熊猫栏随时间的不同

Question

**底部编辑**

我有一个库存数据的数据框架，如下所示：

d = {'product': [a, b, a, b, c], 'amount': [1, 2, 3, 5, 2], 'date': [2020-6-6, 2020-6-6, 2020-6-7, 
2020-6-7, 2020-6-7]}
df = pd.DataFrame(data=d)
df
 product  amount  date
0     a     1      2020-6-6
1     b     2      2020-6-6
2     a     3      2020-6-7
3     b     5      2020-6-7
4     c     2      2020-6-7

我想知道每月的存货差额是多少。输出将如下所示：

df
 product   diff   isnew  date
0     a     nan   nan   2020-6-6
1     b     nan   nan   2020-6-6
2     a     2     False 2020-6-7
3     b     3     False 2020-6-7
4     c     2     True  2020-6-7

不好意思，如果我在第一个例子中不清楚，实际上我有很多个月的数据，所以我不只是在做一个周期与另一个阶段的区别。它需要是一个一般的情况，它查看月份n对n-1，然后n-1和n-2的差值，等等。

在潘达斯做这件事最好的方法是什么？

Answer 1

您可以在列产品上尝试groupby，并为列'diff‘设置diff列数量。然后对列'isnew‘使用duplicated。

df['diff'] = df.groupby('product')['amount'].diff()
df['isnew'] = ~df['product'].duplicated()
print (df)
  product  amount      date  diff  isnew
0       a       1  2020-6-6   NaN   True
1       b       2  2020-6-6   NaN   True
2       a       3  2020-6-7   2.0  False
3       b       5  2020-6-7   3.0  False
4       c       2  2020-6-7   NaN   True

Answer 2

我想这里的关键是找到isnew

# new products by `product`
new_prods = df['date'] != df.date.min()
duplicated = df.duplicated('product')

# first appearance of new products
# or duplicated *old* products
valids = new_prods ^ duplicated
df.loc[valids,'is_new'] = ~ duplicated

# then the difference:
df['diff'] = (df.groupby('product')['amount'].diff()           # normal differences
                  .fillna(df['amount'])         # fill the first value for all product
                  .where(df['is_new'].notna())  # remove the first month
             )

输出：

  product  amount      date is_new  diff
0       a       1  2020-6-6    NaN   NaN
1       b       2  2020-6-6    NaN   NaN
2       a       3  2020-6-7  False   2.0
3       b       5  2020-6-7  False   3.0
4       c       2  2020-6-7   True   2.0