Pandas 列随时间的差异

Question

**在底部编辑**

我有一个包含库存数据的数据框，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

我想知道每个月的库存差异是多少。输出如下所示：

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

抱歉，如果我在第一个示例中不清楚，实际上我有很多个月的数据，所以我不只是在研究一个时期与另一个时期的差异。它需要是一个一般情况，它查看月份 n 与 n-1 的差异，然后是 n-1 和 n-2 等等。

在 Pandas 中最好的方法是什么？

Answer 1

我猜这里的关键是找到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

Answer 2

您可以尝试groupby 列产品和diff 列“差异”列的金额。然后将duplicated 用于“isnew”列。

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