应用条件并计算
True值。
(df == 0).sum(1)ID1 2ID2 0ID3 1dtype: int64df[(df == 0).sum(1) < 2] col0 col1 col2 col3ID2 1 1 2 10ID3 0 1 3 4
# df[(~df.astype(bool)).sum(1) < 2]df[df.astype(bool).sum(1) > len(df.columns)-2] # no inversion needed col0 col1 col2 col3ID2 1 1 2 10ID3 0 1 3 4
为了提高性能,您可以使用
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 col3ID2 1 1 2 10ID3 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)
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