我有以下数据:
物体 | l2a | l2b | l4 | l5 |
---|---|---|---|---|
a | 0.6649 | 0.5916 | 0.033569 | 0.557373 |
b | 0.8421 | 0.5132 | 0.000000 | 0.697193 |
c | 0.6140 | 0.2807 | 0.084217 | 0.650313 |
d | 0.7619 | 0.3810 | 0.000000 | 0.662306 |
e | 0.6957 | 0.3043 | 0.000000 | 0.645135 |
是否可以使用 RMSE 来测量 (a-b)、(a-c)、(a-d)、(a-e)、(b-c)、...、(d,e) 之间的相似度?
例如:
对象a(_a)和对象b(_b)之间的相似性:
diff_l2a = l2a_a - l2a_b
diff_l2b = l2b_a - l2b_b
diff_l4 = l4_a - l4_b
diff_l5 = l5_a - l5_b
然后计算RMSE:
RMSEs = [RMSE(diff_l2a, diff_l2b), RMSE(diff_l2a, diff_l4), RMSE(diff_l2a, diff_l5), ..., RMSE(diff_l4, diff_l5)]
相似之处:
average(RMSEs)
RMSE 相似度 DF 代码:
import pandas as pd
import numpy as np
from sklearn.metrics import mean_squared_error
num_objects = len(df)
similarity_matrix = np.zeros((num_objects, num_objects))
for i in range(num_objects):
for j in range(i + 1, num_objects):
rmse = np.sqrt(mean_squared_error(attributes[i], attributes[j]))
similarity_matrix[i, j] = rmse
similarity_matrix[j, i] = rmse
similarity_df = pd.DataFrame(similarity_matrix, columns=df['object'], index=df['object'])
我现在正在实现测试代码..