如何根据熊猫数据框的条件增加计数器？

Question

我有一堆记录，每个记录都标有一个簇值。

原始数据框， df ：

+-------------+---------+
| measurement | cluster |
+-------------+---------+
| M1          |       6 |
| M2          |       6 |
| M3          |       6 |
| M4          |      12 |
| M5          |      12 |
| M6          |      12 |
| M7          |       2 |
| M8          |       9 |
| M9          |       9 |
| M10         |       9 |
| M11         |       9 |
+-------------+---------+

如何在分配给群集值不等于上一个或下一个的“ x”行时，基于当前群集的值是否等于上一个和下一个来重命名群集？ >

所需的 df ：

+-------------+---------+-------------+
| measurement | cluster | new_cluster |
+-------------+---------+-------------+
| M1          |       6 |           1 |
| M2          |       6 |           1 |
| M3          |       6 |           1 |
| M4          |      12 |           2 |
| M5          |      12 |           2 |
| M6          |      12 |           2 |
| M7          |       2 |           x |
| M8          |       9 |           3 |
| M9          |       9 |           3 |
| M10         |       9 |           3 |
| M11         |       9 |           3 |
+-------------+---------+-------------+

Answer 1

将pandas.factorize用于通过掩码过滤的值：

m = df['cluster'].ne(df['cluster'].shift()).cumsum().duplicated(keep=False)
df.loc[m, 'new_cluster'] =  pd.factorize(df.loc[m, 'cluster'])[0] + 1
print (df)
   measurement  cluster  new_cluster
0           M1        6          1.0
1           M2        6          1.0
2           M3        6          1.0
3           M4       12          2.0
4           M5       12          2.0
5           M6       12          2.0
6           M7        2          NaN
7           M8        9          3.0
8           M9        9          3.0
9          M10        9          3.0
10         M11        9          3.0

如果要将NaN替换为x：

df['new_cluster'] = df['new_cluster'].fillna('x')
print (df)
   measurement  cluster new_cluster
0           M1        6           1
1           M2        6           1
2           M3        6           1
3           M4       12           2
4           M5       12           2
5           M6       12           2
6           M7        2           x
7           M8        9           3
8           M9        9           3
9          M10        9           3
10         M11        9           3

布尔掩码的详细信息-首先为连续值创建辅助程序Series，然后使用keep ='False'用duplicated掩码以返回所有重复项：

print (df['cluster'].ne(df['cluster'].shift()).cumsum())
0     1
1     1
2     1
3     2
4     2
5     2
6     3
7     4
8     4
9     4
10    4
Name: cluster, dtype: int32

print (m)
0      True
1      True
2      True
3      True
4      True
5      True
6     False
7      True
8      True
9      True
10     True
Name: cluster, dtype: bool

Answer 2

设置

df = pd.DataFrame(dict(
    measurement=[f"M{i}" for i in range(1, 12)],
    cluster=[*map(int, '6 6 6 12 12 12 2 9 9 9 9'.split())]
))

---

解决方案

c = df.cluster.values
m0 = c[:-1] == c[1:]

# Test if equal to prior or next
mask = np.append(False, m0) | np.append(m0, False)

f, u = pd.factorize(c[mask])
new = np.array(list('x' * len(c)), dtype=object)
new[mask] = f + 1

df.assign(new_cluster=new)

   measurement  cluster new_cluster
0           M1        6           1
1           M2        6           1
2           M3        6           1
3           M4       12           2
4           M5       12           2
5           M6       12           2
6           M7        2           x
7           M8        9           3
8           M9        9           3
9          M10        9           3
10         M11        9           3