英文:
Interpretation of cosine similarity and jaccard similarity (similarity of histograms)
问题
以下是翻译好的内容：
简介
我想要评估两个“bin counts”数组（与两个直方图相关）之间的相似性，使用Matlab的“pdist2”函数：
% 输入bin_counts_a = [689 430 311 135 66 67 99 23 37 19 8 4 3 4 1 3 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1];bin_counts_b = [569 402 200 166 262 90 50 16 33 12 6 35 49 4 12 8 8 2 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 1];% 将两个“bin counts”向量可视化为条形图：bar(1:length(bin_counts_a),[bin_counts_a;bin_counts_b])
% 相似性计算cosine_similarity = 1 - pdist2(bin_counts_a,bin_counts_b,'cosine')jaccard_similarity = 1 - pdist2(bin_counts_a,bin_counts_b,'jaccard')% 输出cosine_similarity =0.95473215802008jaccard_similarity =0.0769230769230769
问题
如果余弦相似度接近1，这意味着两个向量相似，那么杰卡德相似度不应该也接近1吗？
英文:
Introduction
I would like to assess the similarity between two "bin counts" arrays (related to two histograms), by using the Matlab "pdist2" function:
% Inputbin_counts_a = [689 430 311 135 66 67 99 23 37 19 8 4 3 4 1 3 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 1 0 0 0 0 1];bin_counts_b = [569 402 200 166 262 90 50 16 33 12 6 35 49 4 12 8 8 2 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 2 0 0 0 0 0 0 1];% Visualize the two "bin counts" vectors as bars:bar(1:length(bin_counts_a),[bin_counts_a;bin_counts_b])
% Calculation of similaritiescosine_similarity = 1 - pdist2(bin_counts_a,bin_counts_b,'cosine')jaccard_similarity = 1 - pdist2(bin_counts_a,bin_counts_b,'jaccard')% Outputcosine_similarity =0.95473215802008jaccard_similarity =0.0769230769230769
Question
If the cosine similarity is close to 1, which means the two vectors are similar, shouldn't the jaccard similarity be closer to 1 as well?
答案1
得分: 3
"jaccard" 度量，根据文档，仅考虑"不同的非零坐标的百分比"，但不考虑它们的差异程度。
例如，假设 bin_counts_a 如您的示例所示，以及
bin_counts_b = bin_counts_a + 1;
然后
>> cosine_similarity = 1 - pdist2(bin_counts_a,bin_counts_b,'cosine')cosine_similarity =0.999971577948095
几乎等于 1，因为箱计数非常相似。然而，
>> jaccard_similarity = 1 - pdist2(bin_counts_a,bin_counts_b,'jaccard')jaccard_similarity =0
返回 0，因为bin_counts_b 中的每个条目与bin_counts_a 中的条目(略微)不同。
对于评估直方图之间的相似性，'cosine' 可能是比 'jaccard' 更有意义的选项。您还可以考虑库尔巴克-莱布勒散度，尽管它不对称于两个分布，并且不是由 pdist2 计算的。
英文:
The 'jaccard' measure, according to the documentation, only considers the "percentage of nonzero coordinates that differ", but not by how much they differ.
For instance, assume bin_counts_a as in your example and
bin_counts_b = bin_counts_a + 1;
Then
>> cosine_similarity = 1 - pdist2(bin_counts_a,bin_counts_b,'cosine')cosine_similarity =0.999971577948095
is almost 1 as expected, because the bin counts are very similar. However,
>> jaccard_similarity = 1 - pdist2(bin_counts_a,bin_counts_b,'jaccard')jaccard_similarity =0
gives 0 because each entry in bin_counts_b is (slightly) different from that in bin_counts_a.
For assessing the similarity between the histograms, 'cosine' is probably a more meaningful option than 'jaccard'. You may also want to consider the Kullback-Leibler divergence, although it is not symmetric in the two distributions, and is not computed by pdist2.
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