Stack Overflow Asked by ahbon on January 6, 2021
Given a small dataset df
as follows, I need groupby floor
, find duplicates in room
then return check
column in Pandas:
id floor room
0 1 1 101.0
1 2 1 102.0
2 3 2 201.0
3 4 2 201.0
4 5 2 202.0
5 6 3 NaN
6 7 3 201.0
7 8 3 301.0
I would like to use code as follows since there are many other columns to check:
a = np.where(condition, None, 'duplicates')
# b = np.where(df.area.str.contains('^d+$', na = True), None,
# 'area is not a numbers')
f = (lambda x: ';'.join(y for y in x if pd.notna(y))
if any(pd.notna(np.array(x))) else np.nan )
df['check'] = [f(x) for x in zip(a)]
The expected result will like this:
id floor room check
0 1 1 101.0 NaN
1 2 1 102.0 NaN
2 3 2 201.0 duplicates
3 4 2 201.0 duplicates
4 5 2 202.0 NaN
5 6 3 NaN NaN
6 7 3 201.0 NaN
7 8 3 301.0 NaN
How could I modify the condition code? Thanks for your help at advance.
you can use np.where
with duplicated
. Instead of grouping by floor you can look for duplicates of a subset of ['floor', 'room']
and pass keep=False
to flag both duplicates:
df['check'] = np.where(df.duplicated(['floor', 'room'], keep = False), 'duplicates', np.NaN)
df
Out[1]:
id floor room check
0 1 1 101.0 nan
1 2 1 102.0 nan
2 3 2 201.0 duplicates
3 4 2 201.0 duplicates
4 5 2 202.0 nan
5 6 3 NaN nan
6 7 3 201.0 nan
7 8 3 301.0 nan
Correct answer by David Erickson on January 6, 2021
you can use transform:
df["count"] = df.groupby(["floor", "room"]).transform("count")
result:
id floor room count
0 1 1 101.0 1.0
1 2 1 102.0 1.0
2 3 2 201.0 2.0
3 4 2 201.0 2.0
4 5 2 202.0 1.0
5 6 3 NaN NaN
6 7 3 201.0 1.0
7 8 3 301.0 1.0
Answered by anon01 on January 6, 2021
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