使用数据框计算销售额需要太长时间

Question

I have a problem. I would like to calculate the turnover for a customer in the last 6 months. The methods work on my dummy record, unfortunately the whole thing does not work on my real record as it is too slow. How can I rewrite this so that it performs faster?

Dataframe

   customerId   fromDate  sales
0           1 2022-06-01    100
1           1 2022-05-25     20
2           1 2022-05-25     50
3           1 2022-05-20     30
4           1 2021-09-05     40
5           2 2022-06-02     80
6           3 2021-03-01     50
7           3 2021-02-01     20

Code

from datetime import datetime
from dateutil.relativedelta import relativedelta

import pandas as pd


def find_last_date(date_: datetime) -> datetime:
    six_months = date_ + relativedelta(months=-6)
    return six_months


def sum_func(row: pd.DataFrame, df: pd.DataFrame) -> int :
    return df[
            (df["customerId"] == row["customerId"])
             & (row["fromDate"] + relativedelta(months=-6)<= df["fromDate"])
             & (df["fromDate"]   <= row["fromDate"])
        ]["sales"].sum()

d = {
    "customerId": [1, 1, 1, 1, 1, 2, 3, 3],
    "fromDate": [
        "2022-06-01",
        "2022-05-25",
        "2022-05-25",
        "2022-05-20",
        "2021-09-05",
        "2022-06-02",
        "2021-03-01",
        "2021-02-01",
    ],
    "sales": [100, 20, 50, 30, 40, 80, 50, 20],
}
df = pd.DataFrame(data=d)

df["fromDate"] = pd.to_datetime(df["fromDate"], errors="coerce")
df["last_month"] = df["fromDate"].apply(find_last_date)
df["total_sales"]=df[["customerId", "fromDate"]].apply(lambda x: sum_func(x, df), axis=1)
print(df)

What I want

   customerId   fromDate  sales last_month total_sales
0           1 2022-06-01    100 2022-03-01        200 # 100 + 20 + 50 + 30
1           1 2022-05-25     20 2022-02-25        100 # 20 + 50 + 30
2           1 2022-05-25     50 2022-02-25        100 # 50 + 20 + 30
3           1 2022-05-20     30 2022-02-20        30  # 30
4           1 2021-09-05     40 2021-06-05        40  # 40
5           2 2022-06-02     80 2022-03-02        80  # 80
6           3 2021-03-01     50 2020-12-01        70  # 50 + 20
7           3 2021-02-01     20 2020-11-01        20  # 20

---

print(df['customerId'].value_counts().describe())

count    53979.000
mean        87.404
std       1588.450
min          1.000
25%          2.000
50%          6.000
75%         22.000
max     205284.000

print(df['fromDate'].agg((min, max)))

min   2021-02-22
max   2022-03-26

Answer 1

Use numpy broadcasting per groups with numpy.where for set for True values of Sales and if not match 0 , so possible sum sales to new column:

df["fromDate"] = pd.to_datetime(df["fromDate"], errors="coerce")
df["last_month"] = df["fromDate"] - pd.offsets.DateOffset(months=6)

def f(x):
    d1 = x["fromDate"].to_numpy()
    d2 = x["last_month"].to_numpy()
    mask = (d2[:, None]<=d1) & (d1<=d1[:, None])
    x['total_sales'] = np.dot(mask, x['sales'].to_numpy())
    return x

df = df.groupby('customerId').apply(f)

print(df)
   customerId   fromDate  sales last_month  total_sales
0           1 2022-06-01    100 2021-12-01          200
1           1 2022-05-25     20 2021-11-25          100
2           1 2022-05-25     50 2021-11-25          100
3           1 2022-05-20     30 2021-11-20           30
4           1 2021-09-05     40 2021-03-05           40
5           2 2022-06-02     80 2021-12-02           80
6           3 2021-03-01     50 2020-09-01           70
7           3 2021-02-01     20 2020-08-01           20

EDIT:

df["fromDate"] = pd.to_datetime(df["fromDate"], errors="coerce")
df["last_month"] = df["fromDate"] - pd.offsets.DateOffset(months=6)

