Finding corresponding values in data set

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jrz 2022-9-16

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评论： Star Strider 2022-9-16

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?

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Walter Roberson 2022-9-16

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

are the values in that column sorted?

jrz 2022-9-16

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)

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Answer 1

Star Strider 2022-9-16

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编辑：Star Strider 2022-9-16

在 MATLAB Online 中打开

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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Star Strider 2022-9-16

在 MATLAB Online 中打开

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.

.

jrz 2022-9-16

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

Star Strider 2022-9-16

As always, my pleasure!

No worries!

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Finding corresponding values in data set

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