Python: How to convert Pandas Dataframe Row Values to individual columns?

Question

I have the following dataframe which has the data of different jobs start and end time at different intervals. A small part of dataframe is shown below.

Dataframe(df):

result | job   |  time
START  | JOB0  |  1357  
START  | JOB2  |  2405
END    | JOB2  |  2379
START  | JOB3  |  4010
END    | JOB0  |  5209
END    | JOB3  |  6578
START  | JOB0  |  6000
END    | JOB0  |  6100

(Note - Original Dataframe has 5 Jobs (JOB0 to JOB4) I want to convert the values ( START and END ) of column result as individual columns in the dataframe.

Required Dataframe(df2)

job  |  START  | END
JOB0 |  1357   | 5209
JOB2 |  2405   | 2379
JOB3 |  4010   | 6578
JOB0 |  6000   | 6100

Code
I tried implementing this using a pivot_table but it is giving aggregated values which is not required.

df2 = df.pivot_table('time', 'job','result')

Code Output

result |       END       |      START
job     
JOB0   |    5.000589e+08    5.000636e+08
JOB1   |    4.999141e+08    4.999188e+08
JOB2   |    5.001668e+08    5.001715e+08
JOB3   |    4.995190e+08    4.995187e+08
JOB4   |    5.003238e+08    5.003236e+08

How can I attain the required dataframe?

Answer 1

You have repeated jobs( JOB0 ), So its better to create a unique id for the jobs then pivot it based on id and job like

df['id'] = df.groupby(['job', 'result']).cumcount()
df2 = df.pivot_table(index=['id','job'], columns='result', values='time')

Output:

result    END  START
id job              
0  JOB0  5209   1357
   JOB2  2379   2405
   JOB3  6578   4010
1  JOB0  6100   6000

Have the df sorted on time for sanity (There could be issues if the same jobs overlap each other)

df = df.sort_values(by='time')

Answer 2

You should be able to use pandas.DataFrame.pivot for this as follows:

import pandas as pd

df2 = df.pivot(index="job", columns="result", values="time")