使用 RMSE 测量相似度

Question

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)

Answer 1

RMSE 相似度 DF 代码部分：

num_objects = len(df)
sim_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]))
        sim_matrix[i, j] = rmse
        sim_matrix[j, i] = rmse

代码（带DF）：

import pandas as pd
import numpy as np
from sklearn.metrics import mean_squared_error

data = {
    'object': ['a', 'b', 'c', 'd', 'e'],
    'l2a': [0.6649, 0.8421, 0.6140, 0.7619, 0.6957],
    'l2b': [0.5916, 0.5132, 0.2807, 0.3810, 0.3043],
    'l4': [0.033569, 0.0, 0.084217, 0.0, 0.0],
    'l5': [0.557373, 0.697193, 0.650313, 0.662306, 0.645135]
}
df = pd.DataFrame(data)
attributes = df.iloc[:, 1:].values

num_objects = len(df)
sim_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]))
        sim_matrix[i, j] = rmse
        sim_matrix[j, i] = rmse

sim_df = pd.DataFrame(sim_matrix, columns=df['object'], index=df['object'])

print("Similarity Matrix:")
print(sim_df)

sim = sim_df.values[sim_df.values != 0.0]
average_sim = sim.mean()
print(f"Average Similarity (excluding 0.0): {average_sim:.3f}")

输出：

补充：

如果您想计算基于成对 RMSE 的相似度：

from scipy.spatial.distance import pdist, squareform
sim_matrix = np.sqrt(squareform(pdist(attributes, 'euclidean')))

其他： https://docs.scipy.org/doc/scipy/reference/ generated/scipy.spatial.distance.pdist.html