Take this example from the mathworks help of nanmean():
X = magic(3);
X([1 6:9]) = repmat(NaN,1,5)
X =
NaN 1 NaN
3 5 NaN
4 NaN NaN
>> y = nanmean(X,2)
??? Error using ==> nanmean
Too many input arguments.
Why is it showing error even when the docs say the mean can be taken in any dimension dim of X as y = nanmean(X,dim)? Thanks.
I run exactly the code you have and I get no error. In particlar here is what I ran:
>> X = magic(3);
X([1 6:9]) = repmat(NaN,1,5)
X =
NaN 1 NaN
3 5 NaN
4 NaN NaN
>> y = nanmean(X,2)
y =
1
4
4
>> which nanmean
C:\Program Files\MATLAB\R2010b\toolbox\stats\stats\nanmean.m
The only thing I can think of is that you have a different version of nanmean.m on your path. Try a which nanmean and see if it points into the stats toolbox.
here is the reason:
If X contains a vector of all NaN values along some dimension, the vector is empty once the NaN values are removed, so the sum of the remaining elements is 0. Since the mean involves division by 0, its value is NaN. The output NaN is not a mean of NaN values.
Look at:
http://www.mathworks.com/help/toolbox/stats/nanmean.html
Related
I have this function
where f has values (18 19 20 21 22), and I should compute the value of the function for each value of f and plot each value.
I try to make f as vector 18:22, but it gives result from 1 to 22. Same result when I use for loop. My code is below, how can I modify it to take the values within the range only?
clc
fc=20;
theta=80;
N=16;
f=18:22;
g_m(f)=(sin((N*pi/2).*sin(theta).*(f/fc-1)))./sqrt(N).*(sin(pi/2).*(f/fc-1));
g_p(f)=exp(1j*0.5*(N-1).*pi*sin(theta).*(f/fc-1));
gain(f)=g_m(f).*g_p(f);
figure(1);
plot(f,g_m(f));
I believe this should give you enough information:
f = 5:7;
g(f)= [2, 2, 5]
g =
0 0 0 0 2 2 5
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
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.
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).
I am using the interp1 function to resample (a,b) to (new_a, new_b).
I get NAN for the values of new_b.
a, b are row matrices.
Some values in b are zero. How to get rid of this? Because of NAN I am unable to generate the correct plot for plot(new_a, new_b, 'r*)
c = cat(1,a, b);
[s,i] = sort(c(1,:)); #Sort by the 1st row
sort_ab = c(:,i);
sort_a = sort_ab(1,:);
sort_b = sort_ab(2,:);
new_a = min(sort_a):0.001:max(sort_a);
new_b = interp1(a, b, new_a);
From http://www.mathworks.com/help/techdoc/ref/interp1.html:
For the 'nearest', 'linear', and
'v5cubic' methods,
interp1(x,Y,xi,method) returns NaN for
any element of xi that is outside the
interval spanned by x.
Note that 'linear' is the default interpolation method.
You haven't provided us with the values of a and b, but the above sounds the most likely explanation.