:
df.drop_duplicates(subset=['Company', 'Progress'], keep='first', inplace=True)
# Except for the rows of 'start', calculate the difference in days
df['days_delta'] = np.where((df['Progress'] != '1. Start'), df.Time.diff(-1), 0)
输出：
Company Progress Time days_delta
0 AAA 3. Contract 2020-07-10 123 days
1 AAA 2. Discuss 2020-03-09 36 days
2 AAA 1. Start 2020-02-02 0 days
3 BBB 3. Contract 2019-11-13 144 days
5 BBB 2. Discuss 2019-06-22 68 days
6 BBB 1. Start 2019-04-15 0 days
7 CCC 3. Contract 2020-05-19 41 days
8 CCC 2. Discuss 2020-04-08 98 days
10 CCC 1. Start 2020-01-01 0 days
如果您不想在输出中使用 'days' 单词：
df['days_delta'] = df['days_delta'].dt.days
第一个问题
#Coerce Time to Datetime
df['Time']=pd.to_datetime(df['Time'])
#`groupby().nth[]` `to slice the consecutive order`
df2=(df.merge(df.groupby(['Company'])['Time'].nth([-2,-1]))).sort_values(by=['Company','Time'], ascending=[True, True])
#Apply the universal rule for this problem which is, after groupby nth, drop any agroup with duplicates
df2[~df2.Company.isin(df2[df2.groupby('Company').Progress.transform('nunique')==1].Company.values)]
#Calculate the diff() in Time in each group
df2['diff'] = df2.sort_values(by='Progress').groupby('Company')['Time'].diff().dt.days.fillna(0)#.groupby('Company')['Time'].diff() / np.timedelta64(1, 'D')
#Filter out the groups where start and Discuss Time are in conflict
df2[~df2.Company.isin(df2.loc[df2['diff']<0, 'Company'].unique())]
Company Progress Time diff
1 AAA 1.Start 2020-02-02 0.0
0 AAA 2.Discuss 2020-03-09 36.0
5 CCC 1.Start 2020-01-01 0.0
4 CCC 2.Discuss 2020-03-12 71.0
第二题
#Groupbynth to slice right consecutive groups
df2=(df.merge(df.groupby(['Company'])['Time'].nth([0,1]))).sort_values(by=['Company','Time'], ascending=[True, True])
#Drop any groups after grouping that have duplicates
df2[~df2.Company.isin(df2[df2.groupby('Company').Progress.transform('nunique')==1].Company.values)]
Company Progress Time
1 AAA 2.Discuss 2020-03-09
0 AAA 3.Contract 2020-07-10
5 CCC 2.Discuss 2020-04-08
4 CCC 3.Contract 2020-05-19