python - Pandas: drop row if more than one of multiple columns is zero

Question

I have a dataframe as such:

     col0   col1  col2  col3
ID1    0      2     0     2
ID2    1      1     2     10
ID3    0      1     3     4

I want to remove rows that contain zeros more than once.

I've tried to do:

cols = ['col1', etc]
df.loc[:, cols].value_counts()

But this only works for series and not dataframes.

df.loc[:, cols].count(0) <= 1

Only returns bools.

I feel like I'm close with the 2nd attempt here.

Answer 1

Apply the condition and count the True values.

(df == 0).sum(1)

ID1    2
ID2    0
ID3    1
dtype: int64

df[(df == 0).sum(1) < 2]

     col0  col1  col2  col3
ID2     1     1     2    10
ID3     0     1     3     4

Alternatively, convert the integers to bool and sum that. A little more direct.

# df[(~df.astype(bool)).sum(1) < 2]
df[df.astype(bool).sum(1) > len(df.columns)-2]  # no inversion needed

     col0  col1  col2  col3
ID2     1     1     2    10
ID3     0     1     3     4

---

For performance, you can use np.count_nonzero:

# df[np.count_nonzero(df, axis=1) > len(df.columns)-2]
df[np.count_nonzero(df.values, axis=1) > len(df.columns)-2]

     col0  col1  col2  col3
ID2     1     1     2    10
ID3     0     1     3     4

---

df = pd.concat([df] * 10000, ignore_index=True)

%timeit df[(df == 0).sum(1) < 2]
%timeit df[df.astype(bool).sum(1) > len(df.columns)-2]
%timeit df[np.count_nonzero(df.values, axis=1) > len(df.columns)-2]

7.13 ms ± 161 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
4.28 ms ± 120 μs per loop (mean ± std. dev. of 7 runs, 100 loops each)
997 μs ± 38.2 μs per loop (mean ± std. dev. of 7 runs, 1000 loops each)