Can we use diff() for condition in Python?

Question

Actually I am new in Python and just try to do the algorithm below but I got an error.(ValueError: The truth value of a Series is ambiguous. Use a.empty, a.bool(), a.item(), a.any() or a.all().) I wanna count the time if the values of 'A' are equal.

import numpy as np
import pandas as pd
n=10000000
count=0
df=pd.DataFrame(np.random.randint(0,2,size=(n,1)),columns=['A'])
check =df['A'].diff().eq(0).astype(int)
if check==0:
   count = count+1
   df['B']=count
print(df)

Answer 1

Try creating a boolean index based on where the current value are not equal to the next value.

Sample df:

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

   A
0  1
1  1
2  0
3  0
4  1
5  1
6  1

Index with shift + cumsum:

df['A'].ne(df['A'].shift()).cumsum()

0    1
1    1
2    2
3    2
4    3
5    3
6    3
Name: A, dtype: int32

Use the index in groupby cumcount to enumerate each group:

df['B'] = df.groupby(df['A'].ne(df['A'].shift()).cumsum()).cumcount()

   A  B
0  1  0
1  1  1
2  0  0
3  0  1
4  1  0
5  1  1
6  1  2

---

Complete Code:

import pandas as pd

df = pd.DataFrame({'A': [1, 1, 0, 0, 1, 1, 1]})
df['B'] = df.groupby(df['A'].ne(df['A'].shift()).cumsum()).cumcount()