scipy 中的whiten函式

呼叫kmeans函式，kmeans中呼叫了whited函式。查後，發現whiten是對輸入資料按標準差做歸一化處理。

e​經過whiten後

x i′

=xis

tand

_dev

atio

nx_^ =\frac}

xi′​=s

tand

_dev

atio

nxi​

​與標準化不同的是，白化處理沒有減去均值。

下面是按步驟實現和呼叫函式實現，結果是一樣的。

import numpy as np

from numpy import array
from scipy.cluster.vq import vq, kmeans, whiten
import matplotlib.pyplot as plt
features  = array([[
1.9,
2.3],[
1.5,
2.5],[
0.8,
0.6],[
0.4,
1.8],[
0.1,
0.1],[
0.2,
1.8],[
2.0,
0.5],[
0.3,
1.5],[
1.0,
1.0]])
mean = np.mean(features,axis=0)
# 求每列的均值[0.91111111 1.34444444]
np_square = np.power(features-mean,2)
# # print('mean:\n',mean)
print
('features - mean:\n'
,features-mean)
# print('square:\n',np_square)
mean_sqare = np.mean(np_square,axis =0)
# 方差
# print('mean_square:\n',mean_sqare)
stand_devation = np.sqrt(mean_sqare)
# 標準差
# print('stand_devation:\n',stand_devation)
np_whit = features/stand_devation# scales by devation
print
('******* np_white ******'
)print
(np_whit)
whitened = whiten(features)
# 呼叫scipy中的函式
print
('******* whitened ******'
)print
(whitened)

from numpy import random

import numpy as np
from numpy import array
from scipy.cluster.vq import vq, kmeans, whiten
import matplotlib.pyplot as plt
random.seed(
(1000
,2000))
codes =
3kmeans(whitened,codes)
# (array([[ 2.3110306 ,  2.86287398],    # random
#        [ 1.32544402,  0.65607529],
#        [ 0.40782893,  2.02786907]]), 0.5196582527686241)
# create 50 datapoints in two clusters a and b
pts =
50a = np.random.multivariate_normal([0
,0],
[[4,
1],[
1,4]
], size=pts)
# 中心點、協方差、資料量
b = np.random.multivariate_normal([30
,10],
[[10,
2],[
2,1]
],size=pts)
features = np.concatenate(
(a, b)
)# whiten data
whitened = whiten(features)
# find 2 clusters in the data
codebook, distortion = kmeans(whitened,2)
# 返回聚類中心和誤差
				
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