Replace value in matrix

Hi,

I'm looking at the orientation of lines generating using linear regression.

I create a matrix that increases after each loop, generating a new columns with a number of data higher than the previous columns (because there is more lines).

My problem is that in the end my matrix is [24*17], however the data in the first colum are only 3 values, and the rest (19 cells) filled with zeros.

I'd like to analyse the matrix, but i can'tget rid of the "filling-zeros".

Does anyone have an idea?

Thanks

N.

[Nicolas]

The 17th colum is 24 values, the 1st one : 3 values and 21 zeros, the 2nd: 5 values and 19 zeros; the 3rd: 6 values and 18 zeros... and so on

[Walter Roberson]

Is the matrix the _result_ of linear regression, or is it input being fed into linear regression ? If it is input to linear regression, what would you intend it to mean to the formula -- that the corresponding components are zero ?

[Nicolas]

it is the results of linear regression.

Answer 1

[Oleg Komarov]

You can obtain:

A = [0   0 143 143 152
     0   0   0   0 151
     0   0   0   0 143];
regexprep(evalc('A'), '0', ' ')

A =
               143   143   152
                           151
                           143

But it's a non tractable string array, i.e. what you're trying to do is meaningful only for visualization purposes and privates A of any computational use.

Oleg

[Nicolas]

Oh, yes that's what I should have say since the beginning ! it's to plot the data!

thanks

[Walter Roberson]

But you _did_ say in the beginning: you said "I'd like to analyse the matrix". Not analyze the form of a _printout_ of the matrix. Thus, putting in spaces does not meet your requirements as stated. Did your requirements change along the way?

[Oleg Komarov]

To plot data, i.e. to represent it by means of graphs, follow Paulo's indications.

To display/visualize the matrix (as Walter suggests, to analyze its printout) use the approach I suggested.

Just keep in mind terminology for future questions.

[Nicolas]

regexprep(A, '0', ' ')

[Nicolas]

Sorry, my explanations are a bit confusing.

My first idea was to analyse the matrix, by "analysing" i meant plotting the direction values of each column in a compass plot (all the data, then by sector of 10 degrees) to see preferential orientations. The problem that you helped me to solve was to get rid of the "filling-zeros", because my matrix is growing in each loop both row and column.

I hope it is a bit more clear for you.

Answer 2

[Paulo Silva]

Two options:

1-ignore first column

mat(:,2:end)

2-replace the zeros on the first column

mat(mat(:,1)==0)=inf %replace zeros with inf

you can choose other things besides inf, maybe NaN or a really small value like eps

3-extra option, replace all zeros

mat(mat==0)=inf %again you can choose the value to replace the zeros

Answer 3

[Nicolas]

Hi,

Thank you for your help. It works very well!

this is my matrix

0 0 143 143 152

0 0 0 0 151

0 0 0 0 143

I was trying to use indices.

[r,c,v]=(mat>0)

which gives me

1 3 143

1 4 143

1 5 152

2 5 151

3 5 143

can i rebuilt a matrix like the following using the indices

143 143 152

        151
        143

I'm not very familiar with matlab matrix manipulation, i'm sure i'm missing a clue somewhere !

thanks again for your help

[Walter Roberson]

It is not possible to have a numeric matrix with empty spaces.