If we have a matrix A=[1 2 ; 3 4 ; 5 6] then it will be converted to vectors like x1= [1 3 5] x2= [2 4 6]

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Stephen23
Stephen23 am 11 Mär. 2015
Bearbeitet: Stephen23 am 20 Apr. 2020
If you know how many columns you want, then you can simply assign them in code like this:
>> A = [1,2;3,4;5,6];
>> X1 = A(:,1);
>> X2 = A(:,2);
If you have many columns, or an unknown sized matrix, then you can split it up using num2cell with the second optional argument:
>> B = num2cell(A,1);
>> B{1}
ans =
1
3
5
>> B{2}
ans =
2
4
6
Accessing the contents of the cell array is simple and efficient using indexing:
Accessing dynamically named variables like x1, x2, x3, etc. is not recommended, because it forces you into writing slow, complex, obfuscated, buggy code that is hard to debug. Read this to know why:

5 Kommentare

Thank you Sir
Thank you Sir
Alpha Bravo
Alpha Bravo am 20 Apr. 2020
Bearbeitet: Alpha Bravo am 20 Apr. 2020
x=[1 2 3; 4 5 6; ... ;...; ...]; % some m x n matrix
a = unique(x);
out = [a,histc(x(:),a)];% upto here unique frequencies
x1 = out(:,1);
x2 = out(:,2);
[tbl,chi2stat,pval] = crosstab(x1,x2);
disp(chi2stat);
disp(pval);
chi2stat 1.9866e+04
p 0.3317
is the code above correct? then what to interpret of the matrix
"is the code above correct?"
I don't know. What do you expect that code to do?
kindly;
  1. reduce m x n matrix to its frequency count
  2. crosstab into 2x 2
  3. subject those to compute chi2 and p
  4. interpret that result
  5. that is, to find the chi2, and p of the given matrix; the goodness of fit test

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