Compare values with previous values and count the change : Pandas Groupby

Question

I have a panadas dataframe shown below :

masterid    price
1001        12
1001        12
1001        11
1001        14
1002        15
1003        10
1003        10
1003        10
1004        16
1004        17

I want to count the times when the price changes for a particular master id

Example : For masterid 1001 the price changed from 12 – 11 and from 11 -14 ( 2 times )

Expected output :

masterid    price
1001        2
1002        0
1003        0
1004        1

I have tried using (df['price'].ne(df['price'].shift()) but not sure how to use group by in this case. Thanks in advance

Asked By: Ash

||

Source

Answer 1


#calculate the price difference using diff
# take absolute value and compare with zero
# go another group by and take the count

df.assign(price=
          df.groupby('masterid')['price']
          .diff()
          .abs()>0)
.groupby(['masterid'])['price']
.sum()
.reset_index()

masterid    price
0   1001    2
1   1002    0
2   1003    0
3   1004    1

Answered By: Naveed

Answer 2

You are almost there. The next step is to use pd.groupby.apply with your existing function and sum:

df.groupby('masterid').apply(
    lambda df: (df['price'].ne(df['price'].shift())
).sum())

This stragegy should work pretty generally, though in some cases apply will be slow.

Answered By: philosofool

Compare values with previous values and count the change : Pandas Groupby

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