Need to find the value of x_?, where x_? is x at y=?......

Dear All,

Just a simple question, I need to find x_15, where x_15 is x at y=15. Please imagine x and y as below,

x       y
0.1     20 
0.05    15
0.2     85

Thank you!

Answer 1

[Iman Ansari]

Hi.

x=[0.1; 0.05; 0.2];
y=[20; 15; 85];
x(y==15)

[Ara]

Hi Iman Ansari, Thanks for your help. This works correctly, I just do not know why when I apply this to my code the plot doesn't show anything while I am sure the value is not zero at 15. Do you have any idea? plot(s4_01nov(el1nov==15)/sin(el1nov));

[Iman Ansari]

Try this:

plot(s4_01nov(el1nov==15)./sin(el1nov(el1nov==15)));

[Ara]

Thanks, but doesn't work. It's suppose to show me a sine curve. Matlab read this command but show me nothing. The window doesn't even open it to make me able to see how the figure is look like.

[Iman Ansari]

What are s4_01nov and el1nov?

[Ara]

Two columns (5357x1 double)like the example x and y. Usually Matlab give me an error but I never faced this problem (no error, no figure..)it's weird. When I plot (el1nov,s4_01nov)the figure show me values for el1nov=15 so I am wondering why can not show me for exact value I want.

[Iman Ansari]

See this works:

s4_01nov=rand([5357 1]);
el1nov=randi(20,[5357 1]);
plot(s4_01nov(el1nov==15)./sin(el1nov(el1nov==15)))

but with for example 30, because el1nov doesn't have 30 nothing happened. With nnz(el1nov==15) you can check how many 15 in the el1nov vector:

nnz(el1nov==15)
nnz(el1nov==30)
plot(s4_01nov(el1nov==30)./sin(el1nov(el1nov==30)))

[Ara]

Thank you, yes this works. Ok, now by using "nnz" I realize at that value the vector was zero which is not reasonable based on my observation then I will go to recheck that part. So no wonder then why Matlab shows no figure.

[reetu bharti]

where we can use x_ exactly?

is it known as space variable?