for loop
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Hi,I have created the following loop:
for i = 1:5 x = i*2 end
I wish to be able to record each value of the iteration i.e. 2,4,6,8,10 aswell as arrive at the final value of 10 which the above code dies.Could you tell me how to do this? Thanks
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More Answers (3)
Andrei Bobrov
on 9 Sep 2011
for i1 = 5:-1:1, x(i1) = i1*2;end
for your case
x = (1:5)*2
or
x = 2:2:10
3 Comments
Andrei Bobrov
on 9 Sep 2011
Hi Grzegorz! I begin with last number 'x' and this way preallocate memory. Please read this http://www.mathworks.com/matlabcentral/answers/7039-how-to-multiply-a-vector-with-each-column-of-a-matrix-most-efficiently
mohammad
on 9 Sep 2011
x=[];
for i = 1:5
x =[x;i*2]; %for column matrix
end
OR:
x=[];
for i = 1:5
x =[x,i*2]; %for row matrix
end
2 Comments
Andrei Bobrov
on 9 Sep 2011
Hi mohammad! Please read this http://www.mathworks.com/help/techdoc/matlab_prog/f8-784135.html#f8-793781
Jan
on 9 Sep 2011
This is working and fast for tiny vectors. But it is really worth to pre-allocate arrays to get familiar with good programming methods. Without pre-allocation, this happens:
1. x is created as empty DOUBLE vector. This needs about 100 bytes for the internal header and 0 bytes for the data.
2. Inside the loop another vector x is created, again 100 bytes header, 8 bytes for the data. The former x is released.
3. In the 2nd iteration the next vector is created, again 100 bytes header, 16 bytes of data, 8 bytes of the formar array are copied. The former x is released.
4. etc.
If the loop runs until N, a total of N*100+prod(1:N)*8 bytes are allocated, filled with zeros, partially released later and prod(1:N-1)*8 bytes are copied.
In consequence you apporach is horribly slow for e.g. 1000 elements and might even crash with an out-of-memory error for larger vectors. Therefore *always* pre-allocate.
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