Pandas - Groupby Company and drop rows according to criteria based off the Dates of values being out of order - python-3.x

Question

I have a history data log and want to calculate the number of days between the progress by Company (Timestamp of the early stage must be smaller than the later stage).
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
Expected outputs:
Progress (1. Start --> 2. Discuss)
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
Progress (2. Discuss --> 3. Contract)
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
I did try some stupid ways to do the work but still need manualyl filter in excel, below is my coding:
df_stage1_stage2 = df[(df['Progress']=='1. Start')|(df['Progress']=='2. Discuss ')]
pd.pivot_table(df_stage1_stage2 ,index=['Company','Progress'],aggfunc={'Time':min})
Can anyone help with the problem? thanks

Answer 1

Create some masks to filter out the relevant rows. m1 and m2 filter out groups where 1. Start is not the "first" datetime if looking at in reverse order )since your dates are sorted by Company ascending and date descending). You can create more masks if you need to also check if 2. Discuss and 3. Contract are in order, instead of the current logic which is only checking to make sure that 1. is in order. But, with the data you provided that returns the correct output:
m1 = df.groupby('Company')['Progress'].transform('last')
m2 = np.where((m1 == '1. Start'), 'drop', 'keep')
df = df[m2=='drop']
df
intermediate output:
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
From there, filter as you have indicated by sorting and dropping duplicates based off a subset of the first two columns and keep the 'first' duplicate:
final df1 and df2 output:
df1
df1 = df[df['Progress'] != '3. Contract'] \
.sort_values(['Company', 'Time'], ascending=[True,True]) \
.drop_duplicates(subset=['Company', 'Progress'], keep='first')
df1 output:
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 output:
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

Answer 2

Something like this could work, assuming an already sorted df:
(full example)
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)
Output:
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
If you do not want the 'days' word in output use:
df['days_delta'] = df['days_delta'].dt.days

Answer 3

First Problem
#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
Second Problem
#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