Question

本文的第3页列出了所使用的功能。

数据集将有一个发票条目，每个条目将包含以下字段：

creation_date  payment_date   customer_id
 2016-01-01     2016-01-03       0
 2016-01-02     2016-01-02       1
 2016-01-02     2016-01-02       1 
 2016-01-04     2016-01-05       0

现在，对于每张发票，我需要计算该客户在当前发票创建日期之前已支付的发票数量。所以，结果将是：

 creation_date  payment_date   customer_id    no_invoice_paid 
 2016-01-01     2016-01-03       0                  0
 2016-01-02     2016-01-02       1                  0 
 2016-01-02     2016-01-02       1                  0 
 2016-01-04     2016-01-05       0                  1

我提出了一个天真的解决方案：

data_customer  = data.groupby(by='customer_id')

final_df = pd.DataFrame()

for group , group_data in data_customer:
    group_data = group_data.assign(no_invoice_before=count_paid_invoice)
    final_df  = final_df.append(group_data)

计数付费发票功能如下：

def count_paid_invoice(group_data):
    for index , row in group_data.iterrows() :
     group_data.iloc[index,13] = group_data[(group_data['creation_date'] < row['creation_date']) & (group_data['payment_date'] < row['creation_date'])].shape[0]
    return group_data.iloc[:,13]

但这很慢。有没有办法可以更有效地完成这项工作？

Answer 1

您可以使用cumsum()和pd.concat的组合。

请尝试以下操作：

data['is_invoice_paid'] = 1  # Creating a dummy variable
count_invoice = data.groupby('customer_id')['is_invoice_paid'].cumsum()
count_invoice.name = 'no_invoice_paid'
final_df = pd.concat([data,count_invoice],axis=1)
final_df['no_invoice_paid'] = final_df['no_invoice_paid'] - 1  # to set the count correct
final_df = final_df.drop('is_invoice_paid',axis=1)

我在这里假设两件事：

您的付款日期不能早于创建日期。
对每位客户的付款日期进行了分类。

Answer 2

假设您的数据框名为df，则应该会为您提供所需的结果。

df['no_invoice_paid '] = df.apply(lambda row:
    df[(df.customer_id == row['customer_id']) &
       (df.payment_date < row['creation_date'])].shape[0] ,axis=1)

甚至更短：

df['no_invoice_paid '] = df.apply(lambda row:
    ((df.customer_id == row['customer_id']) &
     (df.payment_date < row['creation_date'])).sum(), axis=1)

另一种方法（类似于Spandan的方法）：

df = df.sort_values(['customer_id', 'payment_date'])
payment_lookup  = df.groupby(('customer_id', 'payment_date')).count().groupby(level=[0]).cumsum()

from functools import lru_cache
@lru_cache(maxsize=1024)
def get_customer_payments(customer_id):
    return payment_lookup.loc[customer_id]

@lru_cache(maxsize=1024)
def lookup_payments(customer_id, payment_date):
    customer_payments = get_customer_payments(customer_id)
    payments_before_current = customer_payments[customer_payments.index <
                                                payment_date]
    try:
        return payments_before_current.values[-1][0]
    except IndexError:
        return 0


df['no_invoice_paid'] = df.apply(lambda row:
                                 lookup_payments(row['customer_id'],
                                                 row['creation_date']), axis=1)

我没有测试性能。如果有效，请告诉我。

熊猫：获取每行满足特定条件的行数

2 个答案: