如何计算熊猫数据框中列值更改的频率

Question

I have a pandas data frame like this:

    id  some_value
0   tag1    v1
1   tag1    v2
2   tag1    v1
3   tag2    v2
4   tag2    v2
5   tag2    v3

and I would like to know how often for each id the value in some_value changed. So for tag1 that would be twice (because it changes first from v1 to v2 and then back), for tag2 it would be once. I have solved the problem like this:

import pandas as pd
df = pd.DataFrame({'id': ['tag1', 'tag1', 'tag1', 'tag2', 'tag2','tag2'], 'some_value': ['v1','v2','v1','v2','v2','v3']})
mask = df['id'] == df['id'].shift(-1)
df['changed'] = df['some_value'] != df['some_value'].shift(-1)
df[mask].groupby('id').sum()

The code works fine in that it returns

    changed
id  
tag1    2.0
tag2    1.0

Is there a more elegant solution to this?

Answer 1

One way to achieve this would be:

def numChanges(x):
    return sum(x.iloc[:-1] != x.shift(-1).iloc[:-1])

df.groupby('id').agg({
    'some_value' : numChanges
})

Please note that if the id column is unsorted, the results would differ, so your solution may produce incorrect results, unless you intend it to be that way.

As an example, below dataset would yield tag2 value as 5 with my solution, but 3 as per yours. Technically, the correct answer would be 5, but if your id variable is sorted, it will not make any difference.

pd.concat([df]*3)  #My solution outputs 5 changes for tag2 and yours will give 3 only