Reputation: 1319
I have a dataset like:
Id Status
1 0
1 0
1 0
1 0
1 1
2 0
1 0
2 0
3 0
3 0
I want to drop all rows of an id after its status became 1, i.e. my new dataset will be:
Id Status
1 0
1 0
1 0
1 0
1 1
2 0
2 0
3 0
3 0
i.e.
1 0 --> gets removed since this row appears after id 1 already had a status of 1
How to implement it efficiently since I have a very large (200 GB+) dataset.
Thanks for your help.
Upvotes: 7
Views: 1930
Reputation: 19885
EDIT: Revisiting this question a month later, there is actually a much simpler way with groupby
and cumsum
: Just group by Id
and take the cumsum
of Status
, then drop the values where the cumsum
is more than 0:
df[df.groupby('Id')['Status'].cumsum() < 1]
The best way I have found is to find the index of the first 1 and slice each group that way. In cases where no 1 exists, return the group unchanged:
def remove(series):
indexless = series.reset_index(drop=True)
ones = indexless[indexless['Status'] == 1]
if len(ones) > 0:
return indexless.iloc[:ones.index[0] + 1]
else:
return indexless
df.groupby('Id').apply(remove).reset_index(drop=True)
Output:
Id Status
0 1 0
1 1 0
2 1 0
3 1 0
4 1 1
5 2 0
6 2 0
7 3 0
8 3 0
Upvotes: 2
Reputation: 2495
Use groupby
with cumsum
to find where status is 1
.
res = df.groupby('Id', group_keys=False).apply(lambda x: x[x.Status.cumsum() > 0])
res
Id Status
4 1 1
6 1 0
Exclude index that Status==0
.
not_select_id = res[res.Status==0].index
df[~df.index.isin(not_select_id)]
Id Status
0 1 0
1 1 0
2 1 0
3 1 0
4 1 1
5 2 0
7 2 0
8 3 0
9 3 0
Upvotes: 1
Reputation: 4670
Here's an idea;
You can create a dict with the first index where the status is 1 for each ID (assuming the DataFrame is sorted by ID):
d = df.loc[df["Status"]==1].drop_duplicates()
d = dict(zip(d["Id"], d.index))
Then you create a column with the first status=1
for each Id:
df["first"] = df["Id"].map(d)
Finally you drop every row where the index is less than than the first
column:
df = df.loc[df.index<df["first"]]
Upvotes: 2