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矩阵标准化怎么求啊?
下面各矩阵要求标准化该怎么求啊?要急用
0.0371 0.0353 0.0334 0.0353 0.0345 0.0369
0.3905 0.1000 0.3415 0.4249 0.3899 0.3037
0.00095 0.00069 0.00057 0.00061 0.00094 0.0113
0.2277 0.1552 0.1901 0.1502 0.2260 0.1477
上面的数据有误
0.0371 0.0353 0.0334 0.0353 0.0345 0.0369
0.3905 0.1000 0.3415 0.4249 0.3899 0.3037
0.00095 0.00069 0.00057 0.00061 0.00094 0.00113
0.2277 0.1552 0.1901 0.1502 0.2260 0.1477
人气:364 ℃ 时间:2020-04-16 07:07:54
解答
用zscore函数.
>> help zscore
ZSCORE Standardized z score.
Z = ZSCORE(X) returns a centered, scaled version of X, the same size as X.
For vector input X, Z is the vector of z-scores (X-MEAN(X)) ./ STD(X).
标准化的意思就是使平均值等于0,标准差等于1.
例如:
>> A=[0.0371 0.0353 0.0334 0.0353 0.0345 0.0369
0.3905 0.1000 0.3415 0.4249 0.3899 0.3037
0.00095 0.00069 0.00057 0.00061 0.00094 0.00113
0.2277 0.1552 0.1901 0.1502 0.2260 0.1477]
>>B=zscore(A)
>>mu=sum(B)
>>sigma=std(B)
B =
-0.7023 -0.5463 -0.6880 -0.6106 -0.7092 -0.6281
1.25250.39631.27491.41471.25491.3329
-0.9022 -1.0505 -0.8972 -0.7909 -0.8947 -0.8911
0.35201.20040.3103 -0.01330.34910.1863
mu=
1.0e-015 *
0.0555 00.0555 -0.3123 00.1110
sigma=
1.00001.00001.00001.00001.00001.0000
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