熊猫按日期条件计数

时间:2021-06-03 14:46:45

标签: python pandas dataframe numpy date

我想统计每个客户在每个订单日期完成的所有订单,以找出每个订单时完成的订单数量。

输入:

customer_id   order_id  order_date
27501235  163595958  2018-12-14
27501235  165810252  2019-01-05

预期输出:

enter image description here

以下代码有效,但速度极慢。 10 万多行需要 10 个小时以上的时间。肯定有更好的方法。

orders_total = []

for y,row in df_dated_filt.iterrows():
    orders_total.append(df_dated_filt[(df_dated_filt["order_id"] != row["order_id"]) & 
    (df_dated_filt["customer_id"] == row["customer_id"]) & 
    (pd.to_datetime(df_dated_filt['order_date'])<pd.to_datetime(row['order_date']))]["order_id"].count())

df_dated_filt["orders_total"] = orders_total

1 个答案:

答案 0 :(得分:2)

尝试 sort_values 按升序获取日期,然后尝试 groupby cumcount 按顺序枚举组:

df['orders_total'] = df.sort_values('order_date').groupby('customer_id').cumcount()

df

   customer_id  order_id order_date  orders_total
0            1        12 2019-01-06             1
1            1        22 2019-01-01             0
2            2        34 2018-05-08             0
3            2        33 2018-05-12             1
4            2        38 2018-05-29             2

完整的工作示例:

import pandas as pd

df = pd.DataFrame({
    'customer_id': [1, 1, 2, 2, 2],
    'order_id': [12, 22, 34, 33, 38],
    'order_date': ['2019-01-06', '2019-01-01', '2018-05-08', '2018-05-12',
                   '2018-05-29']
})
df['order_date'] = pd.to_datetime(df['order_date'])

df['orders_total'] = (
    df.sort_values('order_date')
        .groupby('customer_id')
        .cumcount()
)

print(df)

编辑 假设相同的日期应该通过 rank 每个组具有相同的值:

import pandas as pd

df = pd.DataFrame({
    'customer_id': [1, 1, 1, 2, 2, 2],
    'order_id': [15, 12, 22, 34, 33, 38],
    'order_date': ['2019-01-06', '2019-01-06', '2019-01-01',
                   '2018-05-08', '2018-05-12', '2018-05-29']
})
df['order_date'] = pd.to_datetime(df['order_date'])

df['orders_total'] = (
        df.sort_values('order_date')
        .groupby('customer_id')['order_date']
        .rank(method='dense').astype(int) - 1
)

df

   customer_id  order_id order_date  orders_total
0            1        15 2019-01-06             1
1            1        12 2019-01-06             1
2            1        22 2019-01-01             0
3            2        34 2018-05-08             0
4            2        33 2018-05-12             1
5            2        38 2018-05-29             2