I want to program a counter with python which counts the "1-values" in the column "values". The counter has to start counting after the first 0 value in the rows.
As it is seen in the example:
- the counter value for the first three "1-values" is 0.
- after having found the first 0 and the next value is 1 the count starts working
- for each 0 value the counter is set to 0.
This is just an example. In the reality the table has more than three "1-values" at the beginning. Therefore it desired that the code should bee automated.
| Values | Resulted counter values |
|---|---|
| 1 | 0 |
| 1 | 0 |
| 1 | 0 |
| 0 | 0 |
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
| 1 | 4 |
| 0 | 0 |
| 1 | 1 |
| 1 | 2 |
| 1 | 3 |
CodePudding user response：
You can use groupby.cumcount and a mask with where:
# is the value not a 1?
m = df['Values'].ne(1)
# for each group starting with "not-a-1", cumcount,
# mask the values before the first non-1
df['counter'] = df.groupby(m.cumsum()).cumcount().where(m.cummax(), 0)
output:
Values counter
0 1 0
1 1 0
2 1 0
3 0 0
4 1 1
5 1 2
6 1 3
7 1 4
8 0 0
9 1 1
10 1 2
11 1 3
Intermediates:
Values counter m m.cumsum() cumcount m.cummax()
0 1 0 False 0 0 False
1 1 0 False 0 1 False
2 1 0 False 0 2 False
3 0 0 True 1 0 True
4 1 1 False 1 1 True
5 1 2 False 1 2 True
6 1 3 False 1 3 True
7 1 4 False 1 4 True
8 0 0 True 2 0 True
9 1 1 False 2 1 True
10 1 2 False 2 2 True
11 1 3 False 2 3 True
CodePudding user response：
is_one = df.Values.eq(1)
grouper = is_one.diff().ne(0).cumsum()
counts = df.Values.groupby(grouper).cumcount().add(1)
places = is_one & df.Values.eq(0).cummax()
df["new"] = counts.where(places, other=0)
- Group by consecutive 1s (diff-ne-cumsum trick)
- get a 1, 2, 3... counter per group (cumcount() 1)
- put these counters only where Value is 1 and we're after first 0
- after first 0 is via cumulative maximum of being 0 because once True, then always True
In [71]: df
Out[71]:
Values Resulted counter values new
0 1 0 0
1 1 0 0
2 1 0 0
3 0 0 0
4 1 1 1
5 1 2 2
6 1 3 3
7 1 4 4
8 0 0 0
9 1 1 1
10 1 2 2
11 1 3 3