21

使用内置梯度下降优化器的介绍教程很有意义。然而，k-means 不仅仅是我可以插入梯度下降的东西。似乎我必须编写自己的优化器，但鉴于 TensorFlow 原语，我不太确定如何做到这一点。

我应该采取什么方法？

4

3 回答 3

27

（注意：您现在可以在 github 上获得此代码的更完善的版本作为要点。）

你绝对可以做到，但你需要定义自己的优化标准（对于 k-means，它通常是最大迭代次数和分配稳定的时间）。这是一个示例，说明您可以如何做到这一点（可能有更优化的方法来实现它，当然还有更好的方法来选择初始点）。如果你真的很努力地远离在 python 中迭代地做事情，那么基本上就像你会在 numpy 中做这件事：

import tensorflow as tf
import numpy as np
import time

N=10000
K=4
MAX_ITERS = 1000

start = time.time()

points = tf.Variable(tf.random_uniform([N,2]))
cluster_assignments = tf.Variable(tf.zeros([N], dtype=tf.int64))

# Silly initialization:  Use the first two points as the starting                
# centroids.  In the real world, do this better.                                 
centroids = tf.Variable(tf.slice(points.initialized_value(), [0,0], [K,2]))

# Replicate to N copies of each centroid and K copies of each                    
# point, then subtract and compute the sum of squared distances.                 
rep_centroids = tf.reshape(tf.tile(centroids, [N, 1]), [N, K, 2])
rep_points = tf.reshape(tf.tile(points, [1, K]), [N, K, 2])
sum_squares = tf.reduce_sum(tf.square(rep_points - rep_centroids),
                            reduction_indices=2)

# Use argmin to select the lowest-distance point                                 
best_centroids = tf.argmin(sum_squares, 1)
did_assignments_change = tf.reduce_any(tf.not_equal(best_centroids,
                                                    cluster_assignments))

def bucket_mean(data, bucket_ids, num_buckets):
    total = tf.unsorted_segment_sum(data, bucket_ids, num_buckets)
    count = tf.unsorted_segment_sum(tf.ones_like(data), bucket_ids, num_buckets)
    return total / count

means = bucket_mean(points, best_centroids, K)

# Do not write to the assigned clusters variable until after                     
# computing whether the assignments have changed - hence with_dependencies
with tf.control_dependencies([did_assignments_change]):
    do_updates = tf.group(
        centroids.assign(means),
        cluster_assignments.assign(best_centroids))

sess = tf.Session()
sess.run(tf.initialize_all_variables())

changed = True
iters = 0

while changed and iters < MAX_ITERS:
    iters += 1
    [changed, _] = sess.run([did_assignments_change, do_updates])

[centers, assignments] = sess.run([centroids, cluster_assignments])
end = time.time()
print ("Found in %.2f seconds" % (end-start)), iters, "iterations"
print "Centroids:"
print centers
print "Cluster assignments:", assignments

（请注意，真正的实现需要更加小心初始集群选择，避免所有点都指向一个集群的问题情况等。这只是一个快速演示。我已经更新了我之前的答案以使其有点更清晰和“值得示范”。）

于 2015-11-10T19:04:10.410 回答
4

到目前为止，我看到的大多数答案都只关注 2d 版本（当您需要在 2 维中聚类点时）。这是我在任意维度上的聚类实现。

n dims中k-means 算法的基本思想：

  • 生成随机k个起点
  • 这样做直到你超过耐心或集群分配没有改变：
    • 将每个点分配给最近的起点
    • 通过取其集群中的平均值来重新计算每个起点的位置

为了能够以某种方式验证结果，我将尝试对 MNIST 图像进行聚类。

import numpy as np
import tensorflow as tf
from random import randint
from collections import Counter
from tensorflow.examples.tutorials.mnist import input_data

mnist = input_data.read_data_sets("MNIST_data/")
X, y, k = mnist.test.images, mnist.test.labels, 10

