How to write a simple MATLAB code for simplification of if.. else.. code

My program has a vector x=(x1,x2,...xn) and I want to compare xp with the values of the vector x. I wrote the following code in my program. Can we simplify the following code using any matlab builtin functions

if (xp<=x(1)) s=1;
else if xp>=x(n) s=n;
else
    for i=2:n-1
    if xp>=x(i) && xp<x(i+1) s=i;
    end
    end
    end
end

[madhan ravi]

What’s your input x and your desired output? Give an example .

[PJS KUMAR]

• x=[1 2 3 4 5 6] and the values represent the row numbers of a matrix.
• xp is my input.
• if xp is 1 or below 1, i have to select row 1 or matrix
• if xp is in between any i and i+1 row, i have to select i row. and so on.
• if xp is n or above n, i have to select row n.

The input is any value xp, and the output should be the number of the row in a matrix

[Rik]

Does this code return the result that you want? Is it right that your output is only a scalar? Also, there is a big difference between else if and elseif. Below you find two blocks of code, which is the one you mean?

if (xp<=x(1))
    s=1;
else
    if xp>=x(n)
        s=n;
    else
        for i=2:n-1
            if xp>=x(i) && xp<x(i+1)
                s=i;
            end
        end
    end
end

Or

if (xp<=x(1))
    s=1;
elseif xp>=x(n)
    s=n;
else
    for i=2:n-1
        if xp>=x(i) && xp<x(i+1)
            s=i;
        end
    end
end

[Walter Roberson]

What value is to be selected if xp > x(1) & xp < x(2) ?

Your pseudocode reserves the value 1 for values no greater than x(1), and starts at 2 for values at least as large as x(2), leaving open the question of what should happen for values between the two.

[PJS KUMAR]

for xp > x(1) & xp < x(2) i have to select x(1)

Answer 1

[Walter Roberson]

[~, s] = histc(xp, [-inf x(2:end) inf]);
selected_x = x(s);

This assumes x is in increasing order.

[PJS KUMAR]

what is the purpose of [~,s]

[Walter Roberson]

s is exactly the value you calculate in your pseudocode, 1 for values too low, n for values above the range, and so on.

If you are asking about the construct [~,s] then it is the same as if you had coded

[SoMEVarIaBlEiWILLnoTuSE, s] = histc(xp, [-inf x(2:end) inf]);
clear SoMEVarIaBlEiWILLnoTuSE

That is, it says that you want to ignore the output in that position. Like many functions, histc returns multiple outputs that have different purposes; the first output of histc is the bin counts (because histc was primarily intended to aid in computing histograms.)

[Rik]

A minor edit makes removes the assumption of being ordered.

temp_x=sort(x);
[~, s] = histc(xp, [-inf temp_x(2:end) inf]);
selected_x = temp_x(s);

Answer 2

[Fangjun Jiang]

x=10:10:100;
xp=29;

n=numel(x);
interp1([-realmax,x,realmax],[1,1:n,n],xp,'previous')

[PJS KUMAR]

What is 29

[Fangjun Jiang]

It is an example value of your xp.

[PJS KUMAR]

x=[-1 -1.2 -1.4 -1.6 -1.8 -2]; i.e., the values of x are in decreasing order

• xp=-1.3 (which is in between -1.2 and -1.4) i have to select the row with -1.2, and the output should be row number 2
• xp=-2.5 , (-2.5 < -2) i have to select the row with -2, and the output should be row number 6
• xp=1, (1 > -1) i have to select the row with -1, and the output should be row number 1

[PJS KUMAR]

Example - 1

x=[1 1.2 1.4 1.6 1.8 2]; i.e., the values of x are in increasing order

• xp=1.3 (which is in between 1.2 and 1.4) i have to select the row with 1.2, and the output should be row number 2
• xp=2.5 , (2.5 > 2) i have to select the row with 2, and the output should be row number 6
• xp=0.5, (0.5 < 1) i have to select the row with 1, and the output should be row number 1

Example - 2

x=[-1 -1.2 -1.4 -1.6 -1.8 -2]; i.e., the values of x are in decreasing order

• xp=-1.3 (which is in between -1.2 and -1.4) i have to select the row with -1.2, and the output should be row number 2
• xp=-2.5 , (-2.5 < -2) i have to select the row with -2, and the output should be row number 6
• xp=1, (1 > -1) i have to select the row with -1, and the output should be row number 1

[Fangjun Jiang]

x=[-1 -1.2 -1.4 -1.6 -1.8 -2];
xp=[-1.3, -2.5, 1];

n=numel(x);
index=interp1([realmax,x(2:end),-realmax],[1:n,n],xp,'next')
yp=x(index)
x=[1 1.2 1.4 1.6 1.8 2];
xp=[1.3, 2.5, 0.5];
n=numel(x);
index=interp1([-realmax,x(2:end),realmax],[1:n,n],xp,'previous')
yp=x(index)

Answer 3

[Fangjun Jiang]

Now I think your problem has nothing to do with the increasing or decreasing of the values in x. Rather, it is to lookup and "match" the values in x towards zero, similar to the difference between floor(), ceil() and fix(). To do this, use the 'previous' and 'next' option of interp1() based on the sign of xp value. Please note, you shall add -realmax and realmax to avoid extrapolation, add 0 to avoid nondeterministic when xp fells between two points in x that are on the opposite side of zero.

x=[1 1.2 1.4 1.6 1.8 2 -1 -1.2 -1.4 -1.6 -1.8 -2];
xp=[-1.3, -2.5, -0.5 1 1.3, 2.5, 0.5];

xMod=[x,-realmax,0,realmax];
yp=zeros(size(xp));
for k=1:numel(xp)
    if xp(k)>=0
        yp(k)=interp1(xMod,xMod,xp(k),'previous');
    else
        yp(k)=interp1(xMod,xMod,xp(k),'next');
    end
end