Question

假设您有一个这样的数据框：

>>> df = pd.DataFrame({
        'epoch_minute': [i for i in reversed(range(25090627,25635267))],
        'count': [random.randint(11, 35) for _ in range(25090627,25635267)]})
>>> df.head()
   epoch_minute  count
0      25635266     12
1      25635265     20
2      25635264     33
3      25635263     11
4      25635262     35

和一些相对的纪元分钟变化量，例如：

day = 1440
week = 10080
month = 302400

如何完成此代码块的等效操作：

for i,r in df.iterrows():
    if r['epoch_minute'] - day in df['epoch_minute'].values and \
            r['epoch_minute'] - week in df['epoch_minute'].values and \
            r['epoch_minute'] - month in df['epoch_minute'].values:
        # do stuff

使用以下语法：

valid_rows = df.loc[(df['epoch_minute'] == df['epoch_minute'] - day) &
                    (df['epoch_minute'] == df['epoch_minute'] - week) &
                    (df['epoch_minute'] == df['epoch_minute'] - month]

我了解为什么loc选择不起作用，但是我只是问是否存在一种更优雅的方法来选择有效行而不迭代数据帧的行。

Answer 1

为&添加括号和bitwise AND，为支票成员资格添加isin：

valid_rows = df[(df['epoch_minute'].isin(df['epoch_minute'] - day)) &
                (df['epoch_minute'].isin(df['epoch_minute'] - week)) &
                (df['epoch_minute'].isin(df['epoch_minute'] - month))]

valid_rows = df[((df['epoch_minute'] - day).isin(df['epoch_minute'])) &
                ((df['epoch_minute']- week).isin(df['epoch_minute'] )) &
                ((df['epoch_minute'] - month).isin(df['epoch_minute']))]

熊猫选择DataFrame中存在相对列值的行

1 个答案: