Pandas: Groupby to find first dates for each group

Last update on May 28 2022 13:37:45 (UTC/GMT +8 hours)

Pandas Grouping and Aggregating: Split-Apply-Combine Exercise-31 with Solution

Write a Pandas program to split the following dataset using group by on 'salesman_id' and find the first order date for each group.

Test Data:

    ord_no  purch_amt    ord_date  customer_id  salesman_id
0    70001     150.50  2012-10-05         3002         5002
1    70009     270.65  2012-09-10         3001         5003
2    70002      65.26  2012-10-05         3001         5001
3    70004     110.50  2012-08-17         3003         5003
4    70007     948.50  2012-09-10         3002         5002
5    70005    2400.60  2012-07-27         3002         5001
6    70008    5760.00  2012-09-10         3001         5001
7    70010    1983.43  2012-10-10         3004         5003
8    70003    2480.40  2012-10-10         3003         5003
9    70012     250.45  2012-06-27         3002         5002
10   70011      75.29  2012-08-17         3003         5003
11   70013    3045.60  2012-04-25         3001         5001

Sample Solution:

Python Code :

import pandas as pd
pd.set_option('display.max_rows', None)
#pd.set_option('display.max_columns', None)
df = pd.DataFrame({
'ord_no':[70001,70009,70002,70004,70007,70005,70008,70010,70003,70012,70011,70013],
'purch_amt':[150.5,270.65,65.26,110.5,948.5,2400.6,5760,1983.43,2480.4,250.45, 75.29,3045.6],
'ord_date': ['2012-10-05','2012-09-10','2012-10-05','2012-08-17','2012-09-10','2012-07-27','2012-09-10','2012-10-10','2012-10-10','2012-06-27','2012-08-17','2012-04-25'],
'customer_id':[3005,3001,3002,3009,3005,3007,3002,3004,3009,3008,3003,3002],
'salesman_id': [5002,5005,5001,5003,5002,5001,5001,5004,5003,5002,5004,5001]})
print("Original Orders DataFrame:")
print(df)
print("\nGroupby to find first order date for each group(salesman_id):")
result = df.groupby('salesman_id')['ord_date'].min()
print(result)

Sample Output:

Original Orders DataFrame:
    ord_no  purch_amt    ord_date  customer_id  salesman_id
0    70001     150.50  2012-10-05         3005         5002
1    70009     270.65  2012-09-10         3001         5005
2    70002      65.26  2012-10-05         3002         5001
3    70004     110.50  2012-08-17         3009         5003
4    70007     948.50  2012-09-10         3005         5002
5    70005    2400.60  2012-07-27         3007         5001
6    70008    5760.00  2012-09-10         3002         5001
7    70010    1983.43  2012-10-10         3004         5004
8    70003    2480.40  2012-10-10         3009         5003
9    70012     250.45  2012-06-27         3008         5002
10   70011      75.29  2012-08-17         3003         5004
11   70013    3045.60  2012-04-25         3002         5001

Groupby to find first order date for each group(salesman_id):
salesman_id
5001    2012-04-25
5002    2012-06-27
5003    2012-08-17
5004    2012-08-17
5005    2012-09-10
Name: ord_date, dtype: object

Python Code Editor:

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Previous: Write a Pandas program to split the following dataset using group by on first column and aggregate over multiple lists on second column.
Next: Write a Pandas program to split a given dataset using group by on multiple columns and drop last n rows of from each group.

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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