MATLAB data matrix unitization, normalization, normalization

1. Data matrix unitization

method one:

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1. %% matrix column vector normalization  
2. % The output matrix Y is a unitized matrix  
3. The % method is to divide all elements in the matrix by the second norm of the column vector where the element is located  
4. clc;  
5. clear;  
6. X=[790 3977 849 1294 1927 1105 204 1329  
7.     768 5037 1135 1330 1925 1459 275 1487  
8.     942 2793 820 814 1617 942 155 976  
9.     916 2798 901 932 1599 910 182 1135  
10.     1006 2864 1052 1005 1618 839 196 1081];  
11. %method one  
12. [m,n]=size(X);  
13. for i=1:n  
14.     A(1,i)=norm(X(:,i));  
15. end  
16. A=repmat(A,m,1);  
17. Y=X./A;  

Output result:

Y =
    0.3974 0.4932 0.3959 0.5290 0.4941 0.4601 0.4422 0.4890
    0.3863 0.6247 0.5292 0.5437 0.4936 0.6074 0.5961 0.5471 0.4738
    0.3464 0.3823 0.3327 0.4146 0.3922 0.3360 0.3591 0.4608 0.3470 0.4201
    0.3810 0.4100 0.3789 0.3945 0.4176 0.5060 0.3552
    0.4905 0.4108 0.4149 0.3493 0.4249 0.3977

Method Two:

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1. %% matrix column vector normalization  
2. % The output matrix Y is a unitized matrix  
3. The % method is to divide all elements in the matrix by the second norm of the column vector where the element is located  
4. clc;  
5. clear;  
6. X=[790 3977 849 1294 1927 1105 204 1329  
7.     768 5037 1135 1330 1925 1459 275 1487  
8.     942 2793 820 814 1617 942 155 976  
9.     916 2798 901 932 1599 910 182 1135  
10.     1006 2864 1052 1005 1618 839 196 1081];  
11. %Method Two  
12. [m,n]=size(X);  
13. a=0;  
14. for j=1:n  
15.     for i=1:m  
16.         a=a+X(i,j)^2;  
17.     end  
18.     A(1,j)=sqrt(a);  
19.     a=0;  
20. end  
21. A=repmat(A,m,1);  
22. Y=X./A  

Output result:

= The Y
    0.3974 .4932 .5290 .4941 .4601 0.4422 0.3959 0.4890 0.3863 0.5292 0.5437 0.4936
    .6074 0.5961 .6247 0.5471 .4738 .3464
    .3823 .4146 0.3327 0.3922 0.3360 0.3591
    .4608 0.3470 .4201 .3810 .4100 0.3789 0.3945 0.4176 0.5060 0.3552 0.4905 0.4249 0.3977
    .4108 .4149 .3493
result supra.

2. Data matrix normalization

Normalize, normalize the data of the same dimension of different samples.

Function: mapminmax

Default canonical range (-1,1)

If you want to divide the specification range into (0,1), you can write Y=mapminmax(A,0,1);

This function normalizes the maximum and minimum values ​​in the row vector. If this function is used, each row of the A matrix is ​​a dimension, and each column is a sample.

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1. %% matrix data normalization  
2. % normalization is to deal with singular sample matrices  
3. % Normalize the matrix data into a range to make different dimensions comparable  
4. clc;  
5. clear;  
6. X=[790 3977 849 1294 1927 1105 204 1329  
7.     768 5037 1135 1330 1925 1459 275 1487  
8.     942 2793 820 814 1617 942 155 976  
9.     916 2798 901 932 1599 910 182 1135  
10.     1006 2864 1052 1005 1618 839 196 1081];  
11. Y=mapminmax(X,0,1);  

Output result:

