Question

我写了一段代码，从时间戳列之一中提取月，时，周。我想在我的数据上的其他时间戳列上应用相同的代码，而无需重新编写代码。

<taskdef resource="net/sf/antcontrib/antcontrib.properties">
    <classpath>
        <pathelement location="${env.ANT_HOME}/lib/ant-contrib-0.6.jar"/>
    </classpath>
</taskdef>

Answer 1

首先指定由日期时间填充的列，然后使用f-string s循环创建新列：

cols = ['closed_at', 'another date col']

for x in cols:
    incident_data[x] = pd.to_datetime(incident_data[x], errors='coerce')
    incident_data[f'{x} - Month-Year'] = incident_data[x].dt.to_period('M')
    incident_data[f'{x} - Weekday Num'] = incident_data[x].dt.dayofweek + 1
    incident_data[f'{x} - Weekday'] = incident_data[x].dt.weekday_name
    incident_data[f'{x} - Weekday Combo'] = (incident_data[f'{x} - Weekday Num'].astype(str)+
                                             '-'+incident_data[f'{x} - Weekday'])
    incident_data[f'{x} - Hour Num'] = incident_data[x].dt.hour

Answer 2

您可以像这样在参数中声明具有列名和df的函数：

def transformation(df,column_name):
    df[column_name] = pd.to_datetime(df[column_name], errors='coerce')
    df['closed_at - Month-Year'] = df[column_name].dt.to_period('M')
    df['closed_at - Weekday Num'] = df[column_name].dt.dayofweek + 1
    df['closed_at - Weekday'] = df[column_name].dt.weekday_name
    df['closed_at - Weekday Combo'] = df['closed_at - Weekday Num'].astype(str)+'-'+df['closed_at - Weekday']
    df['closed_at - Hour Num'] = df[column_name].dt.hour
    return df

然后，您可以使用名称列表在不同的列上进行迭代。

df = transformation(df,'closed_at')

如何跨多列循环类似的python pandas代码

2 个答案: