How to write a loop that does not increase by +1?

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Cside
Cside 2020-2-22
编辑： John D'Errico 2020-2-22
Hello :) I currently need to run a loop with the output of "find(collated(1,:) ==0 )". Currently, the output is [2 ; 4 ; 5 ; 6 ; 8], and I need to run the loop with these indexes. I couldn't use a for loop as the number are not increasing by 1. Is there a way I could manage the loop so that the output does not have to run in an +1 order? I read that a while loop may be useful, but am still new to the while loop.
PS
the matrix 'collated' is a 35 x 9 logical, so the idea that I want the code to work is that with each row of collated, I find the 0s i.e in row 1, it will be [2 ; 4 ; 5 ; 6 ; 8], row 2 [2 ;3], row 3 [4; 7; 8; 9] etc...
Appreciate any sort of help! :)
for k = 1:35
while A == find(collated(k,:) ==0)
selectedlocal = find(sesslocal == A);
pfc = dataset_p(1,:);
fef = dataset_f(1,:);
zpfc = zscore(pfc);
zfef = zscore(fef);
zzpfc = zpfc(selectedlocal);
zzfef = zfef(selectedlocal);
[R,P] = corrcoef(zzpfc,zzfef)
end
end
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David Goodmanson
David Goodmanson 2020-2-22
Hi charms, you can do
for k = 2:2:8
(do stuff that depends on k)
end
since the middle 2 sets the increment value
Cside
Cside 2020-2-22
编辑：Cside 2020-2-22
@madhan my expected result would just be [R,P], for the respective A input by the loops
@david - k is different with every experiment I am running. Would be great if there is a way to embedd it!

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采纳的回答

John D'Errico
John D'Errico 2020-2-22
编辑：John D'Errico 2020-2-22
A for loop can have any increment. Even a vector of numbers can be the increment. And it need not be in any order. Some examples:
% standard unit increment
for n = 1:10
stuff
end
% increment of 2
for n = 2:2:12
stuff
end
% only the prime numbers, that do not exceed 100
P = primes(100);
for n = P
stuff
end
% an arbitrary set of numbers
nvals = [1 10 100 1000 10000 100000];
for n = nvals
stuff
end
% an arbitrary set of numbers that need not be increasing, or even be integers
% the values might even be complex numbers.
nvals = [2 1 5 14 pi -1000 1 3 1.273435345 2+3i];
for n = nvals
disp(n)
end
2
1
5
14
3.14159265358979
-1000
1
3
1.273435345
2 + 3i
% the row vector input to for need not be a double precision vector.
% It may even be a cell vector
nvals = {1 pi,'bcde'};
for n = nvals
disp(n)
end
[1]
[3.14159265358979]
'bcde'
% or a character vector.
nvals = 'The quick brown fox jumped over the lazy dog'
for n = nvals
disp(n)
end
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In a few cases, I've reported the value of n at each step through the loop. As you can see, the elements of the row vector may be any number. It can be of class uint8 or double, or even a cell vector.
Be careful of one thing however. The for statement operates as you would expect ONLY when given a ROW vector. It iterates across the COLUMNS of the input. So if the argument to for is a COLUMN vector, then the for loop runs for only one step. If it is a matrix, then the result will be a sequence of vectors!
% A 3x3 matrix, so there will be THREE iterations.
% n will be each of the columns of nvals in turn.
nvals = magic(3)
nvals =
8 1 6
3 5 7
4 9 2
for n = nvals
disp('A column')
disp(n)
end
A column
8
3
4
A column
1
5
9
A column
6
7
2

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