熊猫前一组最小/最大

Question

In Pandas I have dataset like this:

                     Value
2005-08-03 23:15:00   10.5
2005-08-03 23:30:00   10.0
2005-08-03 23:45:00   10.0
2005-08-04 00:00:00   10.5
2005-08-04 00:15:00   10.5
2005-08-04 00:30:00   11.0
2005-08-04 00:45:00   10.5
2005-08-04 01:00:00   11.0
...
2005-08-04 23:15:00   14.0
2005-08-04 23:30:00   13.5
2005-08-04 23:45:00   13.0
2005-08-05 00:00:00   13.5
2005-08-05 00:15:00   14.0
2005-08-05 00:30:00   14.0
2005-08-05 00:45:00   14.5

First I wanted to group data by date and store each group's max value in new column, I used the following code for this task:

df['ValueMaxInGroup'] = df.groupby(pd.TimeGrouper('D'))['Value'].transform(max)

Now I want to create another column to store previous group max value, so the desired data frame would look like:

                     Value  ValueMaxInGroup  ValueMaxInPrevGroup
2005-08-03 23:15:00   10.5             10.5                  NaN
2005-08-03 23:30:00   10.0             10.5                  NaN
2005-08-03 23:45:00   10.0             10.5                  NaN
2005-08-04 00:00:00   10.5             14.0                 10.5
2005-08-04 00:15:00   10.5             14.0                 10.5
2005-08-04 00:30:00   11.0             14.0                 10.5
2005-08-04 00:45:00   10.5             14.0                 10.5
2005-08-04 01:00:00   11.0             14.0                 10.5
...
2005-08-04 23:15:00   14.0             14.0                 10.5
2005-08-04 23:30:00   13.5             14.0                 10.5
2005-08-04 23:45:00   13.0             14.0                 10.5
2005-08-05 00:00:00   13.5             14.5                 14.0
2005-08-05 00:15:00   14.0             14.5                 14.0
2005-08-05 00:30:00   14.0             14.5                 14.0
2005-08-05 00:45:00   14.5             14.5                 14.0

So, to simply get previous row's value, I used

df['ValueInPrevRow'] = df.shift(1)['Value']

Is there any way to get another group's min/max/f(x)? I assumed

df['ValueMaxInPrevGroup'] = df.groupby(pd.TimeGrouper('D')).shift(1)['Value'].transform(max)

but it didn't work.

Answer 1

You could get the desired result by using groupby/agg , shift and merge :

import numpy as np
import pandas as pd
df = pd.DataFrame({'Value': [10.5, 10.0, 10.0, 10.5, 10.5, 11.0, 10.5, 11.0, 14.0, 13.5, 13.0, 13.5, 14.0, 14.0, 14.5]}, index=['2005-08-03 23:15:00', '2005-08-03 23:30:00', '2005-08-03 23:45:00', '2005-08-04 00:00:00', '2005-08-04 00:15:00', '2005-08-04 00:30:00', '2005-08-04 00:45:00', '2005-08-04 01:00:00', '2005-08-04 23:15:00', '2005-08-04 23:30:00', '2005-08-04 23:45:00', '2005-08-05 00:00:00', '2005-08-05 00:15:00', '2005-08-05 00:30:00', '2005-08-05 00:45:00']) 
df.index = pd.DatetimeIndex(df.index)

# This is equivalent to
# df['group'] = pd.to_datetime(df.index.date)
# when freq='D', but the version below works with any freq string, not just `'D'`.
grouped = df.groupby(pd.TimeGrouper('D'))
labels, uniqs, ngroups = grouped.grouper.group_info
df['group'] = grouped.grouper.binlabels[labels]

result = grouped[['Value']].agg(max)
result = result.rename(columns={'Value':'Max'})
result['PreviouMax'] = result['Max'].shift(1)

df = pd.merge(df, result, left_on=['group'], right_index=True)
print(df)

yields

                     Value      group   Max  PreviouMax
2005-08-03 23:15:00   10.5 2005-08-03  10.5         NaN
2005-08-03 23:30:00   10.0 2005-08-03  10.5         NaN
2005-08-03 23:45:00   10.0 2005-08-03  10.5         NaN
2005-08-04 00:00:00   10.5 2005-08-04  14.0        10.5
2005-08-04 00:15:00   10.5 2005-08-04  14.0        10.5
2005-08-04 00:30:00   11.0 2005-08-04  14.0        10.5
2005-08-04 00:45:00   10.5 2005-08-04  14.0        10.5
2005-08-04 01:00:00   11.0 2005-08-04  14.0        10.5
2005-08-04 23:15:00   14.0 2005-08-04  14.0        10.5
2005-08-04 23:30:00   13.5 2005-08-04  14.0        10.5
2005-08-04 23:45:00   13.0 2005-08-04  14.0        10.5
2005-08-05 00:00:00   13.5 2005-08-05  14.5        14.0
2005-08-05 00:15:00   14.0 2005-08-05  14.5        14.0
2005-08-05 00:30:00   14.0 2005-08-05  14.5        14.0
2005-08-05 00:45:00   14.5 2005-08-05  14.5        14.0

---

The main idea here is to use groupby/agg instead of groupby/transform so that we may obtain

result = grouped[['Value']].agg(max)
result = result.rename(columns={'Value':'Max'})
result['PreviouMax'] = result['Max'].shift(1)
#              Max  PreviouMax
# group                       
# 2005-08-03  10.5         NaN
# 2005-08-04  14.0        10.5
# 2005-08-05  14.5        14.0

Then the desired DataFrame can be expressed as the result of merging df with result on the group date.