Y =
    0.1553 1.0000 0.1710 0.2889 0.4567 0.2388 0 0.2982
    0.1035 1.0000 0.1806 0.2215 0.3465 0.2486 0 0.2545
    0.2983 1.0000 0.2521 0.2498 0.5542 0.2983 0 0.3112
    0.2806 1.0000 0.2748 0.2867 0.5417 0.2783 0 0.3643
    0.3036 1.0000 0.3208 0.3032 0.5330 0.3317 0.2410 0

The canonical range is (-1,1)

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1. %% matrix data normalization  
2. % normalization is to deal with singular sample matrices  
3. % Normalize the matrix data into a range to make different dimensions comparable  
4. clc;  
5. clear;  
6. X=[790 3977 849 1294 1927 1105 204 1329  
7.     768 5037 1135 1330 1925 1459 275 1487  
8.     942 2793 820 814 1617 942 155 976  
9.     916 2798 901 932 1599 910 182 1135  
10.     1006 2864 1052 1005 1618 839 196 1081];  
11. Y=mapminmax(X);  

输出结果：

Y =
   -0.6894    1.0000   -0.6581   -0.4222   -0.0867   -0.5224   -1.0000   -0.4037
   -0.7929    1.0000   -0.6388   -0.5569   -0.3070   -0.5027   -1.0000   -0.4910
   -0.4033    1.0000   -0.4958   -0.5004    0.1084   -0.4033   -1.0000   -0.3776
   -0.4388    1.0000   -0.4503   -0.4266    0.0833   -0.4434   -1.0000   -0.2714
   -0.3928    1.0000   -0.3583   -0.3936    0.0660   -0.5180   -1.0000   -0.3366

3.数据矩阵标准化

标准化的数据均值为0，标准差为1

标准化函数zscore(x)

就是原数据减去均值，再除以标准差（无偏估计）

即Z=(x-mean(x))./std(x);

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1. %%矩阵数据标准化  
2. clc;  
3. clear;  
4. X=[790 3977 849 1294 1927 1105 204 1329  
5.     768 5037 1135 1330 1925 1459 275 1487  
6.     942 2793 820 814 1617 942 155 976  
7.     916 2798 901 932 1599 910 182 1135  
8.     1006 2864 1052 1005 1618 839 196 1081];  
9. Y=zscore(X);  

输出结果：

Y =
   -0.9261    0.4840   -0.7522    0.9640    1.1002    0.2177    0.0358    0.6225
   -1.1419    1.5457    1.3487    1.1224    1.0886    1.6449    1.6257    1.3944
    0.5651   -0.7020   -0.9653   -1.1488   -0.6967   -0.4395   -1.0614   -1.1023
    0.3100   -0.6969   -0.3702   -0.6294   -0.8011   -0.5685   -0.4568   -0.3254
    1.1929   -0.6308    0.7390   -0.3081   -0.6909   -0.8547   -0.1433   -0.5892

也可以按照上面的公式：

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1. %%矩阵数据标准化  
2. clc;  
3. clear;  
4. X=[790 3977 849 1294 1927 1105 204 1329  
5.     768 5037 1135 1330 1925 1459 275 1487  
6.     942 2793 820 814 1617 942 155 976  
7.     916 2798 901 932 1599 910 182 1135  
8.     1006 2864 1052 1005 1618 839 196 1081];  
9.   
10. Y=(X-repmat(mean(X),5,1))./repmat(std(X),5,1);  

输出结果：

= The Y
   -0.9261 -0.7522 0.9640 1.1002 .2177 .4840 0.0358 0.6225 -1.1419
   1.5457 1.3487 1.1224 1.0886 1.6449 1.6257 1.3944
    0.5651 -0.7020 -0.9653 -1.1488 -0.6967 -0.4395 -1.0614 -1.1023
    0.3100 -0.6969 -0.3702 -0.6294 -0.8011 -0.5685 -0.4568 - 0.3254
    1.1929 -0.6308 0.7390 -0.3081 -0.6909 -0.8547 -0.1433 -0.5892
are consistent with the above results.