Selecting column with smallest value in pandas dataframe

user_2751829 0
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I have a pandas dataframe with this structure:

df_test

V       |  A    | B   |  C      |  D
-10     |  nan  | nan |  nan    |  nan  
-9.9    |  10   | 1   |  -2200  |  100
-9.8    |  11   | 2   |  -2211  |  1

I would like to add a new column minimum_difference containing the column name with the smallest absolute value for that specific row (ignoring V), like this:

V       |  A    | B   |  C      |  D    | minimum_difference
-10     |  nan  | nan |  nan    |  nan  | nan
-9.9    |  10   | 1   |  -2200  |  100  | B
-9.8    |  11   | 2   |  -2211  |  1    | D
pandas dataframe python-3.x python
3 Answers

You can use df.drop to skip the V column, and df.abs to convert the dataframe's values to their absolute equivalents, and use df.idxmin with axis=1 to find the index (which will be a column name) of the smallest (absolute) value:

df_test['minimum_difference'] = df_test.drop('V', axis=1).abs().idxmin(axis=1)

Output:

>>> df
      V     A    B       C      D minimum_difference
0 -10.0   NaN  NaN     NaN    NaN                NaN
1  -9.9  10.0  1.0 -2200.0  100.0                  B
2  -9.8  11.0  2.0 -2211.0    1.0                  D
Admin1

You can check if a row contains NaN, then put NaN, otherwise calculate the min of the row and put the index.

In [101]: import numpy as np

In [102]: df['minimum_difference'] = np.where(df.isna().any(1), np.nan, df.abs().idxmin(1))

In [103]: df
Out[103]: 
      V     A    B       C      D minimum_difference
0 -10.0   NaN  NaN     NaN    NaN                NaN
1  -9.9  10.0  1.0 -2200.0  100.0                  B
2  -9.8  11.0  2.0 -2211.0    1.0                  D
user_58208140

Here I am sharing Mayank's snippet with little update.

In [84]: df['minimum_difference'] = df.ix[:, 1:].abs().idxmin(1)

In [85]: df
Out[85]: 
      V     A    B       C      D minimum_difference
0 -10.0   NaN  NaN     NaN    NaN                  NaN
1  -9.9  10.0  1.0 -2200.0  100.0                  B
2  -9.8  11.0  2.0 -2211.0  200.0                  B
user_84763710
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