Question

我创建了一个函数，用于在计算的时间内按给定的列值（特定供应商或商店）计算已售商品的数量。编写下面的函数的更有效方法是什么？

在大型数据集上，多次调用函数非常耗时。我在下面附上预期的结果。

我在示例中使用此数据：


df = pd.DataFrame({
    'CreatedDate' : ['2019-03-02 18:03:08.690000',
    '2019-02-01 21:22:43.109000',
    '2019-03-09 21:21:44.337000',
    '2019-04-04 21:45:59.988000',
    '2019-03-15 18:53:17.683000',
    '2019-03-03 00:24:28.003000',
    '2019-04-14 22:01:20.759000',
    '2019-02-26 19:28:01.800000',
    '2019-03-21 21:23:09.629000',
    '2019-02-03 00:55:51.756000'], 

    'Status' : ['resignation', 'sold','in delivery',
    'sold','in delivery','resignation', 
    'sold','sold','sold','sold'],

    'Vendor' :
    ['A','B','C','A','A','A','A','A','A','B'],

    'Shop' : 
    ['a','a','b','a','a','a','a','a','a','a']})

#fun
def add_count_values(df, month_start, month_stop, grouping_column, statuses, new_col):
    df['CreatedDate'] = pd.to_datetime(df['CreatedDate'])
    df['DateStartBack'] = df['CreatedDate']-pd.DateOffset(months=month_start)
    df['DateStopBack'] = df['CreatedDate'] - pd.DateOffset(months=month_stop)

    col = []

    for i in df.index:
        check_start = df.iloc[i]['DateStartBack']  
        check_stop = df.iloc[i]['DateStopBack']  
        check_group = df.iloc[i][grouping_column]  # grouping column (shop or vendor)

        count = len(df.loc[(df['CreatedDate'] > check_stop) &
                                (df['CreatedDate'] < check_start) &
                                (df.iloc[:][grouping_column] == check_group) &
                                (df['Status'].isin(statuses))]
                         )
        col.append(count)

    df = df.drop(columns=['DateStartBack', 'DateStopBack'])
    df2 = pd.DataFrame({new_col: col})

    return df.join(df2)


add_count_values(df, 0, 1, 'Vendor', ['sold','in delivery'], 'vendor_last_30days_sold')

I expect result like this

基于多个日期条件选择和计数行的最快方法是什么？

0 个答案: