Identifying closest value in a column for each filter using Pandas

Question

I have a data frame with categories and values. I need to find the value in each category closest to a value. I think I'm close but I can't really get the right output when applying the results of argsort to the original dataframe.

For example, if the input was defined in the code below the output should have only (a, 1, True), (b, 2, True), (c, 2, True) and all other isClosest Values should be False.

If multiple values are closest then it should be the first value listed marked.

Here is the code I have which works but I can't get it to reapply to the dataframe correctly. I would love some pointers.

df = pd.DataFrame()
df['category'] = ['a', 'b', 'b', 'b', 'c', 'a', 'b', 'c', 'c', 'a']
df['values'] = [1, 2, 3, 4, 5, 4, 3, 2, 1, 0]
df['isClosest'] = False

uniqueCategories = df['category'].unique()
for c in uniqueCategories:
    filteredCategories = df[df['category']==c]    
    sortargs = (filteredCategories['value']-2.0).abs().argsort()
    #how to use sortargs so that we set column in df isClosest=True if its the closest value in each category to 2.0?

Answer 1

You can create a column of absolute differences:

df['dif'] = (df['values'] - 2).abs()

df
Out: 
  category  values  dif
0        a       1    1
1        b       2    0
2        b       3    1
3        b       4    2
4        c       5    3
5        a       4    2
6        b       3    1
7        c       2    0
8        c       1    1
9        a       0    2

And then use groupby.transform to check whether the minimum value of each group is equal to the difference you calculated:

df['is_closest'] = df.groupby('category')['dif'].transform('min') == df['dif']

df
Out: 
  category  values  dif is_closest
0        a       1    1       True
1        b       2    0       True
2        b       3    1      False
3        b       4    2      False
4        c       5    3      False
5        a       4    2      False
6        b       3    1      False
7        c       2    0       True
8        c       1    1      False
9        a       0    2      False

df.groupby('category')['dif'].idxmin() would also give you the indices of the closest values for each category. You can use that for mapping too.

For selection:

df.loc[df.groupby('category')['dif'].idxmin()]
Out: 
  category  values  dif
0        a       1    1
1        b       2    0
7        c       2    0

For assignment:

df['is_closest'] = False
df.loc[df.groupby('category')['dif'].idxmin(), 'is_closest'] = True
df
Out: 
  category  values  dif is_closest
0        a       1    1       True
1        b       2    0       True
2        b       3    1      False
3        b       4    2      False
4        c       5    3      False
5        a       4    2      False
6        b       3    1      False
7        c       2    0       True
8        c       1    1      False
9        a       0    2      False

The difference between these approaches is that if you check equality against the difference, you would get True for all rows in case of ties. However, with idxmin it will return True for the first occurrence (only one for each group).

Answer 2

Solution with DataFrameGroupBy.idxmin - get indexes of minimal values per group and then assign boolean mask by Index.isin to column isClosest:

idx = (df['values'] - 2).abs().groupby([df['category']]).idxmin()
print (idx)
category
a    0
b    1
c    7
Name: values, dtype: int64

df['isClosest'] = df.index.isin(idx)
print (df)
  category  values isClosest
0        a       1      True
1        b       2      True
2        b       3     False
3        b       4     False
4        c       5     False
5        a       4     False
6        b       3     False
7        c       2      True
8        c       1     False
9        a       0     False