Numpy - Clustering - Distance - Vectorisation

Question

我用sklearn Kmeans聚类了一个数据样本（400 k个样本，维数= 205,200个星团）。

对于每个群集，我想知道群集中心与群集最远的样本之间的最大距离，以便了解有关群集“大小”的信息。 这里是我的代码：

import numpy as np 
import scipy.spatial.distance as spd 
diam = np.empty([200]) 
for i in range(200): 
    diam[i] = spd.cdist(seed[np.newaxis, i, 1:], data[data[:, 0]==i][:,1:]).max() 

“种子” 是聚类中心（200x206）。 “种子”的第一列包含群集内的样本数量（此处不相关）。

“数据”是样本（400kx206）。数据的第一列包含集群编号。

问题：这是使用循环（不是“numpy”）完成的。是否有可能“引导”它？

Answer 1

这里有一个量化的方法 -

# Sort data w.r.t. col-0 
data_sorted = data[data[:, 0].argsort()] 

# Get counts of unique tags in col-0 of data and repeat seed accordingly. 
# Thus, we would have an extended version of seed that matches data's shape. 
seed_ext = np.repeat(seed,np.bincount(data_sorted[:,0]),axis=0) 

# Get euclidean distances between extended seed version and sorted data 
dists = np.sqrt(((data_sorted[:,1:] - seed_ext[:,1:])**2).sum(1)) 

# Get positions of shifts in col-0 of sorted data 
shift_idx = np.append(0,np.nonzero(np.diff(data_sorted[:,0]))[0]+1) 

# Final piece of puzzle is to get tag based maximum values from dists, 
# where each tag is unique number in col-0 of data 
diam_out = np.maximum.reduceat(dists,shift_idx) 

运行测试和验证输出 -

定义功能：

def loopy_cdist(seed,data): # from OP's solution 
    N = seed.shape[0] 
    diam = np.empty(N) 
    for i in range(N): 
     diam[i]=spd.cdist(seed[np.newaxis,i,1:], data[data[:,0]==i][:,1:]).max() 
    return diam 

def vectorized_repeat_reduceat(seed,data): # from this solution 
    data_sorted = data[data[:, 0].argsort()] 
    seed_ext = np.repeat(seed,np.bincount(data_sorted[:,0]),axis=0) 
    dists = np.sqrt(((data_sorted[:,1:] - seed_ext[:,1:])**2).sum(1)) 
    shift_idx = np.append(0,np.nonzero(np.diff(data_sorted[:,0]))[0]+1) 
    return np.maximum.reduceat(dists,shift_idx) 

def vectorized_indexing_maxat(seed,data): # from @moarningsun's solution 
    seed_repeated = seed[data[:,0]] 
    dist_to_center = np.sqrt(np.sum((data[:,1:]-seed_repeated[:,1:])**2, axis=1)) 
    diam = np.zeros(len(seed)) 
    np.maximum.at(diam, data[:,0], dist_to_center) 
    return diam 

验证输出：

In [417]: # Inputs 
    ...: seed = np.random.rand(20,20) 
    ...: data = np.random.randint(0,20,(40000,20)) 
    ...: 

In [418]: np.allclose(loopy_cdist(seed,data),vectorized_repeat_reduceat(seed,data)) 
Out[418]: True 

In [419]: np.allclose(loopy_cdist(seed,data),vectorized_indexing_maxat(seed,data)) 
Out[419]: True 

运行时：

In [420]: %timeit loopy_cdist(seed,data) 
10 loops, best of 3: 35.9 ms per loop 

In [421]: %timeit vectorized_repeat_reduceat(seed,data) 
10 loops, best of 3: 28.9 ms per loop 

In [422]: %timeit vectorized_indexing_maxat(seed,data) 
10 loops, best of 3: 24.1 ms per loop 

Answer 2

非常相似想法@Divakar，但不必进行排序：

seed_repeated = seed[data[:,0]] 
dist_to_center = np.sqrt(np.sum((data[:,1:]-seed_repeated[:,1:])**2, axis=1)) 

diam = np.zeros(len(seed)) 
np.maximum.at(diam, data[:,0], dist_to_center) 

ufunc.at被称为是缓慢的，所以这将是有趣的，看看这是更快。

Answer 3

我们可以在索引方面稍微聪明些，节省大约4倍的成本。

首先让建立正确的形状的一些数据：

seed = np.random.randint(0, 100, (200,206)) 
data = np.random.randint(0, 100, (4e5,206)) 
seed[:, 0] = np.arange(200) 
data[:, 0] = np.random.randint(0, 200, 4e5) 
diam = np.empty(200) 

原答复的时间：

%%timeit 
for i in range(200): 
    diam[i] = spd.cdist(seed[np.newaxis, i, 1:], data[data[:, 0]==i][:,1:]).max() 

1 loops, best of 3: 1.35 s per loop 

moarningsun的回答是：

%%timeit 
seed_repeated = seed[data[:,0]] 
dist_to_center = np.sqrt(np.sum((data[:,1:]-seed_repeated[:,1:])**2, axis=1)) 
diam = np.zeros(len(seed)) 
np.maximum.at(diam, data[:,0], dist_to_center) 

1 loops, best of 3: 1.33 s per loop 

Divakar的回答是：

%%timeit 
data_sorted = data[data[:, 0].argsort()] 
seed_ext = np.repeat(seed,np.bincount(data_sorted[:,0]),axis=0) 
dists = np.sqrt(((data_sorted[:,1:] - seed_ext[:,1:])**2).sum(1)) 
shift_idx = np.append(0,np.nonzero(np.diff(data_sorted[:,0]))[0]+1) 
diam_out = np.maximum.reduceat(dists,shift_idx) 

1 loops, best of 3: 1.65 s per loop 

正如我们所看到的，除了更大的内存占用之外，还没有真正获得任何矢量化解决方案。为了避免这种情况，我们需要返回到原来的答案，这是真的做这些事情的正确方法，而是试图减少索引量：

%%timeit 
idx = data[:,0].argsort() 
bins = np.bincount(data[:,0]) 
counter = 0 
for i in range(200): 
    data_slice = idx[counter: counter+bins[i]] 
    diam[i] = spd.cdist(seed[None, i, 1:], data[data_slice, 1:]).max() 
    counter += bins[i] 

1 loops, best of 3: 281 ms per loop 

仔细检查答案：

np.allclose(diam, dam_out) 
True 

这是假设python循环不好的问题。他们往往是，但不是在所有情况下。