1 

Index a b   c   d   e   f   g   h   i   j   zeros
0   9   -8  -3  0   0   3   0   0   0   3   5
1   0   0   8   2   -5  0   0   -5  4   12  4
2   0   0   3   13  0   -9  11  -2  5   0   4
3   0   0   1   12  1   0   11  -6  1   0   4
4   0   0   1   11  0   -3  0   -2  -6  12  4
5   0   0   1   0   6   -5  0   -5  3   7   4
6   0   0   0   12  -6  0   2   7   -1  2   4
7   0   0   0   0   5   -5  5   0   8   7   5
8   0   0   0   0   0   0   0   0   0   0   10
9   0   0   -2  12  0   -1  0   3   6   3   4
10  0   0   -4  3   -4  0   0   4   -1  1   4

This is my input dataframe

I want to sort my dataframe based on highest zeros which can be done easily by sorting using zeros column. but in case of 1st adn 7 row i have the same number of zeros. In this case i am suppose to look at the order of occurance of 0's from left of the dataframe to the right (from column a to column j) which should give me the output as shown below: basically, when you look at the order of zeros you will see that in the result i have 4 columns containing zeros being placed first out of two rows contianing zeros = 5.

This is my output dataframe

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  • the index of the second column has gotten reset. please dont get confused by it. if there is more information that is required please let me know Commented Jun 4, 2022 at 12:34

1 Answer 1

Reset to default
1

You can create a weight column:

df['weight'] = (df.loc[:, 'a':'j'].eq(0)
                  .mul((2 ** np.arange(len(df.columns) - 1))[::-1])
                  .sum(axis=1))

Output:

>>> df.sort_values(['zeros', 'pri'], ascending=False)
    a  b  c   d  e  f   g  h  i   j  zeros  weight
0   9 -8 -3   0  0  3   0  0  0   3      5     110
1   0  0  8   2 -5  0   0 -5  4  12      4     792
2   0  0  3  13  0 -9  11 -2  5   0      4     801
3   0  0  1  12  1  0  11 -6  1   0      4     785
4   0  0  1  11  0 -3   0 -2 -6  12      4     808
5   0  0  1   0  6 -5   0 -5  3   7      4     840
6   0  0  0  12 -6  0   2  7 -1   2      4     912
7   0  0  0   0  5 -5   5  0  8   7      5     964
8   0  0  0   0  0  0   0  0  0   0     10    1023
9   0  0 -2  12  0 -1   0  3  6   3      4     808
10  0  0 -4   3 -4  0   0  4 -1   1      4     792

Step by step:

>>> out = df.loc[:, 'a':'j'].eq(0)
        a      b      c      d      e      f      g      h      i      j
0   False  False  False   True   True  False   True   True   True  False
1    True   True  False  False  False   True   True  False  False  False
2    True   True  False  False   True  False  False  False  False   True
3    True   True  False  False  False   True  False  False  False   True
4    True   True  False  False   True  False   True  False  False  False
5    True   True  False   True  False  False   True  False  False  False
6    True   True   True  False  False   True  False  False  False  False
7    True   True   True   True  False  False  False   True  False  False
8    True   True   True   True   True   True   True   True   True   True
9    True   True  False  False   True  False   True  False  False  False
10   True   True  False  False  False   True   True  False  False  False

>>> out = out.mul((2 ** np.arange(len(df.columns) - 1))[::-1])
      a    b    c   d   e   f  g  h  i  j
0     0    0    0  64  32   0  8  4  2  0
1   512  256    0   0   0  16  8  0  0  0
2   512  256    0   0  32   0  0  0  0  1
3   512  256    0   0   0  16  0  0  0  1
4   512  256    0   0  32   0  8  0  0  0
5   512  256    0  64   0   0  8  0  0  0
6   512  256  128   0   0  16  0  0  0  0
7   512  256  128  64   0   0  0  4  0  0
8   512  256  128  64  32  16  8  4  2  1
9   512  256    0   0  32   0  8  0  0  0
10  512  256    0   0   0  16  8  0  0  0

>>> out = out.sum(axis=1)
0      110
1      792
2      801
3      785
4      808
5      840
6      912
7      964
8     1023
9      808
10     792
dtype: int64
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