使用 pandas/python 的基於優先級的分類

Question

我在下面的數據框和代碼列表中開具發票相關數據

df = pd.DataFrame({
    'invoice':[1,1,2,2,2,3,3,3,4,4,4,5,5,6,6,6,7],
    'code':[101,104,105,101,106,106,104,101,104,105,111,109,111,110,101,114,112],
    'qty':[2,1,1,3,2,4,7,1,1,1,1,4,2,1,2,2,1]
})

+---------+------+-----+
| invoice | code | qty |
+---------+------+-----+
|    1    |  101 |  2  |
+---------+------+-----+
|    1    |  104 |  1  |
+---------+------+-----+
|    2    |  105 |  1  |
+---------+------+-----+
|    2    |  101 |  3  |
+---------+------+-----+
|    2    |  106 |  2  |
+---------+------+-----+
|    3    |  106 |  4  |
+---------+------+-----+
|    3    |  104 |  7  |
+---------+------+-----+
|    3    |  101 |  1  |
+---------+------+-----+
|    4    |  104 |  1  |
+---------+------+-----+
|    4    |  105 |  1  |
+---------+------+-----+
|    4    |  111 |  1  |
+---------+------+-----+
|    5    |  109 |  4  |
+---------+------+-----+
|    5    |  111 |  2  |
+---------+------+-----+
|    6    |  110 |  1  |
+---------+------+-----+
|    6    |  101 |  2  |
+---------+------+-----+
|    6    |  114 |  2  |
+---------+------+-----+
|    7    |  104 |  2  |
+---------+------+-----+

代碼列表是，

Soda =  [101,102]
Hot =  [103,109]
Juice =  [104,105]
Milk =  [106,107,108]
Dessert =  [110,111]

我的任務是根據下面指定的Order of Priority添加一個新的category列。

1. 優先級 1：如果任何發票的數量超過 10 個，則應歸類為Mega 。 例如： invoice 3 is 12的qty總和invoice 3 is 12

2. Priority No.2：來自rest of the invoice的rest of the invoice 。 如果invoice任何code在Milk列表中，則類別應為Healthy 。 例如： invoice 2 code 106是Milk 。 因此，完整發票被歸類為Healthy 。 不管發票中是否存在其他項目（ code 101 & 105 ）。 由於優先級適用於full發票。

3. 重點三：從rest of the invoice ，如果有的話code中的invoice是在Juice列表中，則這有2 parts

(3.1) 如果果汁數量的總和equal to 1 ，則類別應為OneJuice 。 例如： invoice 1有code 104和qty 1無論發票中是否存在其他項目（ code 101 ），此發票invoice 1都將獲得OneJuice 。 由於優先級適用於full發票。

(3.2) 如果果汁數量的總和greater than 1 ，則類別應為ManyJuice 。 例如： invoice 4 code 104 & 105 ， qty 1 + 1 = 2 。

4. 優先級4：從rest of the invoice的rest of the invoice ，如果rest of the invoice任何code在Hot列表中，則應將其歸類為HotLovers 。 無論發票中是否存在其他項目。

5. 優先級 5：從rest of the invoice的rest of the invoice ，如果rest of the invoice任何code在Dessert列表中，則應將其歸類為DessertLovers 。

6. 最后，其余的發票應歸類為Others 。

我想要的輸出如下。

+---------+------+-----+---------------+
| invoice | code | qty |    category   |
+---------+------+-----+---------------+
|    1    |  101 |  2  |    OneJuice   |
+---------+------+-----+---------------+
|    1    |  104 |  1  |    OneJuice   |
+---------+------+-----+---------------+
|    2    |  105 |  1  |    Healthy    |
+---------+------+-----+---------------+
|    2    |  101 |  3  |    Healthy    |
+---------+------+-----+---------------+
|    2    |  106 |  2  |    Healthy    |
+---------+------+-----+---------------+
|    3    |  106 |  4  |      Mega     |
+---------+------+-----+---------------+
|    3    |  104 |  7  |      Mega     |
+---------+------+-----+---------------+
|    3    |  101 |  1  |      Mega     |
+---------+------+-----+---------------+
|    4    |  104 |  1  |   ManyJuice   |
+---------+------+-----+---------------+
|    4    |  105 |  1  |   ManyJuice   |
+---------+------+-----+---------------+
|    4    |  111 |  1  |   ManyJuice   |
+---------+------+-----+---------------+
|    5    |  109 |  4  |   HotLovers   |
+---------+------+-----+---------------+
|    5    |  111 |  2  |   HotLovers   |
+---------+------+-----+---------------+
|    6    |  110 |  1  | DessertLovers |
+---------+------+-----+---------------+
|    6    |  101 |  2  | DessertLovers |
+---------+------+-----+---------------+
|    6    |  114 |  2  | DessertLovers |
+---------+------+-----+---------------+
|    7    |  104 |  2  |     ManyJuice |
+---------+------+-----+---------------+