#https://stackoverflow.com/a/27670190/2901002
def chunking_dot(big_matrix, small_matrix, chunk_size=10000):
    # Make a copy if the array is not already contiguous
    small_matrix = np.ascontiguousarray(small_matrix)
    R = np.empty((big_matrix.shape[0], small_matrix.shape[1]))
    for i in range(0, R.shape[0], chunk_size):
        end = i + chunk_size
        R[i:end] = np.dot(big_matrix[i:end], small_matrix)
    return R

def f(x):
    d1 = x["fromDate"].to_numpy()
    d2 = x["last_month"].to_numpy()
    mask = (d2[:, None]<=d1) & (d1<=d1[:, None])
    # print (mask)

    x['total_sales'] = chunking_dot(mask, x[['sales']].to_numpy())

    return x

df = df.groupby('customerId').apply(f)

---

print(df)
   customerId   fromDate  sales last_month  total_sales
0           1 2022-06-01    100 2021-12-01        200.0
1           1 2022-05-25     20 2021-11-25        100.0
2           1 2022-05-25     50 2021-11-25        100.0
3           1 2022-05-20     30 2021-11-20         30.0
4           1 2021-09-05     40 2021-03-05         40.0
5           2 2022-06-02     80 2021-12-02         80.0
6           3 2021-03-01     50 2020-09-01         70.0
7           3 2021-02-01     20 2020-08-01         20.0

Answer 2

Using multiprocessing and consider 6 months as 180 days to reduce the memory size and the time computing.

Copy the following code to a python file and run it from the console (not from a Jupyter Notebook)

import pandas as pd
import numpy as np
import multiprocessing as mp
import time

def sum_sales(customer, df):
    # 1st pass: sum sales of same days, reduce the row numbers
    df1 = df.groupby('fromDate')['sales'].sum()

    # Generate all missing dates
    df1 = df1.reindex(pd.date_range(df1.index.min(), df1.index.max(), freq='D'), fill_value=0)

    # 2nd pass: use a sliding window of 180 days to sum
    df1 = df1.rolling(90, min_periods=0).sum().astype(int)

    # Restore original index for the group
    df1 = df1.reindex(df['fromDate']).reset_index(drop=True).to_frame().set_index(df.index)

    return df1


if __name__ == '__main__':  # Do not remove this line! Mandatory
    # Setup a minimal reproducible example
    N = 3_000_000
    D = pd.to_datetime('2021-1-1')
    rng = np.random.default_rng(2022)
    dti = D + pd.to_timedelta(rng.integers(0, 365*2, N), unit='D')
    cust = rng.integers(0, 75000, N)
    sales = rng.integers(1, 200, N)
    df = pd.DataFrame({'customerId': cust, 'fromDate': dti, 'sales': sales})

    # Ensure your dataframe is sorted by fromDate for rolling window
    df.sort_values(['customerId', 'fromDate'], ignore_index=True)

    start = time.time()
    with mp.Pool(mp.cpu_count() - 1) as p:
        results = p.starmap(sum_sales, df.groupby('customerId'))
    df['total_sales'] = pd.concat(results)
    end = time.time()
    print(f"Elapsed time: {end - start:.2f} seconds")

For 3mio records and 75k different customers on 2 years (730 days)

[...]$ python mp.py
Elapsed time: 24.36 seconds

However the number of sales per customer is well balanced than your:

>>> df['customerId'].value_counts().describe(percentiles=np.linspace(0, 1, 11)
count    75000.000000
mean        40.000000
std          6.349157
min         15.000000
0%          15.000000
10%         32.000000
20%         35.000000
30%         37.000000
40%         38.000000
50%         40.000000
60%         41.000000
70%         43.000000
80%         45.000000
90%         48.000000  # <- check the 90th percentile of your data
100%        73.000000
max         73.000000  # <- max transactions for a single customer
Name: customerId, dtype: float64

Because the sales are properly distributed per customer, my sample takes advantage of multiprocessing. In your case, I don't think it will be the case (check the 90th percentile).

The check with your dataframe:

>>> df
   customerId   fromDate  sales  total_sales
0           1 2022-06-01    100          200
1           1 2022-05-25     20          100
2           1 2022-05-25     50          100
3           1 2022-05-20     30           30
4           1 2021-09-05     40           40
5           2 2022-06-02     80           80
6           3 2021-03-01     50           70
7           3 2021-02-01     20           20

If you decide to choose to keep a variable moving window of 6 months instead of a fixed moving window of 180 days, the algorithm will me the same. The important point in the code is to reduce the number of rows per customer. In your sample, you can group the sales for a same (customer, date). The customer 1 have 2 rows for 2022-05-25 so you can sum them immediately.

IIUC, in your real data, you have a customer with 205284 sales between 2021-02-22 and 2022-03-26 (397 days), so this user has an average of 517 transactions per day (?). If you sum sales of same days, you reduce the number of records from 205284 to 397...