Finding corresponding values in data set

jrz
2022 9 16
1 个回答
回答已采纳
更新时间：2022 9 16
3 次查看（30 天）

0 个投票

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?

2 个评论
显示 无 隐藏 无

Walter Roberson
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
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)

请先登录，再进行评论。

请先登录，再回答此问题。

 采纳的回答

Star Strider
Star Strider 2022-9-16
编辑：Star Strider 2022-9-16
Ran in:

0 个投票

Ran in:
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.
.

4 个评论
显示 2更早的评论 隐藏 2更早的评论

jrz
jrz 2022-9-16
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
Star Strider 2022-9-16
Ran in:
Ran in:
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
jrz 2022-9-16
so sorry for the confusion!, i misspoke in my reply. Thanks again for your help, i understand now
Star Strider
Star Strider 2022-9-16
As always, my pleasure!
No worries!

请先登录，再进行评论。

更多回答（0 个）

类别

在 帮助中心 和 File Exchange 中查找有关 Dates and Time 的更多信息

标签

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by