python - 条件下的 DataFrameGroupBy diff()

Question

假设我有一个 DataFrame:

df = pd.DataFrame({'CATEGORY':['a','b','c','b','b','a','b'],
                   'VALUE':[pd.np.NaN,1,0,0,5,0,4]})

看起来像

    CATEGORY    VALUE
0      a         NaN
1      b         1
2      c         0
3      b         0
4      b         5
5      a         0
6      b         4

我把它分组:

df = df.groupby(by='CATEGORY')

现在，让我借助一组“b”上的示例展示我想要的东西:

df.get_group('b')

b组:

    CATEGORY    VALUE
1      b          1
3      b          0
4      b          5
6      b          4

我需要:在每组范围内，统计VALUE之间的diff()值，跳过所有 NaN s 和 0秒。所以结果应该是:

    CATEGORY    VALUE  DIFF
1      b          1      - 
3      b          0      -
4      b          5      4
6      b          4     -1

Answer 1

您可以使用 diff在删除 0 和 NaN 值后减去值:

df = pd.DataFrame({'CATEGORY':['a','b','c','b','b','a','b'],
               'VALUE':[pd.np.NaN,1,0,0,5,0,4]})

grouped = df.groupby("CATEGORY")

# define diff func
diff = lambda x: x["VALUE"].replace(0, np.NaN).dropna().diff()
df["DIFF"] = grouped.apply(diff).reset_index(0, drop=True)

print(df)

  CATEGORY  VALUE  DIFF
0        a    NaN   NaN
1        b    1.0   NaN
2        c    0.0   NaN
3        b    0.0   NaN
4        b    5.0   4.0
5        a    0.0   NaN
6        b    4.0  -1.0