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
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