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Pandas: GroupBy with condition of two labels and ranges

Pandas Grouping and Aggregating: Split-Apply-Combine Exercise-29 with Solution

Write a Pandas program to split a given dataset using group by on specified column into two labels and ranges.

Split the group on 'salesman_id',
Ranges:
1) (5001...5006)
2) (5007..5012)

Test Data:

    salesman_id  sale_jan
0          5001    150.50
1          5002    270.65
2          5003     65.26
3          5004    110.50
4          5005    948.50
5          5006   2400.60
6          5007   1760.00
7          5008   2983.43
8          5009    480.40
9          5010   1250.45
10         5011     75.29
11         5012   1045.60

Sample Solution:

Python Code :

import pandas as pd
import numpy as np
pd.set_option('display.max_rows', None)
pd.set_option('display.max_columns', None)
df = pd.DataFrame({
'salesman_id': [5001,5002,5003,5004,5005,5006,5007,5008,5009,5010,5011,5012],
'sale_jan':[150.5, 270.65, 65.26, 110.5, 948.5, 2400.6, 1760, 2983.43, 480.4,  1250.45, 75.29,1045.6]})
print("Original Orders DataFrame:")
print(df)
result = df.groupby(pd.cut(df['salesman_id'], 
                  bins=[0,5006,np.inf],  
                  labels=['S1', 'S2']))['sale_jan'].sum().reset_index()
print("\nGroupBy with condition of  two labels and ranges:")
print(result)

Sample Output:

Original Orders DataFrame:
    salesman_id  sale_jan
0          5001    150.50
1          5002    270.65
2          5003     65.26
3          5004    110.50
4          5005    948.50
5          5006   2400.60
6          5007   1760.00
7          5008   2983.43
8          5009    480.40
9          5010   1250.45
10         5011     75.29
11         5012   1045.60

GroupBy with condition of  two labels and ranges:
  salesman_id  sale_jan
0          S1   3946.01
1          S2   7595.17

Python Code Editor:


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Previous: Write a Pandas program to split a given dataset, group by one column and remove those groups if all the values of a specific columns are not available.
Next: Write a Pandas program to split the following dataset using group by on first column and aggregate over multiple lists on second column.

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Python: Tips of the Day

Understanding slice notation:

It's pretty simple really:

a[start:stop]  # items start through stop-1
a[start:]      # items start through the rest of the array
a[:stop]       # items from the beginning through stop-1
a[:]           # a copy of the whole array

There is also the step value, which can be used with any of the above:

a[start:stop:step] # start through not past stop, by step

The key point to remember is that the :stop value represents the first value that is not in the selected slice. So, the difference between stop and start is the number of elements selected (if step is 1, the default).

The other feature is that start or stop may be a negative number, which means it counts from the end of the array instead of the beginning. So:

a[-1]    # last item in the array
a[-2:]   # last two items in the array
a[:-2]   # everything except the last two items

Similarly, step may be a negative number:

a[::-1]    # all items in the array, reversed
a[1::-1]   # the first two items, reversed
a[:-3:-1]  # the last two items, reversed
a[-3::-1]  # everything except the last two items, reversed

Python is kind to the programmer if there are fewer items than you ask for. For example, if you ask for a[:-2] and a only contains one element, you get an empty list instead of an error. Sometimes you would prefer the error, so you have to be aware that this may happen.

Relation to slice() object

The slicing operator [] is actually being used in the above code with a slice() object using the : notation (which is only valid within []), i.e.:

a[start:stop:step]

is equivalent to:

a[slice(start, stop, step)]

Slice objects also behave slightly differently depending on the number of arguments, similarly to range(), i.e. both slice(stop) and slice(start, stop[, step]) are supported. To skip specifying a given argument, one might use None, so that e.g. a[start:] is equivalent to a[slice(start, None)] or a[::-1] is equivalent to a[slice(None, None, -1)].

While the : -based notation is very helpful for simple slicing, the explicit use of slice() objects simplifies the programmatic generation of slicing.

Ref: https://bit.ly/2MHaTp7