Question

我想创建一个列，以显示在部门99进行的先前购买（每个客户）的累计计数（滚动总和）

我的数据框看起来像这样；每行是一个单独的事务。

    id  chain   dept    category    company     brand   date    productsize     productmeasure  purchasequantity    purchaseamount  sale
0   86246   205     7   707     1078778070  12564   2012-03-02  12.00   OZ  1   7.59    268.90
1   86246   205     63  6319    107654575   17876   2012-03-02  64.00   OZ  1   1.59    268.90
2   86246   205     97  9753    1022027929  0   2012-03-02  1.00    CT  1   5.99    268.90
3   86246   205     25  2509    107996777   31373   2012-03-02  16.00   OZ  1   1.99    268.90
4   86246   205     55  5555    107684070   32094   2012-03-02  16.00   OZ  2   10.38   268.90
5   86246   205     97  9753    1021015020  0   2012-03-02  1.00    CT  1   7.80    268.90
6   86246   205     99  9909    104538848   15343   2012-03-02  16.00   OZ  1   2.49    268.90
7   86246   205     59  5907    102900020   2012    2012-03-02  16.00   OZ  1   1.39    268.90
8   86246   205     9   921     101128414   9209    2012-03-02  4.00    OZ  2   1.50    268.90

我这样做了：


shopdata6['transactions_99'] = 0
shopdata6['transactions_99'] = shopdata6[shopdata6['dept'] == 99].groupby(['id', 'dept'])['transaction_99'].cumsum()

更新：

id dept  date   purchase purchase_count_dept99(desired)

id1 199  date1  $10       0    

id1 99  date1  $10       1

id1 100 date1  $50       1

id1 99  date2  $30       2

id2 100 date1  $10       0

id2 99  date1  $10       1

id3 99 date3  $10        1

应用了此：

shopdata6['transaction_99'] = np.where(shopdata6['dept']==99, 1, 0)
shopdata6['transaction_99'] = shopdata6.groupby(['id'])['transaction_99'].transform('cumsum')

结果看起来还不错，但这是正确的吗？

Answer 1

您的代码应简化：

s = (shopdata6['dept']==99).astype(int)
shopdata6['transaction_99'] = s.groupby(shopdata6['id']).cumsum()
print (shopdata6)
    id  dept   date purchase  purchase_count_dept99(desired)  transaction_99
0  id1   199  date1      $10                               0               0
1  id1    99  date1      $10                               1               1
2  id1   100  date1      $50                               1               1
3  id1    99  date2      $30                               2               2
4  id2   100  date1      $10                               0               0
5  id2    99  date1      $10                               1               1
6  id3    99  date3      $10                               1               1

Answer 2

如果我正确理解您的问题，则需要.cumcount()

df["transaction_99"] = df[df["dept"] == 99].groupby("id").cumcount()

要使计数从1开始，只需添加即可。

df["transaction_99"] = df["transaction_99"] + 1

Answer 3

shopdata6['transaction_99'] = np.where(shopdata6['dept']==99, 1, 0)
shopdata6['transaction_99'] = shopdata6.groupby(['id'])['transaction_99'].transform('cumsum')

获取以其他列为条件的累计熊猫总和

3 个答案: