从 groupby 中每个组的上一列获取值

Question

这是我的df -

Site	Product	Period	Inflow	Outflow	Production	Opening Inventory	New Opening Inventory	Closing Inventory	Production Needed
California	Apples	1	0	3226	4300	1213	1213	0	0
California	Apples	2	0	3279	3876	0	0	0	0
California	Apples	3	0	4390	4530	0	0	0	0
California	Apples	4	0	4281	3870	0	0	0	0
California	Apples	5	0	4421	4393	0	0	0	0
California	Oranges	1	0	505	400	0	0	0	0
California	Oranges	2	0	278	505	0	0	0	0
California	Oranges	3	0	167	278	0	0	0	0
California	Oranges	4	0	124	167	0	0	0	0
California	Oranges	5	0	106	124	0	0	0	0
Montreal	Maple Syrup	1	0	445	465	293	293	0	0
Montreal	Maple Syrup	2	0	82	398	0	0	0	0
Montreal	Maple Syrup	3	0	745	346	0	0	0	0
Montreal	Maple Syrup	4	0	241	363	0	0	0	0
Montreal	Maple Syrup	5	0	189	254	0	0	0	0

如图所示，按Site 和Product 分组时，共有三个组。对于三个组中的每一个，我都想执行以下操作（第 2 到第 5 阶段）-

• 将New Opening Inventory设置为上一期的Closing Inventory
• 使用公式计算下一个周期的Closing Inventory，Closing Inventory = Production + Inflow + New Opening Inventory - Outflow

我正在尝试使用groupby 和for loop 的组合来解决这个问题

这是我目前所拥有的 -

如果df 是一个单独的组，我可以简单地做

# calculate closing inventory of period 1
df['Closing Inventory'] = np.where(df['PeriodNo']==1, <formula>, 0)

for i in range(1, len(df)):
    df.loc[i, 'New Opening Inventory'] = df.loc[i-1, 'Closing Inventory']
    df.loc[i, 'Closing Inventory'] = df.loc[i, 'Production'] + df.loc[i, 'Inflow'] + df.loc[i, 'New Opening Inventory'] - df.loc[i, 'Outflow']

当我尝试将此for loop 嵌套在groups 上的循环中时

# calculate closing inventory of period 1 for all groups
df['Closing Inventory'] = np.where(df['PeriodNo']==1, <formula>, 0)

g = df.groupby(['Site', 'Product']

alist = []

for k in g.groups.keys():
    temp = g.get_group(k).reset_index(drop=True)
    for i in range(1, len(temp)):
        temp.loc[i, 'New Opening Inventory'] = temp.loc[i-1, 'Closing Inventory']
        temp.loc[i, 'Closing Inventory'] = temp.loc[i, 'Production'] + temp.loc[i, 'Inflow'] + temp.loc[i, 'New Opening Inventory'] - temp.loc[i, 'Outflow']
    alist.append(temp)

df2 = pd.concat(alist, ignore_index=True)
df2

此解决方案有效，但使用嵌套循环似乎非常低效。有没有更好的方法来做到这一点？

Answer 1

您的新期初库存始终是之前的期末库存。

所以我可以修改这个公式

期末库存 = 生产 + 流入 + 新期初库存 - 流出

到

期末库存 = 生产 + 流入 + 以前的期末库存 - 流出

对于第一行，您没有期末库存。但是从第 2 行开始，您计算期末库存并将期末库存结转到下一行。

在获取Closing Inventory之前，首先使用列表推导计算“Production” + “Inflow” - “Overflow”。列表推导比 for 循环执行得更好。

df['Closing Inventory'] = [x + y - z if p > 1 else 0 for p, x, y, z in zip(df['Period'], df['Production'], df['Inflow'], df['Outflow'])]

# df[['Site', 'Product', 'Closing Inventory']]
#         Site  Product Closing Inventory
# 0 California  Apples                  0
# 1 California  Apples                597
# 2 California  Apples                140
# 3 California  Apples               -411
# 4 California  Apples                -28
# 5 California  Oranges                 0
# 6 California  Oranges               227
# 7 California  Oranges               111
# ...

然后，公式的其余部分是添加之前计算的期末库存，这意味着您可以cumsum 这个结果。

For row 1, Previous Closing (0) + calculated part (597) = 597
For row 2, Previous Closing (597) + calculated part (140) = 737
...

df['Closing Inventory'] = df.groupby(['Site', 'Product'])['Closing Inventory'].cumsum()

# df[['Site', 'Product', 'Closing Inventory']]
#         Site  Product Closing_Inventory
# 0 California  Apples                  0
# 1 California  Apples                597
# 2 California  Apples                737
# 3 California  Apples                326
# 4 California  Apples                298
# 5 California  Oranges                 0
# 6 California  Oranges               227
# 7 California  Oranges               338
# ...

同样，新的期初库存始终是之前的期末库存，除非期间为 1。因此，首先，移动期末库存，然后在期间为 1 时选择新的期初库存。

我使用combine_first 从新的期初或期末库存中挑选价值。

df['New Opening Inventory'] = (df['New Opening Inventory'].replace(0, np.nan)
                               .combine_first(
                                   df.groupby(['Site', 'Product'])['Closing Inventory']
                                   .shift()
                                   .fillna(0)
                               ).astype(int))

结果

          Site  Product Period  New Opening Inventory Closing Inventory
0   California  Apples       1                   1213                 0
1   California  Apples       2                      0               597
2   California  Apples       3                    597               737
3   California  Apples       4                    737               326
4   California  Apples       5                    326               298
5   California  Oranges      1                      0                 0
6   California  Oranges      2                      0               227
7   California  Oranges      3                    227               338
...

使用我笔记本电脑上的示例数据，

Original solution: 8.44 ms ± 280 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)
This solution: 2.95 ms ± 23.7 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

我认为通过列表理解和矢量化功能，这个解决方案可以执行得更快。