Finding corresponding values in data set

I have a set of matrices each with 4 columns. I want to extract the value of the 1st column corresponding to the 0 in the second column and plot that point. How can I do this? and for cases where there is no exact zero, interpolate between the two values that cross 0?

[Walter Roberson]

Is there always exactly one 0 or zero crossing, or could there be several?

are the values in that column sorted?

[jrz]

yes there is only a single zero crossing or single 0 for each. The values in the columns are in ascending order, e.g. -5 at (1,) and 2 at (1,50)

Answer 1

[Star Strider]

I would just do the interpolation using interp1 since it will interpolate to 0 or the closest value to it.

Try this —

M = randn(10,4)

M = 10×4

-0.9697 -0.7712 -1.3959 1.3028 -1.7855 -0.0497 0.0510 -0.6687 -0.3479 0.4030 -0.8218 0.5535 0.6310 1.3745 -2.0030 -0.6301 -0.3321 -0.3106 0.8438 -0.4688 -2.2774 -0.0997 0.3758 -2.0011 -0.0985 0.6001 0.4742 0.0693 -1.8270 1.2099 0.1478 -0.5050 0.4731 -0.3345 1.4999 0.3136 -0.4747 -0.7732 0.6202 0.6346

L = size(M,1);
idx = find(diff(sign(M(:,2))))

idx = 4×1

2 4 6 8

for k = 1:numel(idx)
    idxrng = max(1,idx(k)-1) : min(L,idx(k)+1);
    Result(k,:) = interp1(M(idxrng,2), M(idxrng,:),0);
end

Result

Result = 4×4

-1.6276 -0.0000 -0.0449 -0.5344 -0.3390 -0.0000 0.1189 -0.0239 -1.9668 0 0.3898 -1.7060 0.2685 -0.0000 1.1328 0.2262

EDIT — Aesthetic tweaks.

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[jrz]

Thanks for your reply, but i dont really undertand whats going on here. For your M i would just want a single value from the second column that corresponds to where the values in the 1st column cross zero. is that what the result matrix is saying in some form?

[Star Strider]

My pleasure!

‘I want to extract the value of the 1st column corresponding to the 0 in the second column and plot that point.’

I am slightly confused now, since this is different from the original question.

This retrieves the values for the first two columns when the second (or first) column crosses (or equals) zero —

M = randn(10,4)+0.7

M = 10×4

-0.2369 1.8739 0.0779 0.7713 0.6648 0.8674 1.3288 0.6227 2.6427 -0.9644 -0.0458 1.4343 -0.7491 0.3339 1.1961 0.5221 1.7124 0.8753 2.4023 1.4774 0.0145 2.6902 1.0609 0.9595 1.0549 1.1404 1.3261 2.5583 -1.0573 1.9985 2.2614 -2.3209 0.5639 -0.0326 -0.1718 -0.6114 0.9802 0.0070 0.6125 -0.5356

L = size(M,1);
idx = find(diff(sign(M(:,2))))

idx = 4×1

2 3 8 9

for k = 1:numel(idx)
    idxrng = max(1,idx(k)-1) : min(L,idx(k)+1);
    Result2(k,:) = interp1(M(idxrng,2), M(idxrng,1:2),0);
end

Result2                                                     % First & Second Columns When Second Column Crosses Zero

Result2 = 4×2

1.6014 -0.0000 0.1233 -0.0000 0.5775 0 0.9062 0.0000

idx = find(diff(sign(M(:,1))))

idx = 5×1

1 3 4 7 8

for k = 1:numel(idx)
    idxrng = max(1,idx(k)-1) : min(L,idx(k)+1);
    Result1(k,:) = interp1(M(idxrng,1), M(idxrng,1:2),0);
end

Result1                                                     % First & Second Columns When First Column Crosses Zero

Result1 = 5×2

0.0000 1.6094 -0.0000 0.6166 -0.0000 0.4987 0.0000 2.6808 0 0.6739

I do not have your data, so I am not certain how you want the points plotted.

Other interpolation methods (for example 'nearest') are possible if you do not want a linear (default) interpolation, and only the nearest point to the zero-crossing.

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[jrz]

so sorry for the confusion!, i misspoke in my reply. Thanks again for your help, i understand now

[Star Strider]

As always, my pleasure!

No worries!