I am trying to calculate the Hamming weight of a vector in Matlab.
function Hamming_weight (vet_dec)
Ham_Weight = sum(dec2bin(vet_dec) == '1')
endfunction
The vector is:
Hamming_weight ([208 15 217 252 128 35 50 252 209 120 97 140 235 220 32 251])
However, this gives the following result, which is not what I want:
Ham_Weight =
10 10 9 9 9 5 5 7
I would be very grateful if you could help me please.
You are summing over the wrong dimension!
sum(dec2bin(vet_dec) == '1',2).'
ans =
3 4 5 6 1 3 3 6 4 4 3 3 6 5 1 7
dec2bin(vet_dec) creates a matrix like this:
11010000
00001111
11011001
11111100
10000000
00100011
00110010
11111100
11010001
01111000
01100001
10001100
11101011
11011100
00100000
11111011
As you can see, you're interested in the sum of each row, not each column. Use the second input argument to sum(x, 2), which specifies the dimension you want to sum along.
Note that this approach is horribly slow, as you can see from this question.
EDIT
For this to be a valid, and meaningful MATLAB function, you must change your function definition a bit.
function ham_weight = hamming_weight(vector) % Return the variable ham_weight
ham_weight = sum(dec2bin(vector) == '1', 2).'; % Don't transpose if
% you want a column vector
end % endfunction is not a MATLAB command.
Related
i´m doing the statistical evaluation for my master´s thesis. the levene test was significant so i did the welch anova which was significant. now i tried the games-howell post hoc test but it didn´t work.
can anybody help me sending me the exact functions which i have to run in R to do the games-howell post hoc test and to get kind of a compact letter display, where it shows me which treatments are not significantly different from each other? i also wanted to ask if i did the welch anova the right way (you can find the output of R below)
here it the output which i did till now for the statistical evalutation:
data.frame': 30 obs. of 3 variables:
$ Dauer: Factor w/ 6 levels "0","2","4","6",..: 1 2 3 4 5 6 1 2 3 4 ...
$ WH : Factor w/ 5 levels "r1","r2","r3",..: 1 1 1 1 1 1 2 2 2 2 ...
$ TSO2 : num 107 86 98 97 88 95 93 96 96 99 ...
> leveneTest(TSO2~Dauer, data=TSO2R)
`Levene's Test for Homogeneity of Variance (center = median)
Df F value Pr(>F)
group 5 3.3491 0.01956 *
24
Signif. codes: 0 ‘’ 0.001 ‘’ 0.01 ‘’ 0.05 ‘.’ 0.1 ‘ ’ 1`
`> oneway.test (TSO2 ~Dauer, data=TSO2R, var.equal = FALSE) ###Welch-ANOVA
One-way analysis of means (not assuming equal variances)
data: TSO2 and Dauer
F = 5.7466, num df = 5.000, denom df = 10.685, p-value = 0.00807
'''`
Thank you very much!
Suppose I have the following script, which constructs a symbolic array, A_known, and a symbolic vector x, and performs a matrix multiplication.
clc; clearvars
try
pkg load symbolic
catch
error('Symbolic package not available!');
end
syms V_l k s0 s_mean
N = 3;
% Generate left-hand-side square matrix
A_known = sym(zeros(N));
for hI = 1:N
A_known(hI, 1:hI) = exp(-(hI:-1:1)*k);
end
A_known = A_known./V_l;
% Generate x vector
x = sym('x', [N 1]);
x(1) = x(1) + s0*V_l;
% Matrix multiplication to give b vector
b = A_known*x
Suppose A_known was actually unknown. Is there a way to deduce it from b and x? If so, how?
Til now, I only had the case where x was unknown, which normally can be solved via x = b \ A.
Mathematically, it is possible to get a solution, but it actually has infinite solutions.
Example
A = magic(5);
x = (1:5)';
b = A*x;
A_sol = b*pinv(x);
which has
>> A
A =
17 24 1 8 15
23 5 7 14 16
4 6 13 20 22
10 12 19 21 3
11 18 25 2 9
but solves A as A_sol like
>> A_sol
A_sol =
3.1818 6.3636 9.5455 12.7273 15.9091
3.4545 6.9091 10.3636 13.8182 17.2727
4.4545 8.9091 13.3636 17.8182 22.2727
3.4545 6.9091 10.3636 13.8182 17.2727
3.1818 6.3636 9.5455 12.7273 15.9091
Out of bound error occured.
This is Octave language.
for ii=1:1:10
m(ii)=ii*8
q=m(ii)
if (ii>=2)
q(ii).xdot=(q(ii).x-q(ii-1).x)/Ts;
end
end
But error says
q(2): out of bound 1
How can I fixed it?
For this type of assignment you do not need a loop and anyway you need to define Ts.
To calculate differential increase you can use diff
x=(1:1:10)*8
x =
8 16 24 32 40 48 56 64 72 80
octave:5> Ts=2
Ts = 2
octave:6> xdot=diff(x)/Ts
xdot =
4 4 4 4 4 4 4 4 4
octave:7> size(x)
ans =
1 10
octave:8> size(xdot)
ans =
1 9
I'm a CS freshman and I find the division way of finding a binary number to be a pain. Is it possible to use log to quickly find 24, for instance, in binary?
