Pandas: Split a given dataframe into groups with bin counts
Pandas Grouping and Aggregating: Split-Apply-Combine Exercise-18 with Solution
Write a Pandas program to split a given dataframe into groups with bin counts.
Test Data:
ord_no purch_amt customer_id sales_id 0 70001 150.50 3005 5002 1 70009 270.65 3001 5003 2 70002 65.26 3002 5004 3 70004 110.50 3009 5003 4 70007 948.50 3005 5002 5 70005 2400.60 3007 5001 6 70008 5760.00 3002 5005 7 70010 1983.43 3004 5007 8 70003 2480.40 3009 5008 9 70012 250.45 3008 5004 10 70011 75.29 3003 5005 11 70013 3045.60 3002 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],
'customer_id':[3005,3001,3002,3009,3005,3007,3002,3004,3009,3008,3003,3002],
'sales_id':[5002,5003,5004,5003,5002,5001,5005,5007,5008,5004,5005,5001]})
print("Original DataFrame:")
print(df)
groups = df.groupby(['customer_id', pd.cut(df.sales_id, 3)])
result = groups.size().unstack()
print(result)
Sample Output:
Original DataFrame: ord_no purch_amt customer_id sales_id 0 70001 150.50 3005 5002 1 70009 270.65 3001 5003 2 70002 65.26 3002 5004 3 70004 110.50 3009 5003 4 70007 948.50 3005 5002 5 70005 2400.60 3007 5001 6 70008 5760.00 3002 5005 7 70010 1983.43 3004 5007 8 70003 2480.40 3009 5008 9 70012 250.45 3008 5004 10 70011 75.29 3003 5005 11 70013 3045.60 3002 5001 sales_id (5000.993, 5003.333] (5003.333, 5005.667] (5005.667, 5008.0] customer_id 3001 1.0 NaN NaN 3002 1.0 2.0 NaN 3003 NaN 1.0 NaN 3004 NaN NaN 1.0 3005 2.0 NaN NaN 3007 1.0 NaN NaN 3008 NaN 1.0 NaN 3009 1.0 NaN
Python Code Editor:
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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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