Replace value in matrix

2011 3 14
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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.

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Nicolas
Nicolas 2011-3-14
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
Walter Roberson 2011-3-14
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
Nicolas 2011-3-16
it is the results of linear regression.

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Oleg Komarov
Oleg Komarov 2011-3-16

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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

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Nicolas
Nicolas 2011-3-16
Oh, yes that's what I should have say since the beginning ! it's to plot the data!
thanks
Walter Roberson
Walter Roberson 2011-3-16
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
Oleg Komarov 2011-3-16
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
Nicolas 2011-3-16
regexprep(A, '0', ' ')
Nicolas
Nicolas 2011-3-16
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.

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更多回答（2 个）

Paulo Silva
Paulo Silva 2011-3-14

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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
Nicolas
Nicolas 2011-3-16

0 个投票

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

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Walter Roberson
Walter Roberson 2011-3-16
It is not possible to have a numeric matrix with empty spaces.

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