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Python 带更新的滚动平均值_Python_Pandas_Dataframe - Fatal编程技术网

Python 带更新的滚动平均值

Python 带更新的滚动平均值,python,pandas,dataframe,Python,Pandas,Dataframe,以dataframe为例: df = pd.DataFrame({ "a": [None, None, None, None, 1, 2, -1, 0, 1], "b": [5, 4, 6, 7, None, None, None, None, None] }) >> a b 0 NaN 5.0 1 NaN 4.0 2 NaN 6.0 3 NaN 7.0 4 1.0 NaN 5 2.0 NaN 6 -1.0 NaN 7

以dataframe为例:

df = pd.DataFrame({
    "a": [None, None, None, None, 1, 2, -1, 0, 1],
    "b": [5, 4, 6, 7, None, None, None, None, None]
})

>>  a    b
0   NaN  5.0
1   NaN  4.0
2   NaN  6.0
3   NaN  7.0
4   1.0  NaN
5   2.0  NaN
6  -1.0  NaN
7   0.0  NaN
8   1.0  NaN
对于b中的每个缺失值,我想取前4个值的平均值加上a中具有相同索引的值。例如,在7之后:

4: (5   + 4 + 6 + 7) / 4 + 1 = 6.5
5: (6.5 + 4 + 6 + 7) / 4 + 2 = 7.88
   ...
结果数据帧应为:

>>  a    b
0   NaN  5.00
1   NaN  4.00
2   NaN  6.00
3   NaN  7.00
4   1.0  6.50
5   2.0  7.88
6  -1.0  5.84
7   0.0  6.80
8   1.0  7.76

如何实现这一点?

在这里使用for循环,panda不是按行的,他们不能在将来的计算中使用以前的计算值。(矢量化)

l=[]
对于zip中的x,y(*df.values.T.tolist()):

如果len(l)部分“加上a中具有相同索引的值”,则没有真正意义。我看不到a的值在平均值中,但它在平均值中denominator@d_kennetz这不是分母。我明白了,这是最后一个4+值的平均值!你可以查一下你的名字。。。犯错误W-B.WenB+1.
l=[]
for x ,y in zip(*df.values.T.tolist()):
    if len(l)<4:
        l.append(y)
    else:
        l.append(sum(l[-4:])/4+x)

l
Out[188]: [5.0, 4.0, 6.0, 7.0, 6.5, 7.875, 5.84375, 6.8046875, 7.755859375]
df.b=l
df
Out[190]: 
     a         b
0  NaN  5.000000
1  NaN  4.000000
2  NaN  6.000000
3  NaN  7.000000
4  1.0  6.500000
5  2.0  7.875000
6 -1.0  5.843750
7  0.0  6.804688
8  1.0  7.755859