If you want to use logarithms, you can.
Define log2(b) as log(b) / log(2) or ln(b) / ln(2) (they are the same).
Repeat the following:
Define n as the integer part of log2(b). There is a 1 in the nth position in the binary representation of b.
Set b = b - 2n
Repeat first step until b = 0.
Worked example: Converting 2835 to binary
log2(2835) = 11.47.. => n = 11
The binary representation has a 1 in the 211 position.
2835 - (211 = 2048) = 787
log2(787) = 9.62... => n = 9
The binary representation has a 1 in the 29 position.
787 - (29 = 512) = 275
log2(275) = 8.10... => n = 8
The binary representation has a 1 in the 28 position.
275 - (28 = 256) = 19
log2(19) = 4.25... => n = 4
The binary representation has a 1 in the 24 position.
19 - (24 = 16) = 3
log2(3) = 1.58.. => n = 1
The binary representation has a 1 in the 21 position.
3 - (21 = 2) = 1
log2(1) = 0 => n = 0
The binary representation has a 1 in the 20 position.
We know the binary representation has 1s in the 211, 29, 28, 24, 21, and 20 positions:
2^ 11 10 9 8 7 6 5 4 3 2 1 0
binary 1 0 1 1 0 0 0 1 0 0 1 1
so the binary representation of 2835 is 101100010011.
From a CS perspective, binary is quite easy because you usually only need to go up to 255. Or 15 if using HEX notation. The more you use it, the easier it gets.
How I do it on the fly, is by remembering all the 2 powers up to 128 and including 1. (The presence of the 1 instead of 1.4xxx possibly means that you can't use logs).
128,64,32,16,8,4,2,1
Then I use the rule that if the number is bigger than each power in descending order, that is a '1' and subtract it, else it's a '0'.
So 163
163 >= 128 = '1' R 35
35 !>= 64 = '0'
35 >= 32 = '1' R 3
3 !>= 16 = '0'
3 !>= 8 = '0'
3 !>= 4 = '0'
3 >= 2 = '1' R 1
1 >= 1 = '1' R 0
163 = 10100011.
It may not be the most elegant method, but when you just need to convert something ad-hoc thinking of it as comparison and subtraction may be easier than division.
Yes, you have to loop through 0 -> power which is bigger than you need and then take the remainder and do the same, which is a pain too.
I would suggest you trying recursion approach of division called 'Divide and Conquer'.
http://web.stanford.edu/class/archive/cs/cs161/cs161.1138/lectures/05/Small05.pdf
But again, since you need a binary representation, I guess unless you use ready utils, division approach is the simplest one IMHO.
This is an example in Section 6.3.1 Comma Separated Lists Generated from Cell Arrays of the Octave documentation (I browsed it through the doc command on the Octave prompt) which I don't quite understand.
in{1} = [10, 20, 30, 40, 50, 60, 70, 80, 90];
in{2} = inf;
in{3} = "last";
in{4} = "first";
out = cell(4, 1);
[out{1:3}] = find(in{1 : 3}); % line which I do not understand
So at the end of this section, we have in looking like:
in =
{
[1,1] =
10 20 30 40 50 60 70 80 90
[1,2] = Inf
[1,3] = last
[1,4] = first
}
and out looking like:
out =
{
[1,1] =
1 1 1 1 1 1 1 1 1
[2,1] =
1 2 3 4 5 6 7 8 9
[3,1] =
10 20 30 40 50 60 70 80 90
[4,1] = [](0x0)
}
Here, find is called with 3 output parameters (forgive me if I'm wrong on calling them output parameters, I am pretty new to Octave) from [out{1:3}], which represents the first 3 empty cells of the cell array out.
When I run find(in{1 : 3}) with 3 output parameters, as in:
[i,j,k] = find(in{1 : 3})
I get:
i = 1 1 1 1 1 1 1 1 1
j = 1 2 3 4 5 6 7 8 9
k = 10 20 30 40 50 60 70 80 90
which kind of explains why out looks like it does, but when I execute in{1:3}, I get:
ans = 10 20 30 40 50 60 70 80 90
ans = Inf
ans = last
which are the 1st to 3rd elements of the in cell array.
My question is: Why does find(in{1 : 3}) drop off the 2nd and 3rd entries in the comma separated list for in{1 : 3}?
Thank you.
The documentation for find should help you answer your question:
When called with 3 output arguments, find returns the row and column indices of non-zero elements (that's your i and j) and a vector containing the non-zero values (that's your k). That explains the 3 output arguments, but not why it only considers in{1}. To answer that you need to look at what happens when you pass 3 input arguments to find as in find (x, n, direction):
If three inputs are given, direction should be one of "first" or
"last", requesting only the first or last n indices, respectively.
However, the indices are always returned in ascending order.
so in{1} is your x (your data if you want), in{2} is how many indices find should consider (all of them in your case since in{2} = Inf) and {in3}is whether find should find the first or last indices of the vector in{1} (last in your case).