For each element in a group determine if it is present in the next group (in order as these groups appear - not necessarily numerical). For the last group - all False
.
Example:
df = pd.DataFrame({'group': [ 0, 1, 1, 0, 2 ],
'val': ['a', 'b', 'a', 'c', 'c']})
grouped = df.groupby('group')
print(result)
0 True
1 False
2 False
3 False
4 False
Name: val, dtype: bool
What is the best way to do it? I can accomplish it like this, but it seems too hacky:
keys = list(grouped.groups.keys())
iterator_keys = iter(keys[1:])
def f(ser):
if ser.name == keys[-1]:
return ser.isin([])
next_key = next(iterator_keys)
return ser.isin(grouped.get_group(next_key)['val'])
result = grouped['val'].apply(f)
CodePudding user response:
Try:
g = df.groupby("group")
m = g["val"].agg(set).shift(-1, fill_value=set())
x = g["val"].transform(lambda x: x.isin(m[x.name]))
print(x)
Prints:
0 True
1 False
2 False
3 False
4 False
Name: val, dtype: bool
Note:
If you want to replace values of the last group with any values
(not necessarily with False
), you can do this:
m = g["val"].agg(set).shift(-1)
x = g["val"].transform(lambda x: x.isin(m[x.name])
if not pd.isnull(m[x.name])
else values)
For example, if you set values = True
, the x
will be:
0 True
1 False
2 False
3 False
4 True
Name: val, dtype: bool