如何根據閾值對多列進行分組並在Python中創建新列

Question

我有如下所示的數據框

輸入

Invoice No  Date    Text            Vendor    Days
1000001     1/1/2020    Rent Payment    A   0
1000003     2/1/2020    Rent Payment    A   1
1000005     4/1/2020    Rent Payment    A   2
1000007     6/1/2020    Water payment   A   2
1000008     9/2/2020    Rep Payment     A   34
1000010     9/2/2020    Car Payment     A   0
1000011     10/2/2020   Car Payment     A   1
1000012     15/2/2020   Car Payment     A   5
1000013     16/2/2020   Car Payment     A   1
1000015     17/2/2020   Car Payment     A   1
1000002     1/1/2020    Rent Payment    B   -47
1000004     4/1/2020    Con Payment     B   3
1000006     6/1/2020    Con Payment     B   2
1000009     9/2/2020    Water payment   B   34
1000014    17/2/2020    Test Payment    B   8
1000016    19/2/2020    Test Payment    B   2

健康）狀況

如何編寫python條件來檢查描述、供應商名稱和天數列，如果描述、供應商名稱相同並且天數<=2，那么這些行應該在通用組名下分組在一起，比如（G1）所有的其他行可以分配一個唯一的組名。所有分組的行都應該有唯一的組名，如輸出所示

預期產出

Invoice No  Date        Text          Vendor   Days    Group
1000001     1/1/2020    Rent Payment    A       0        G1
1000003     2/1/2020    Rent Payment    A       1        G1
1000005     4/1/2020    Rent Payment    A       2        G1
1000007     6/1/2020    Water payment   A       2        G2
1000008     9/2/2020    Rep Payment     A       34       G3
1000010     9/2/2020    Car Payment     A       0        G4
1000011    10/2/2020    Car Payment     A       1        G4
1000012    15/2/2020    Car Payment     A       5        G5
1000013    16/2/2020    Car Payment     A       1        G5
1000015    17/2/2020    Car Payment     A       1        G5
1000002    1/1/2020     Rent Payment    B      -47       G6
1000004    4/1/2020     Con Payment     B       3        G7
1000006    6/1/2020     Con Payment     B       2        G7
1000009    9/2/2020     Water payment   B      34        G8
1000014    17/2/2020    Test Payment    B       8        G9
1000016    19/2/2020    Test Payment    B       2        G9

Answer 1

您需要在三個項目上使用groupby ： 'Text' 、 'Vendor' ，以及一個布爾表示，表示在由['Text', 'Vendor']單獨定義的組內'Days'變化是否超過2 。

之后，您需要命名唯一的組。 我在下面提供了兩種方法。

ngroup

f = lambda x: x.diff().fillna(0).gt(2).cumsum()
d = df.groupby(['Text', 'Vendor']).Days.transform(f)
g = df.groupby(['Text', 'Vendor', d], sort=False).ngroup()
df.assign(Group=g.add(1).astype(str).radd('G'))

    Invoice No       Date           Text Vendor  Days Group
0      1000001   1/1/2020   Rent Payment      A     0    G1
1      1000003   2/1/2020   Rent Payment      A     1    G1
2      1000005   4/1/2020   Rent Payment      A     2    G1
3      1000007   6/1/2020  Water payment      A     2    G2
4      1000008   9/2/2020    Rep Payment      A    34    G3
5      1000010   9/2/2020    Car Payment      A     0    G4
6      1000011  10/2/2020    Car Payment      A     1    G4
7      1000012  15/2/2020    Car Payment      A     5    G5
8      1000013  16/2/2020    Car Payment      A     1    G5
9      1000015  17/2/2020    Car Payment      A     1    G5
10     1000002   1/1/2020   Rent Payment      B   -47    G6
11     1000004   4/1/2020    Con Payment      B     3    G7
12     1000006   6/1/2020    Con Payment      B     2    G7
13     1000009   9/2/2020  Water payment      B    34    G8
14     1000014  17/2/2020   Test Payment      B     8    G9
15     1000016  19/2/2020   Test Payment      B     2    G9

---

factorize

f = lambda x: x.diff().fillna(0).gt(2).cumsum()
d = df.groupby(['Text', 'Vendor']).Days.transform(f)
g = pd.factorize([*zip(df.Text, df.Vendor, d)])[0]
df.assign(Group=[f'G{i + 1}' for i in g])

    Invoice No       Date           Text Vendor  Days Group
0      1000001   1/1/2020   Rent Payment      A     0    G1
1      1000003   2/1/2020   Rent Payment      A     1    G1
2      1000005   4/1/2020   Rent Payment      A     2    G1
3      1000007   6/1/2020  Water payment      A     2    G2
4      1000008   9/2/2020    Rep Payment      A    34    G3
5      1000010   9/2/2020    Car Payment      A     0    G4
6      1000011  10/2/2020    Car Payment      A     1    G4
7      1000012  15/2/2020    Car Payment      A     5    G5
8      1000013  16/2/2020    Car Payment      A     1    G5
9      1000015  17/2/2020    Car Payment      A     1    G5
10     1000002   1/1/2020   Rent Payment      B   -47    G6
11     1000004   4/1/2020    Con Payment      B     3    G7
12     1000006   6/1/2020    Con Payment      B     2    G7
13     1000009   9/2/2020  Water payment      B    34    G8
14     1000014  17/2/2020   Test Payment      B     8    G9
15     1000016  19/2/2020   Test Payment      B     2    G9

---

一些細節

#        The first element of group    Cumulatively summing True/False
#        will get NaN so we fill it    will create a new value every time
#        in with 0         ║           we see a True.  This creates groups
#                          ║               ║     
#         adjacent differences   Should be obvious
#               ╭─┴──╮ ╭───╨───╮ ╭─┴─╮ ╭───╨──╮
f = lambda x: x.diff().fillna(0).gt(2).cumsum()

Answer 2

您可以將您的條件組合到groupby並使用ngroup 。

df['Group'] = df['Group'] = (df.groupby([df['Description'].ne(df['Description'].shift()).cumsum(), 
                             df['Vendor'].ne(df['Vendor'].shift()).cumsum(), 
                             df['Days']<=2]).ngroup()+1)
                            .astype(str).str.pad(2, 'left','G') 

# same description : df['Description'].ne(df['Description'].shift()).cumsum()
# same vendor : df['Vendor'].ne(df['Vendor'].shift()).cumsum()
# Days<=2 : df['Days']<=2

輸出：

    Invoice No  Date    Description Vendor  Days    Group
0   123456  2020-01-01  Rent Payment    A   0   G1
1   123457  2020-02-01  Rent Payment    A   1   G1
2   123458  2020-04-01  Rent Payment    A   2   G1
3   123459  2020-06-01  Water Payment   A   2   G2
4   123460  2020-09-02  Rent Payment    A   34  G3
5   123461  2020-09-02  Rep Payment     A   0   G4
6   123462  2020-10-02  Rep Payment     A   1   G4
7   123463  2020-11-02  Rep Payment     A   2   G4
8   123464  2020-02-20  Water Payment   A   11  G5