calculate mean using while and iteration?

Hi,

I have data of ~3500x2. I want to calulate mean of second column for a particular condition in first column using 'while'.

let data (a, b) be like

0.5 1.8

0.6 1.5

0.9 1.8

1.0 1.5

1.1 1.4

1.2 1.4

1.5 1.6

1.8 1.2

2.1 1.2

2.3 1.3

2.4 1.5

2.6 1.8

2.9 2.0

3.0 3.0

3.12 3.2

3.15 1.9

3.16 1.7

3.18 2.2

I need to calculate mean of b, if a> 0.5 and a<1.5. Then increase 'a' by 1 and calculate mean of b (i.e for a > 1.5 and a<2.5) and so on. It may be a silly question but I am stuck with it. My code is

del=0.5;
k=1;
a(k)=1;
while(a(k) >(a(k)-del) && a(k)< (a(k)+del))
xn(k)=mean(b(k));
 k= k+1;
a(k)=a(k)+1;
end
but it shows error Index exceeds array bounds.

Error in untitled (line 12)
a(k)=a(k)+1;

Thank you for your help.

[David Wilson]

My code below is a bit ugly, but I think it does what you want:

cutoff = [0.5 1.5]; % band of interest
maxA = ceil(max(a))+0.5; 
bmean = []; 
for i=1:maxA
   idx = find(a>cutoff(1) & a<cutoff(2)); 
   bmean(i) = mean(b(idx));
   cutoff = cutoff+1; 
end 

The means of column "b" are in variable bmean.

I note that you specified strict < as opposed to <= which may, or may not be what you really want.

Note that column a need not be sorted in increasing order.

[Madan Kumar]

Thank you...

Answer 1

[Rik]

So you want to calculate these values?

xn(1)=mean(b(a>0.5 & a<1.5));
xn(2)=mean(b(a>1.5 & a<2.5));
xn(3)=mean(b(a>2.5 & a<3.5));
etc?

You don't need a while loop for that:

a=[0.5 0.6 0.9 1.0 1.1 1.2 1.5 1.8 2.1 2.3 2.4 2.6 2.9 3.0 3.12 3.15 3.16 3.18];
b=[1.8 1.5 1.8 1.5 1.4 1.4 1.6 1.2 1.2 1.3 1.5 1.8 2.0 3.0 3.2 1.9 1.7 2.2];
del=0.5;
xn=zeros(1,ceil(max(a-del)));
for k=1:size(xn,2)
    xn(k)=mean(b(a>(k-del) & a<(k+del)));
end

[Madan Kumar]

Thank you...

[Madan Kumar]

Suppose, I need

xn(0.5)=mean(b(a>0.25 & a<0.75));

xn(1)=mean(b(a>0.75 & a<1.25));

xn(1.5)=mean(b(a>1.25 & a<1.75));

xn(2)=mean(b(a>1.75 & a<2.25)); etc.

How do I need to change the loop? because if I do like below, there is error

del=0.25;
xn=zeros(1,ceil(max(a-del)));
for k=1:0.5:size(xn,2)
 xn(k)=mean(b(a>(k-del) & a<(k+del)));   
end
array indices must be positive integers or logical values.

[Stephen23]

"Suppose, I need xn(0.5)=... "

You can't. Indices must be whole integers greater than zero.

Either change del to 0.25 and leave it at that, or write a function which lets you have any input values that you desire. But you certainly cannot have indexing with non-integer values.

[Madan Kumar]

Ok...thank you...

Answer 2

[KSSV]

Why loop? YOu can use inbuilt in mean. Let a,b be your columns.

idx = a>0.5 & a<1.5 ; 
mean(b(idx))

[Madan Kumar]

Thank you...