创建余弦相似度矩阵 numpy

Question

假设我有一个如下所示的 numpy 矩阵：

array([array([ 0.0072427 ,  0.00669255,  0.00785213,  0.00845336,  0.01042869]),
   array([ 0.00710799,  0.00668831,  0.00772334,  0.00777796,  0.01049965]),
   array([ 0.00741872,  0.00650899,  0.00772273,  0.00729002,  0.00919407]),
   array([ 0.00717589,  0.00627021,  0.0069514 ,  0.0079332 ,  0.01069545]),
   array([ 0.00617369,  0.00590539,  0.00738468,  0.00761699,  0.00886915])], dtype=object)

如何生成一个 5 x 5 矩阵，其中矩阵的每个索引都是原始矩阵中两个对应行的余弦相似度？

例如第 0 行第 2 列的值将是原始矩阵中第 1 行和第 3 行之间的余弦相似度。

这是我尝试过的方法：

from sklearn.metrics import pairwise_distances
from scipy.spatial.distance import cosine
import numpy as np

#features is a column in my artist_meta data frame
#where each value is a numpy array of 5 floating point values, similar to the
#form of the matrix referenced above but larger in volume

items_mat = np.array(artist_meta['features'].values)

dist_out = 1-pairwise_distances(items_mat, metric="cosine")

上面的代码给我以下错误：

ValueError: 使用序列设置数组元素。

不确定为什么我会得到这个，因为每个数组的长度相同 (5)，我已经验证过。

Answer 1

让x成为你的数组

from scipy.spatial.distance import cosine

m, n = x.shape
distances = np.zeros((m,n))
for i in range(m):
    for j in range(n):
        distances[i,j] = cosine(x[i,:],x[:,j])

Answer 2

设m为数组

m = np.array([
        [ 0.0072427 ,  0.00669255,  0.00785213,  0.00845336,  0.01042869],
        [ 0.00710799,  0.00668831,  0.00772334,  0.00777796,  0.01049965],
        [ 0.00741872,  0.00650899,  0.00772273,  0.00729002,  0.00919407],
        [ 0.00717589,  0.00627021,  0.0069514 ,  0.0079332 ,  0.01069545],
        [ 0.00617369,  0.00590539,  0.00738468,  0.00761699,  0.00886915]
    ])

per wikipedia: Cosine_Similarity

我们可以用

计算我们的分子

d = m.T @ m

我们的‖A‖是

norm = (m * m).sum(0, keepdims=True) ** .5

那么相同点是

d / norm / norm.T

[[ 1.      0.9994  0.9979  0.9973  0.9977]
 [ 0.9994  1.      0.9993  0.9985  0.9981]
 [ 0.9979  0.9993  1.      0.998   0.9958]
 [ 0.9973  0.9985  0.998   1.      0.9985]
 [ 0.9977  0.9981  0.9958  0.9985  1.    ]]

距离是

1 - d / norm / norm.T

[[ 0.      0.0006  0.0021  0.0027  0.0023]
 [ 0.0006  0.      0.0007  0.0015  0.0019]
 [ 0.0021  0.0007  0.      0.002   0.0042]
 [ 0.0027  0.0015  0.002   0.      0.0015]
 [ 0.0023  0.0019  0.0042  0.0015  0.    ]]

Answer 3

如前所述，您可以使用 sklearn 中的 pairwise 函数。这是一个完整的实现以及它与 sklearn 和 scipy 版本匹配的验证。我在这个例子中四舍五入到小数点后四位。

import numpy as np
from scipy.spatial.distance import cosine
from sklearn.metrics import pairwise_distances

def cosine_distance_matrix(column: pd.Series, decimals: int = 4):
    """
    Calculate cosine distance of column against itself (pairwise)
    
    Args:
        column:
            pandas series containing np.array values
        decimals:
            how many places to round the output
            
    Returns:
        distance matrix of shape (len(column), len(column))
    """
    M = np.vstack(column.values)
    
    # Perform division by magnitude of pairs first
    # M / (||A|| * ||B||)
    M_norm = M / np.sqrt(np.square(M).sum(1, keepdims=True))
    
    # Perform dot product
    similarity = M_norm @ M_norm.T
    
    # Convert from similarity to distance
    return (1 - similarity).round(decimals)

# Example for testing
sample_column = pd.Series([
    np.array([3, 4]),
    np.array([7, 24]),
    np.array([1, 1])
])

# Try our own fast implementation
custom_version = cosine_distance_matrix(sample_column, decimals=4)

# Use pairwise function from sklearn
pairwise_version = pairwise_distances(
    np.vstack(sample_column.values),
    metric="cosine"
).round(4)

# Equals pairwise version
assert (custom_version == pairwise_version).all()

# Check single element
assert custom_version[0, 1] == cosine(sample_column[0], sample_column[1]).round(4)

创建余弦相似度矩阵 numpy

create cosine similarity matrix numpy

python

numpy

matrix

cosine-similarity