## for loop help

### James (view profile)

on 7 Aug 2011

Hi

I want to be able to select c = 2 3 5 6 8 9 12 13 15 16 18 19 22 23 25 26 28 29 32....

I need to do this all the way up to 89. Can i do this in a loop?

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### Neels (view profile)

on 7 Aug 2011

hi James, can you tell if the series follows a specific sequence or just some random values

### Tim Mottram (view profile)

on 7 Aug 2011

Hi James,

So you want all numbers accept those ending in a 0,1 or 4? If this is correct then put an IF statement inside your FOR loop with something like: if a~= 0 && a~= 1 && a~=4, where a is the last digit of the current step being used by the for loop. Put the for loop incrementer (i = i+1) after the END of this IF statement. How you calculate, a, is up to you...

Regards

Tim

James

### James (view profile)

on 7 Aug 2011

i suppose when you look at it that way then yes i need to skip all numbers ending in 0,1,4 and 7

how do i select the last digit of the current step?

Tim Mottram

### Tim Mottram (view profile)

on 7 Aug 2011

see Jan's comment for selecting the last digit, or transform the number into a string,(string = num2str(current for loop)) then ask for its length, (digit = length(string)) and if digit (this will always be the last digit) of string is not equal to 0,1,4 or 7, then proceed. some rough code..
for i = 0:YourEndPoint
s = num2str(i);
d = length(s);
if s(d) ~= 0 && ~=1 && ~= 4 && ~= 7
your code for what you want
end
i = i+1
end

### Jan Simon (view profile)

on 7 Aug 2011
```for i = 0:90
if all(mod(i, 10) ~= [0,1,4,7])
disp(i)
end
end
```

James

### James (view profile)

on 7 Aug 2011

how does the if line work?

all(mod(i, 10) ~= [0,1,4,7])

Jan Simon

### Jan Simon (view profile)

on 7 Aug 2011

@James: Read it loud: if all (reminder of i divided by 10) are not equal to an element of [0, 1, 4, 7].
Example: i = 18, mod(i, 10)=8, (8~=[0,1,4,7]) = [1,1,1,1], all([1,1,1,1]) = TRUE.
i = 17, mod(i, 10)=7, (7~=[0,1,4,7])=[1,1,1,0], all([1,1,1,0]) = FALSE

### Jan Simon (view profile)

on 7 Aug 2011

Another method:

```for a = 0:10:80
for b = [2,3,5,6,8,9]
k = a + b;
disp(k);
end
end
```

### Jan Simon (view profile)

on 7 Aug 2011

And if I'm on the way:

```ind = bsxfun(@plus, 0:10:80, [2,3,5,6,8,9]');
for i = reshape(ind, 1, [])
disp(i)
end
```

### James (view profile)

on 7 Aug 2011

Thanks. I was also just wondering if you can help me with something else similar:

```for c = 11:20
if c~=11 && c~=14 && c~=17 && c~=20
break
end
```
```% how can I loop through for all values of r, which are r=11,14,17,20????
```
```end
```

Jan Simon

### Jan Simon (view profile)

on 7 Aug 2011

@James: ??? Do you mean: for r = [11, 14, 17, 20], ... end

Tim Mottram

### Tim Mottram (view profile)

on 7 Aug 2011

for i = 11:3:20 ??

### Paulo Silva (view profile)

on 7 Aug 2011

yet another way

```r=9;              %number of repetitions of the sequence, ex 9, max is 89
b=[2 3 5 6 8 9];  %original sequence
c=repmat(b,r,1);  %repeat the sequence in each row
d=0:10:10*r-1;    %create vector with sum values
e=repmat(d,6,1)'; %create array from vector
f=(c+e)';         %now all in the right place do the sum
f(:)'             %put the values in just a vector
```

### Andrei Bobrov (view profile)

on 7 Aug 2011
```a=1:90;
k = reshape(a,10,[]);
k(1:3:end,:)=[];
c = k(:)'
```

```a = reshape(1:90,10,[])';
c = [];
for j1 = 1:size(a,1)
for j2 = 1:3
c = [c a(j1,j2*3-[1 0])];
end
end
```

MORE

```a = reshape(1:90,10,[]);
c=reshape(a(bsxfun(@minus,(3:3:9),[1 0]'),:),1,[])
```

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