Pandas: Split the specified dataframe into groups, group by month and year based on order date and find the total purchase amount year wise, month wise
Pandas Grouping and Aggregating: Split-Apply-Combine Exercise-12 with Solution
Write a Pandas program to split the following dataframe into groups, group by month and year based on order date and find the total purchase amount year wise, month wise.
Test Data:
ord_no purch_amt ord_date customer_id salesman_id 0 70001 150.50 05-10-2012 3001 5002 1 70009 270.65 09-10-2012 3001 5005 2 70002 65.26 05-10-2012 3005 5001 3 70004 110.50 08-17-2012 3001 5003 4 70007 948.50 10-09-2012 3005 5002 5 70005 2400.60 07-27-2012 3001 5001 6 70008 5760.00 10-09-2012 3005 5001 7 70010 1983.43 10-10-2012 3001 5006 8 70003 2480.40 10-10-2012 3005 5003 9 70012 250.45 06-17-2012 3001 5002 10 70011 75.29 07-08-2012 3005 5007 11 70013 3045.60 04-25-2012 3005 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': ['05-10-2012','09-10-2012','05-10-2013','08-17-2013','10-09-2013','07-27-2014','10-09-2012','10-10-2012','10-10-2012','06-17-2014','07-08-2012','04-25-2012'],
'customer_id':[3001,3001,3005,3001,3005,3001,3005,3001,3005,3001,3005,3005],
'salesman_id': [5002,5005,5001,5003,5002,5001,5001,5006,5003,5002,5007,5001]})
print("Original Orders DataFrame:")
print(df)
df['ord_date']= pd.to_datetime(df['ord_date'])
print("\nYear wise Month wise purchase amount:")
result = df.groupby([df['ord_date'].dt.year, df['ord_date'].dt.month]).agg({'purch_amt':sum})
print(result)
Sample Output:
Original Orders DataFrame: ord_no purch_amt ord_date customer_id salesman_id 0 70001 150.50 05-10-2012 3001 5002 1 70009 270.65 09-10-2012 3001 5005 2 70002 65.26 05-10-2013 3005 5001 3 70004 110.50 08-17-2013 3001 5003 4 70007 948.50 10-09-2013 3005 5002 5 70005 2400.60 07-27-2014 3001 5001 6 70008 5760.00 10-09-2012 3005 5001 7 70010 1983.43 10-10-2012 3001 5006 8 70003 2480.40 10-10-2012 3005 5003 9 70012 250.45 06-17-2014 3001 5002 10 70011 75.29 07-08-2012 3005 5007 11 70013 3045.60 04-25-2012 3005 5001 Year wise Month wise purchase amount: purch_amt ord_date ord_date 2012 4 3045.60 5 150.50 7 75.29 9 270.65 10 10223.83 2013 5 65.26 8 110.50 10 948.50 2014 6 250.45 7 2400.60
Python Code Editor:
Have another way to solve this solution? Contribute your code (and comments) through Disqus.
Previous: Write a Pandas program to split the following dataframe into groups and calculate monthly purchase amount.
Next: Write a Pandas program to split the following dataframe into groups based on first column and set other column values into a list of values.
What is the difficulty level of this exercise?
Test your Programming skills with w3resource's quiz.
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
- New Content published on w3resource:
- HTML-CSS Practical: Exercises, Practice, Solution
- Java Regular Expression: Exercises, Practice, Solution
- Scala Programming Exercises, Practice, Solution
- Python Itertools exercises
- Python Numpy exercises
- Python GeoPy Package exercises
- Python Pandas exercises
- Python nltk exercises
- Python BeautifulSoup exercises
- Form Template
- Composer - PHP Package Manager
- PHPUnit - PHP Testing
- Laravel - PHP Framework
- Angular - JavaScript Framework
- Vue - JavaScript Framework
- Jest - JavaScript Testing Framework