在Python中计算Jaccard相似度

Question

我有20,000个文档,我想计算真正的Jaccard相似度,以便稍后我可以检查MinWise散列的准确度是近似的.

每个文档都表示为numpy矩阵中的一列,其中每一行都是出现在document(entry = 1)或不出现(entry = 0)的单词.有大约600个单词(行).

因此,例如,列1将是[1 0 0 0 0 0 1 0 0 0 1 0],这意味着在其中出现单词1,7,11而没有其他单词.

除了我的元素比较方法之外,还有更有效的方法来计算相似性吗？我不知道如何使用集合来提高速度,因为集合刚刚变为(0,1),但是现在代码的速度非常慢.

import numpy as np

#load file into python
rawdata = np.loadtxt("myfile.csv",delimiter="\t")
#Convert the documents from rows to columns
rawdata = np.transpose(rawdata)
#compute true jacard similarity
ndocs = rawdata.shape[1]
nwords = rawdata.shape[0]
tru_sim = np.zeros((ndocs,ndocs))

#computes jaccard similarity of 2 documents
def jaccard(c1, c2):
    n11 = sum((c1==1)&(c2==1))
    n00 = sum((c1==0)&(c2==0))
    jac = n11 / (nfeats-n00)
    return (jac)

for i in range(0,ndocs):
    tru_sim[i,i]=1
    for j in range(i+1,ndocs):
        tru_sim[i,j] = jaccard(rawdata[:,i],rawdata[:,j])

Answer 1

这是一种矢量化方法 -

# Get the row, col indices that are to be set in output array        
r,c = np.tril_indices(ndocs,-1)

# Use those indicees to slice out respective columns 
p1 = rawdata[:,c]
p2 = rawdata[:,r]

# Perform n11 and n00 vectorized computations across all indexed columns
n11v = ((p1==1) & (p2==1)).sum(0)
n00v = ((p1==0) & (p2==0)).sum(0)

# Finally, setup output array and set final division computations
out = np.eye(ndocs)
out[c,r] = n11v / (nfeats-n00v)

计算n11v和n00v使用的替代方法np.einsum-

n11v = np.einsum('ij,ij->j',(p1==1),(p2==1).astype(int))
n00v = np.einsum('ij,ij->j',(p1==0),(p2==0).astype(int))

如果rawdata由0s且1s仅组成，获得它们的更简单方法是 -

n11v = np.einsum('ij,ij->j',p1,p2)
n00v = np.einsum('ij,ij->j',1-p1,1-p2)

基准测试

函数定义 -

def original_app(rawdata, ndocs, nfeats):
    tru_sim = np.zeros((ndocs,ndocs))
    for i in range(0,ndocs):
        tru_sim[i,i]=1
        for j in range(i+1,ndocs):
            tru_sim[i,j] = jaccard(rawdata[:,i],rawdata[:,j])
    return tru_sim

def vectorized_app(rawdata, ndocs, nfeats):
    r,c = np.tril_indices(ndocs,-1)
    p1 = rawdata[:,c]
    p2 = rawdata[:,r]
    n11v = ((p1==1) & (p2==1)).sum(0)
    n00v = ((p1==0) & (p2==0)).sum(0)
    out = np.eye(ndocs)
    out[c,r] = n11v / (nfeats-n00v)
    return out

验证和时间 -

In [6]: # Setup inputs
   ...: rawdata = (np.random.rand(20,10000)>0.2).astype(int)
   ...: rawdata = np.transpose(rawdata)
   ...: ndocs = rawdata.shape[1]
   ...: nwords = rawdata.shape[0]
   ...: nfeats = 5
   ...: 

In [7]: # Verify results
   ...: out1 = original_app(rawdata, ndocs, nfeats)
   ...: out2 = vectorized_app(rawdata, ndocs, nfeats)
   ...: print np.allclose(out1,out2)
   ...: 
True

In [8]: %timeit original_app(rawdata, ndocs, nfeats)
1 loops, best of 3: 8.72 s per loop

In [9]: %timeit vectorized_app(rawdata, ndocs, nfeats)
10 loops, best of 3: 27.6 ms per loop

300x+那里有一些神奇的加速！

那么，为什么这么快呢？嗯，涉及很多因素，最重要的一个因素是 NumPy 数组是为性能而构建的，并针对矢量化计算进行了优化。使用提议的方法，我们很好地利用了它，因此看到了这样的加速。

这是related Q&A详细讨论这些性能标准的一个。