到目前為止，我已經在下面嘗試過。 有用。 但非常天真，根本不是pythonic。 同樣，當我將其應用於原始數據集時，代碼非常非常慢。

# Calculating Priority No.1 
L = df.groupby(['invoice'])['qty'].transform('sum') >= 10
df_Large = df[L]['invoice'].to_frame()
df_Large['category'] = 'Mega'
df_Large.drop_duplicates(['invoice'], inplace=True)


# Calculating Priority No.2
df_1 = df[~L] # removing Priority No.1 calculated above
M = (df_1['code'].isin(Milk)
.groupby(df_1['invoice'])
.transform('any'))
df_Milk = df_1[M]['invoice'].to_frame()
df_Milk['category'] = 'Healthy'
df_Milk.drop_duplicates(['invoice'], inplace=True)

# Calculating Priority No.3

# 3.a Part -1

df_2 = df[~L & ~M]  # removing Priority No.1 & 2 calculated above
J_1 = (df_2['code'].isin(Juice)
.groupby(df_2['invoice'])
.transform('sum') == 1)
df_SM = df_2[J_1]['invoice'].to_frame()
df_SM['category'] = 'OneJuice'
df_SM.drop_duplicates(['invoice'], inplace=True)


# 3.b Part -2
J_2 = (df_2['code'].isin(Juice)
.groupby(df_2['invoice'])
.transform('sum') > 1)
df_MM = df_2[J_2]['invoice'].to_frame()
df_MM['category'] = 'ManyJuice'
df_MM.drop_duplicates(['invoice'], inplace=True)


# Calculating Priority No.4
df_3 = df[~L & ~M & ~J_1 & ~J_2]  # removing Priority No.1, 2 & 3 (a & b) calculated above
H = (df_3['code'].isin(Hot)
.groupby(df_3['invoice'])
.transform('any'))
df_Hot = df_3[H]['invoice'].to_frame()
df_Hot['category'] = 'HotLovers'
df_Hot.drop_duplicates(['invoice'], inplace=True)


# Calculating Priority No.5
df_4 = df[~L & ~M & ~J_1 & ~J_2 & ~H ] # removing Priority No.1, 2, 3 (a & b) and 4 calculated above
D = (df_4['code'].isin(Dessert)
.groupby(df_4['invoice'])
.transform('any'))
df_Dessert = df_4[D]['invoice'].to_frame()
df_Dessert['category'] = 'DessertLovers'
df_Dessert.drop_duplicates(['invoice'], inplace=True)

# merge all dfs
category = pd.concat([df_Large,df_Milk,df_SM,df_MM,df_Hot,df_Dessert], axis=0,sort=False, ignore_index=True)

# Final merge to the original dataset
df = df.merge(category,on='invoice', how='left').fillna(value='Others')

因此需要幫助來清理此代碼以提高速度/效率和 pythonic 方式。

Answer 1

你可以嘗試使用np.select

df['category'] = np.select([
    df.groupby('invoice')['qty'].transform('sum') >= 10,
    df['code'].isin(Milk).groupby(df.invoice).transform('any'),
    (df['qty']*df['code'].isin(Juice)).groupby(df.invoice).transform('sum') == 1,
    (df['qty']*df['code'].isin(Juice)).groupby(df.invoice).transform('sum') > 1,
    df['code'].isin(Hot).groupby(df.invoice).transform('any'),
    df['code'].isin(Dessert).groupby(df.invoice).transform('any')
],
    ['Mega','Healthy','OneJuice','ManyJuice','HotLovers','DessertLovers'],
    'Other'
)
print(df)

輸出

    invoice  code  qty       category
0         1   101    2       OneJuice
1         1   104    1       OneJuice
2         2   105    1        Healthy
3         2   101    3        Healthy
4         2   106    2        Healthy
5         3   106    4           Mega
6         3   104    7           Mega
7         3   101    1           Mega
8         4   104    1      ManyJuice
9         4   105    1      ManyJuice
10        4   111    1      ManyJuice
11        5   109    4      HotLovers
12        5   111    2      HotLovers
13        6   110    1  DessertLovers
14        6   101    2  DessertLovers
15        6   114    2  DessertLovers
16        7   104    2      ManyJuice

---

微基准

pd.show_versions()

commit           : None
python           : 3.7.5.final.0
python-bits      : 64
OS               : Linux
OS-release       : 4.4.0-18362-Microsoft
machine          : x86_64
processor        : x86_64
byteorder        : little
LC_ALL           : None
LANG             : C.UTF-8
LOCALE           : en_US.UTF-8

pandas           : 0.25.3
numpy            : 1.17.4

數據是用

def make_data(n):
     return pd.DataFrame({
    'invoice':np.arange(n)//3,
    'code':np.random.choice(np.arange(101,112),n),
    'qty':np.random.choice(np.arange(1,8), n, p=[10/25,10/25,1/25,1/25,1/25,1/25,1/25])
})

結果

perfplot.show(
    setup=make_data,
    kernels=[get_category, get_with_np_select],
    n_range=[2**k for k in range(8, 20)],
    logx=True,
    logy=True,
    equality_check=False,
    xlabel='len(df)')