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

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PJS KUMAR
PJS KUMAR 2018-9-13
编辑： Fangjun Jiang 2018-9-17
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
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Walter Roberson
Walter Roberson 2018-9-14
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.

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回答（3 个）

Walter Roberson
Walter Roberson 2018-9-14
[~, s] = histc(xp, [-inf x(2:end) inf]);
selected_x = x(s);
This assumes x is in increasing order.
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Walter Roberson
Walter Roberson 2018-9-14
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
Rik 2018-9-14
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);

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Fangjun Jiang
Fangjun Jiang 2018-9-13
编辑：Fangjun Jiang 2018-9-13
x=10:10:100;
xp=29;
n=numel(x);
interp1([-realmax,x,realmax],[1,1:n,n],xp,'previous')
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PJS KUMAR
PJS KUMAR 2018-9-14
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
Fangjun Jiang 2018-9-14
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)

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Fangjun Jiang
Fangjun Jiang 2018-9-17
编辑：Fangjun Jiang 2018-9-17
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

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