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Let's say I have DataFrames df1 and df2:

>>> df1 = pd.DataFrame({'A': [0, 2, 4], 'B': [2, 17, 7], 'C': [4, 9, 11]})
>>> df1
   A   B   C
0  0   2   4
1  2  17   9
2  4   7  11
>>> df2 = pd.DataFrame({'A': [9, 2, 32], 'B': [1, 3, 8], 'C': [6, 2, 41]})
>>> df2
    A  B   C
0   9  1   6
1   2  3   2
2  32  8  41

What I want is the 3rd DataFrame that will have minimal rows (min is calculated based on column B), that is:

>>> df3
   A  B   C
0  9  1   6
1  2  3   2
2  4  7  11

I really don't want to do this by iterating over all rows and comparing them one by one, is there a faster and compact way to do this?

Let's say I have DataFrames df1 and df2:

>>> df1 = pd.DataFrame({'A': [0, 2, 4], 'B': [2, 17, 7], 'C': [4, 9, 11]})
>>> df1
   A   B   C
0  0   2   4
1  2  17   9
2  4   7  11
>>> df2 = pd.DataFrame({'A': [9, 2, 32], 'B': [1, 3, 8], 'C': [6, 2, 41]})
>>> df2
    A  B   C
0   9  1   6
1   2  3   2
2  32  8  41

What I want is the 3rd DataFrame that will have minimal rows (min is calculated based on column B), that is:

>>> df3
   A  B   C
0  9  1   6
1  2  3   2
2  4  7  11

I really don't want to do this by iterating over all rows and comparing them one by one, is there a faster and compact way to do this?

  • python
  • pandas
  • dataframe
  • min
Share asked Nov 19, 2024 at 18:19 k1r1t0k1r1t0 7671 gold badge5 silver badges20 bronze badges
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1 Answer 1

Reset to default 2

You can mask df1 with df2 when df2['B'] < df1['B']:

out = df1.mask(df2['B'].lt(df1['B']), df2)

Output:

   A  B   C
0  9  1   6
1  2  3   2
2  4  7  11

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