﻿ Pandas: Relative frequency within each group - w3resource

# Pandas: Relative frequency within each group

## Pandas Grouping and Aggregating: Split-Apply-Combine Exercise-24 with Solution

Write a Pandas program to split the following datasets into groups on customer_id to summarize purch_amt and calculate percentage of purch_amt in each group.

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','05-10-2012','08-17-2012','10-09-2012','07-27-2012','10-09-2012','10-10-2012','10-10-2012','06-17-2012','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)
gr_data = df.groupby(['customer_id','salesman_id']).agg({'purch_amt': 'sum'})
gr_data["% (Purch Amt.)"] = gr_data.apply(lambda x:  100*x / x.sum())
print("\nPercentage of purch_amt in each group of customer_id:")
print(gr_data)
``````

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  05-10-2012         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

Percentage of purch_amt in each group of customer_id:
purch_amt  % (Purch Amt.)
customer_id salesman_id
3001        5001           2400.60       13.685510
5002            400.95        2.285764
5003            110.50        0.629946
5005            270.65        1.542941
5006           1983.43       11.307278
3005        5001           8870.86       50.571626
5002            948.50        5.407276
5003           2480.40       14.140440
5007             75.29        0.429219
```

Python Code Editor:

Have another way to solve this solution? Contribute your code (and comments) through Disqus.

Next: Write a Pandas program to split a dataset, group by one column and get mean, min, and max values by group, also change the column name of the aggregated metric. Using the following dataset find the mean, min, and max values of purchase amount (purch_amt) group by customer id (customer_id).

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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