plotting two-vectors on same axis

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Raghavendra Reddy P
Raghavendra Reddy P 2015 年 6 月 1 日
I have vector x contains 33-elements
x = [
1.0000
0.9974
0.9853
0.9793
0.9735
0.9588
0.9560
0.9461
0.9421
0.9385
0.9380
0.9372
0.9343
0.9333
0.9319
0.9306
0.9286
0.9280
0.9969
0.9933
0.9926
0.9920
0.9817
0.9751
0.9718
0.9572
0.9552
0.9462
0.9397
0.9362
0.9321
0.9291
0.9288]
>> t=[1:33];
>> plot(t,x)
gives you a graph, now i have a y vector which contains only 13 elements
y = [
1.0000
0.9974
0.9853
0.9712
0.9588
0.9360
0.9333
0.9295
0.9928
0.9763
0.9483
0.9397
0.9279]
now i wanted to draw graph on existing one such that elements values of y, which are near to x values should come same t-axis points. example:- y(4)=0.9712 is almost equal equal to x(5)=0.9735 at t=5, y(end)=0.9279 is equal to x(end)=0.9288 at t=33... hence i want y(4),y(end) on t=5,t=33.

採用された回答

Walter Roberson
Walter Roberson 2015 年 6 月 1 日
y(end)=0.9279 is equal to x(end)=0.9288 at t=33
0.9729 is smaller than any x value, and is closest to x(18) which is 0.9280.
y(4)=0.9712 is almost equal equal to x(5)=0.9735 at t=5
0.9712 is closer to x(25) = 0.9718
The descriptive part about "near to x values" leads to this code:
plot(1:length(x),x,'r',interp1(x, 1:length(x), y, 'nearest', 'extrap'),y,'g.-')
which has each y value being drawn at the t value that corresponds to the x value that is closest to the y value.
Your rules about what is near enough do not appear to be well defined.
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Raghavendra Reddy P
Raghavendra Reddy P 2015 年 6 月 1 日
Thank you sir, since the above values are small fraction numbers i couldn't differentiate exactly the way i wanted. your coding is fine. i just wanted to draw y graph on same x-axis of x graph such that elements values of y and x are closer.

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