Numpy - 聚类 - 距离 - 向量化

Question

我使用 sklearn Kmeans 对数据样本（400 k 样本，维度 = 205、200 个集群）进行了聚类。

我想知道，对于每个集群，集群中心与集群最远样本之间的最大距离，以便了解集群的“大小”。 这是我的代码：

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”的第一列包含集群内的样本数（此处无关）。

“数据”是样本（400kx206）。数据的第一列包含集群编号。

问题：这是使用循环完成的（不是那么“numpy”）。是否可以“矢量化”它？

Answer 1

我们可以更智能地编制索引并节省约 4 倍的成本。

首先让我们构建一些正确形状的数据：

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)

原答案时间：

%%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：

%%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

迪瓦卡的回答：

%%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

正如我们所见，除了更大的内存占用之外，矢量化解决方案并没有真正获得任何好处。为了避免这种情况，我们需要回到原来的答案，这确实是做这些事情的正确方法，而是尝试减少索引量：

%%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。它们通常是，但并非在所有情况下都如此。

Answer 2

这是一种矢量化方法 -

# 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)

运行时测试和验证输出 -

定义函数：

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 3

与@Divakar 非常相似的想法，但无需排序：

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 虽然速度很慢，所以看看哪个更快会很有趣。