这是我的代码
df = df[df['scorecard_version'] != '9.0']
df = df[df['scorecard_version'] != '8.0']
df = df[df['scorecard_version'] != '10.0']
df = df[df['scorecard_version'] != '11.0']
df = df[df['scorecard_version'] != '11.1']
有没有更短的替代方案？
最佳答案
使用isin使用 ~ 的反转 bool 掩码:
df[~df['scorecard_version'].isin(['9.0','8.0','10.0','11.0','11.1'])]
替代解决方案 numpy.in1d :
df[~np.in1d(df['scorecard_version'].values, ['9.0','8.0','10.0','11.0','11.1'])]
关于python - 如何有效地进行多个 `is not`过滤，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/50755079/