In [10]: df
Out[10]:
PART AVAILABLE_INVENTORY DEMAND
1 A 12 6
2 A 12 2
3 A 12 1
4 B 24 1
5 B 24 1
6 B 24 4
7 B 24 3
想要的输出:
PART AVAILABLE_INVENTORY DEMAND AI AI_AFTER
1 A 12 6 12 6
2 A 12 2 6 4
3 A 12 1 4 3
4 B 24 1 24 23
5 B 24 1 23 22
6 B 24 4 22 18
7 B 24 3 18 15
到目前为止,我拥有的代码如下,但它没有给出我正在寻找的输出:
def retain(df):
df['PREV_PART'] = df['PART'].shift()
df['PREV_AI_AFTER'] = df['AI'].shift() - df['DEMAND'].shift()
df['AI'] = np.where(df['PART'] != df['PREV_PART'], df['AI'], df['PREV_AI_AFTER'])
df['AI_AFTER'] = df['AI'] - df['DEMAND']
df['AI'] = df['AVAILABLE_INVENTORY']
retain(df)
在考虑性能的情况下,最快的方法是什么?
发布于 2020-04-22 00:32:35
您可以使用groupby
来完成此操作,其中cumsum
位于'DEMAND‘列,shift
位于前面创建的'AI_AFTER’列:
df['AI_AFTER'] = df['AVAILABLE_INVENTORY'] - df.groupby('PART')['DEMAND'].cumsum()
df['AI'] = df.groupby('PART')['AI_AFTER'].shift().fillna(df['AVAILABLE_INVENTORY'])
print (df)
PART AVAILABLE_INVENTORY DEMAND AI_AFTER AI
1 A 12 6 6 12.0
2 A 12 2 4 6.0
3 A 12 1 3 4.0
4 B 24 1 23 24.0
5 B 24 1 22 23.0
6 B 24 4 18 22.0
7 B 24 3 15 18.0
发布于 2020-04-22 00:47:06
VERRRY类似于Ben.T's Answer。如果你喜欢这种方法,请选择他们的答案。这就是我安排这个过程的方式。
def f(d):
i = d.AVAILABLE_INVENTORY
c = d.DEMAND.cumsum()
return pd.concat({'AI': i - c.shift(fill_value=0), 'AI_AFTER': i - c}, axis=1)
df.join(df.groupby('PART').apply(f))
PART AVAILABLE_INVENTORY DEMAND AI AI_AFTER
1 A 12 6 12 6
2 A 12 2 6 4
3 A 12 1 4 3
4 B 24 1 24 23
5 B 24 1 23 22
6 B 24 4 22 18
7 B 24 3 18 15
https://stackoverflow.com/questions/61348293
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