如何根据阈值在 Python 中对多列进行分组并创建新列

Posted 2023-03-11

【中文标题】如何根据阈值在 Python 中对多列进行分组并创建新列

【英文标题】：How to groupby multiple columns and create a new column in Python based on thresholds

【发布时间】：2020-06-24 16:14:11

【问题描述】：

我有如下数据框

输入

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

条件

如何编写检查描述、供应商名称和天数列的python条件，如果描述、供应商名称相同且天数为

预期输出

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

【参考方案1】：

您需要在三个项目上使用groupby：'Text'、'Vendor'，以及在仅由['Text', 'Vendor'] 定义的组中'Days' 的变化是否超过2 的布尔表示。

之后，您需要命名唯一组。下面我提供了两种方法。

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'Gi + 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

---

一些细节

#        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()

【讨论】：

@piSquared，我提供的预期输入和输出中存在小错误，刚刚更正。

我的意思是要求检查供应商和描述以及天列，如果供应商和描述相同并且相邻行之间的天差为

【参考方案2】：

您可以将您的条件组合成groupby 并使用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

输出：

    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

【讨论】：

到第 4 行是它的 aggining 正确，从第 5 行开始它不正确。

你为什么把Rent Payment和Rep Payment当作同一个Description？ @Rahulrajan

抱歉打错了，我的意思是要求检查供应商和描述以及天列，如果供应商和描述相同并且相邻行之间的天差是

这正是它的作用。 @Rahulrajan 在您的示例中似乎有一个错误。 Rent Payment 和 Rep Payment 在同一个组中，即使不一样。