circshift function working explanation needed

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Completely new to matlab. Studying some sample codes.
% bitget and num2str and circular shift
x = 0b10011010u8 % x is 10011010
value3= bitget(x, 8:-1:1) % x's binary representation is 10011010
formatSpec4= '%d'
s4= num2str(value3, formatSpec4);
s5= s4;
s5(1:4) = circshift(s5(1:4),-1);
s6= s5;
Not able to understand the syntax and functionality of circshift. Thank in advance
Please explain me the functionality of circshift.

Accepted Answer

Voss
Voss on 16 Jan 2022
circshift(s,d) for a vector s and positive integer d shifts the elements of s to the right by d amount, wrapping back to the beginning when they go off the end (thats why it's called circular). If d is negative, elements are shifted to the left by -d amount.
It's easier to see what it does with a few examples in MATLAB rather than a few sentences in English.
s = [10 20 30 40 50] % original vector s
s = 1×5
10 20 30 40 50
circshift(s,1) % shift s right by 1, last element wraps around to become first
ans = 1×5
50 10 20 30 40
circshift(s,-1) % shift s left by 1, first element wraps around to become last
ans = 1×5
20 30 40 50 10
circshift(s,2) % right by 2
ans = 1×5
40 50 10 20 30
circshift(s,-4) % left by 4
ans = 1×5
50 10 20 30 40
circshift(s,5) % shifting by the length of the vector returns the original vector
ans = 1×5
10 20 30 40 50
In your code, the first 4 elements of s5 are shifted left by one place and stored back into s5 as the first 4 elements (and the remaining elements are untouched):
s5 = [1 2 3 4 5 6 7 8]
s5 = 1×8
1 2 3 4 5 6 7 8
s5(1:4) = circshift(s5(1:4),-1)
s5 = 1×8
2 3 4 1 5 6 7 8

More Answers (1)

Walter Roberson
Walter Roberson on 16 Jan 2022
M = (10:10:40).' + (1:9)
M = 4×9
11 12 13 14 15 16 17 18 19 21 22 23 24 25 26 27 28 29 31 32 33 34 35 36 37 38 39 41 42 43 44 45 46 47 48 49
circshift(M, -1)
ans = 4×9
21 22 23 24 25 26 27 28 29 31 32 33 34 35 36 37 38 39 41 42 43 44 45 46 47 48 49 11 12 13 14 15 16 17 18 19
You can see that this is the same as
[M(2:end,:); M(1,:)]
And more generally, circshift(M, -K) would be
[M(K+1:end,:); M(1:K,:)]
  2 Comments
Walter Roberson
Walter Roberson on 16 Jan 2022
For vectors:
M = 1 : 9
M = 1×9
1 2 3 4 5 6 7 8 9
circshift(M, -1)
ans = 1×9
2 3 4 5 6 7 8 9 1
which is [M(2:end),M(1)] .
When you specify a scalar for the shift, then circshift operates on the first non-singleton dimension.

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