CODEWITHSUNDEEP
python-3.x
Shanoo
Shanoo

Reputation: 1255

Pandas dataframe if else condition based on previous rows

I have a pandas dataframe as below:

df = pd.DataFrame({'X':[1,1,1, 0, 0]})
df

    X
0   1
1   1
2   1
3   0
4   0

Now I want to create another variable 'Y' and Values for Y should be based on the below condition:

If X = 1 , Y=1
If X = 0 and previous X = 1, Y = 2
If X = 0 and previous x = 0, Y = 0

So, my final output should look like below:

    X    Y
0   1    1    
1   1    1
2   1    1
3   0    2
4   0    0

This can be achieved by iterating over rows and setting up a current and previous row and using iloc but I want a more efficient way of doing this faster

Upvotes: 3

Views: 344

Answers (2)

PMende
PMende

Reputation: 5460

Celius provided an answer with nested calls to np.where. This can become unfeasible if the number of conditions grow. You can use np.select instead to achieve the same result:

import numpy as np
import pandas as pd


df = pd.DataFrame({
    'X': [1, 1, 1, 0, 0]
})
conditions = [
    df["X"] == 1,
    (df["X"] == 0) & (df["X"].shift() == 1),
    (df["X"] == 0) & (df["X"].shift() == 0)
]
values = [1, 2, 0]
df['Y'] = np.select(conditions, values, default=np.nan)

Upvotes: 0

Celius Stingher
Celius Stingher

Reputation: 18367

You can try using np.where and shift:

import pandas as pd
import numpy as np
df = pd.DataFrame({'X':[1,1,1, 0, 0]})
df['Y'] = np.where(df['X'] == 1,1,np.where(df['X'].shift(periods=1) == 1,2,0))
print(df)

Output:

   X  Y
0  1  1
1  1  1
2  1  1
3  0  2
4  0  0

Upvotes: 1

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