There’s cumcount, for precisely this case: df[‘col_c’] = g.cumcount() As it says in the docs: Number each item in each group from 0 to the length of that group – 1. Original answer (before cumcount was defined). You could create a helper function to do this: def add_col_c(x): x[‘col_c’] = np.arange(len(x)) return x First sort … Read more
I stumbled upon the answer which was embarrassingly simple. The groupby statement has a ‘cumcount()’ option which will enumerate group items. df[‘sequence’]=df.groupby(‘patient’).cumcount() The caveat is that the records have to be in the order you want them enumerated.
You can use the compare-cumsum-groupby pattern (which I really need to getting around to writing up for the documentation), with a final cumcount: >>> df = pd.DataFrame({“binary”: [0,1,1,1,0,0,1,1,0]}) >>> df[“consec”] = df[“binary”].groupby((df[“binary”] == 0).cumsum()).cumcount() >>> df binary consec 0 0 0 1 1 1 2 1 2 3 1 3 4 0 0 5 0 … Read more
use cumcount(), see docs here In [4]: df.groupby([‘c1’, ‘c2’]).cumcount() Out[4]: 0 0 1 1 2 0 3 1 4 0 5 1 6 2 7 0 8 0 9 0 10 1 11 2 dtype: int64 If you want orderings starting at 1 In [5]: df.groupby([‘c1’, ‘c2’]).cumcount()+1 Out[5]: 0 1 1 2 2 1 3 … Read more
The ave function computes a function by group. > id <- c(1,2,3,2,2,1,2,3) > data.frame(id,count=ave(id==id, id, FUN=cumsum)) id count 1 1 1 2 2 1 3 3 1 4 2 2 5 2 3 6 1 2 7 2 4 8 3 2 I use id==id to create a vector of all TRUE values, which get … Read more