# Get occurences of a value in column for current row and above

I got a task i normally do in excel, and can’t figure out how to convert in a smart way to Python.
I have a dataframe roughly 700 rows and 30 columns, though only one column is relevant for this issue:

``````Column X
---------
A          1
A          1
A          1
A          1
A          1
A          1
A          1
A          1
A          1
A          1
A          1
A          0
A          0
A          0
A          0
A          0
B          1
B          1
B          1
B          1
B          1
B          1
B          1
B          1
B          1
B          1
B          1
B          0
B          0
B          0
C          1
C          1
``````

I need to create a new column with the value 1, for the first 11 occurences of whatever is in column X, So the first 11 rows of "A" gets value 1 and the remaning ones gets 0.
So i guess i can make a column that count occurences of df.at[idx, "Column X"], but only for current row and rows above.
Once i have this i can make another column that writes 1 if the newly created column has a value of 11 or lower, and otherwise returns 0.

NB: Data is always sorted by X, so will follow the pattern shown above.

Any suggestions how this can be done?

### >Solution :

Use `GroupBy.cumcount` for counter starting by `0`, so for first `11` values compare by `Series.lt` for less like `11` and cast output to inteegrs for `Tru/False` to `1,0` mapping:

``````df['new'] = df.groupby('Column X').cumcount().lt(11).astype(int)
print (df)
Column X  new
0         A    1
1         A    1
2         A    1
3         A    1
4         A    1
5         A    1
6         A    1
7         A    1
8         A    1
9         A    1
10        A    1
11        A    0
12        A    0
13        A    0
14        A    0
15        A    0
16        B    1
17        B    1
18        B    1
19        B    1
20        B    1
21        B    1
22        B    1
23        B    1
24        B    1
25        B    1
26        B    1
27        B    0
28        B    0
29        B    0
30        C    1
31        C    1
``````