所以这里X是我要聚类的数据(10000, 784)，y是实数，k是聚类数（与位数相同。现在实际算法：

# select random points as a starting position. You can do better by randomly selecting k points.
start_pos = tf.Variable(X[np.random.randint(X.shape[0], size=k),:], dtype=tf.float32)
centroids = tf.Variable(start_pos.initialized_value(), 'S', dtype=tf.float32)

# populate points
points           = tf.Variable(X, 'X', dtype=tf.float32)
ones_like        = tf.ones((points.get_shape()[0], 1))
prev_assignments = tf.Variable(tf.zeros((points.get_shape()[0], ), dtype=tf.int64))

# find the distance between all points: http://stackoverflow.com/a/43839605/1090562
p1 = tf.matmul(
    tf.expand_dims(tf.reduce_sum(tf.square(points), 1), 1),
    tf.ones(shape=(1, k))
)
p2 = tf.transpose(tf.matmul(
    tf.reshape(tf.reduce_sum(tf.square(centroids), 1), shape=[-1, 1]),
    ones_like,
    transpose_b=True
))
distance = tf.sqrt(tf.add(p1, p2) - 2 * tf.matmul(points, centroids, transpose_b=True))

# assign each point to a closest centroid
point_to_centroid_assignment = tf.argmin(distance, axis=1)

# recalculate the centers
total = tf.unsorted_segment_sum(points, point_to_centroid_assignment, k)
count = tf.unsorted_segment_sum(ones_like, point_to_centroid_assignment, k)
means = total / count

# continue if there is any difference between the current and previous assignment
is_continue = tf.reduce_any(tf.not_equal(point_to_centroid_assignment, prev_assignments))

with tf.control_dependencies([is_continue]):
    loop = tf.group(centroids.assign(means), prev_assignments.assign(point_to_centroid_assignment))

sess = tf.Session()
sess.run(tf.global_variables_initializer())

# do many iterations. Hopefully you will stop because of has_changed is False
has_changed, cnt = True, 0
while has_changed and cnt < 300:
    cnt += 1
    has_changed, _ = sess.run([is_continue, loop])

# see how the data is assigned
res = sess.run(point_to_centroid_assignment)

现在是时候检查我们的集群有多好了。为此，我们将集群中出现的所有实数组合在一起。之后，我们将看到该集群中最受欢迎的选择。在完美聚类的情况下，我们将在每组中只有一个值。在随机集群的情况下，每个值将在组中大致相等。

nums_in_clusters = [[] for i in xrange(10)]
for cluster, real_num in zip(list(res), list(y)):
    nums_in_clusters[cluster].append(real_num)

for i in xrange(10):
    print Counter(nums_in_clusters[i]).most_common(3)

这给了我这样的东西：

[(0, 738), (6, 18), (2, 11)]
[(1, 641), (3, 53), (2, 51)]
[(1, 488), (2, 115), (7, 56)]
[(4, 550), (9, 533), (7, 280)]
[(7, 634), (9, 400), (4, 302)]
[(6, 649), (4, 27), (0, 14)]
[(5, 269), (6, 244), (0, 161)]
[(8, 646), (5, 164), (3, 125)]
[(2, 698), (3, 34), (7, 14)]
[(3, 712), (5, 290), (8, 110)]

这非常好，因为大多数计数都在第一组中。您会看到聚类混淆了 7 和 9、4 和 5。但是 0 的聚类效果非常好。

一些改进方法：

  • 运行算法几次并选择最好的一个（基于到集群的距离）
  • 处理没有分配给集群的情况。在我的例子中，你会得到 Nanmeans变量，因为count它是 0。
  • 随机点初始化。
于 2017-05-12T05:43:03.913 回答
1

如今，您可以直接使用（或从中获得灵感）KMeansClustering Estimator。您可以在 GitHub 上查看它的实现。

于 2019-02-11T12:32:58.943 回答