python - 将 pandas 移动窗口与列表进行比较，以找到错误最少的窗口

Question

我已将我的数据集缩减到最后几个步骤。我的 Pandas 数据框看起来像这样

    FAC
0   1
1   2
2   1
3   3
4   2
5   1
6   2
7   1
8   1
9   3
10  2
11  1
12  2
13  3
14  1

我还有一个我确定匹配的列表。

match_list = [1, 2, 1, 1, 3]

我正在寻找的是滑过(5 项窗口)数据框列并找到与列表模式匹配的行。最终结果看起来像这样。如果有任何帮助，我将不胜感激。

    FAC Error
0   1   some val
1   2   some val
2   1   some val
3   3   some val
4   2   some val
5   1   some val
6   2   some val
7   1   0
8   1   some val
9   3   some val
10  2   some val
11  1   some val
12  2   some val
13  3   some val
14  1   some val

Answer 1

这可以通过rolling来完成:

match_list = [1, 2, 1, 1, 3]
match_list = np.array(match_list)

def match(x):
    return (len(x)==len(match_list) and (x==match_list).all())


df['error'] = np.where(df.FAC.rolling(5, center=True).apply(match)==1, 0, 'some value')

输出:

    FAC       error
0     1  some value
1     2  some value
2     1  some value
3     3  some value
4     2  some value
5     1  some value
6     2  some value
7     1           0
8     1  some value
9     3  some value
10    2  some value
11    1  some value
12    2  some value
13    3  some value
14    1  some value

---

更新:要对匹配进行计数，您可以在函数内简单地执行 mean 而不是 all:

def count_match(x):
    return (len(x)==len(match_list))* (x==match_list).mean()

df['error'] = df.FAC.rolling(5,center=True).apply(count_match)

输出:

    FAC  error
0     1    NaN
1     2    NaN
2     1    0.6
3     3    0.0
4     2    0.4
5     1    0.4
6     2    0.2
7     1    1.0
8     1    0.2
9     3    0.2
10    2    0.4
11    1    0.6
12    2    0.0
13    3    NaN
14    1    NaN