Pandas 轧制 function 有重叠

Question

I would like to apply a function to one pandas dataframe column which does the following task:

• I have a cycle counter that starts from a value but sometimes restarts.
• I would like to have the counter continue and increase its value.

The function I use at the moment is the following one:

Code

import pandas as pd

d = {'Cycle':[100,100,100,100,101,101,101,102,102,102,102,102,102,103,103,103,100,100,100,100,101,101,101,101]}
df = pd.DataFrame(data=d)

df.loc[:,'counter'] = df['Cycle'].to_numpy()
df.loc[:,'counter'] = df['counter'].rolling(2).apply(lambda x: x[0] if (x[0] == x[1]) else x[0]+1, raw=True)

print(df)

Output

    Cycle  counter
0     100      NaN
1     100    100.0
2     100    100.0
3     100    100.0
4     101    101.0
5     101    101.0
6     101    101.0
7     102    102.0
8     102    102.0
9     102    102.0
10    102    102.0
11    102    102.0
12    102    102.0
13    103    103.0
14    103    103.0
15    103    103.0
16    100    104.0
17    100    100.0
18    100    100.0
19    100    100.0
20    101    101.0
21    101    101.0
22    101    101.0
23    101    101.0

My goal is to get a dataframe similar to this one:

    Cycle  counter
0     100      NaN
1     100    100.0
2     100    100.0
3     100    100.0
4     101    101.0
5     101    101.0
6     101    101.0
7     102    102.0
8     102    102.0
9     102    102.0
10    102    102.0
11    102    102.0
12    102    102.0
13    103    103.0
14    103    103.0
15    103    103.0
16    100    104.0
17    100    104.0
18    100    104.0
19    100    104.0
20    101    105.0
21    101    105.0
22    101    105.0
23    101    105.0

• How do I use the rolling function with one overlap?
• Do you have any recommendation to reach my goal?

Best regards,

Matteo

Answer 1

We can use shift and ne (same as != ) to check where the Cycle column changes.

Then we use cumsum to make a counter which changes each time Cycle changes.

We add the first value of Cycle to the counter -1 , to let it start at 100 :

groups = df['Cycle'].ne(df['Cycle'].shift()).cumsum()
df['counter'] = groups + df['Cycle'].iat[0] - 1

    Cycle  counter
0     100      100
1     100      100
2     100      100
3     100      100
4     101      101
5     101      101
6     101      101
7     102      102
8     102      102
9     102      102
10    102      102
11    102      102
12    102      102
13    103      103
14    103      103
15    103      103
16    100      104
17    100      104
18    100      104
19    100      104
20    101      105
21    101      105
22    101      105
23    101      105

Details: groups gives us a counter starting at 1 :

print(groups)

0     1
1     1
2     1
3     1
4     2
5     2
6     2
7     3
8     3
9     3
10    3
11    3
12    3
13    4
14    4
15    4
16    5
17    5
18    5
19    5
20    6
21    6
22    6
23    6
Name: Cycle, dtype: int64

Answer 2

Another approach would be to identify the points in the Cycle column where the value changes using.diff(). Then at those points increment from the original initial cycle value and merge to the original dataframe forward filling the new values.

df2 = df[df['Cycle'].diff().apply(lambda x: x!=0)].reset_index()
df2['Target Count'] = df[df['Cycle'].diff().apply(lambda x: x!=0)].reset_index().reset_index().apply(lambda x: df.iloc[0,0] + x['level_0'], axis = 1)
df = df.merge(df2.drop('Cycle', axis = 1), right_on = 'index', left_index = True, how = 'left').ffill().set_index('index', drop = True)
def df.index.name
df

Cycle  Target Count
0     100         100.0
1     100         100.0
2     100         100.0
3     100         100.0
4     101         101.0
5     101         101.0
6     101         101.0
7     102         102.0
8     102         102.0
9     102         102.0
10    102         102.0
11    102         102.0
12    102         102.0
13    103         103.0
14    103         103.0
15    103         103.0
16    100         104.0
17    100         104.0
18    100         104.0
19    100         104.0
20    101         105.0
21    101         105.0
22    101         105.0
23    101         105.0