I have a vector containing the values 0, 1, 2 and 3. What I want to do is take the lower two bits from each set of 16 elements drawn from this vector and append them all together to get one uint32. Anyone know an easy way to do this?
Follow-up: What if the number of elements in the vector isn't an integer multiple of 16?
Here's a vectorized version:
v = floor(rand(64,1)*4);
nWord = size(v,1)/16;
sum(reshape([bitget(v,2) bitget(v,1)]',[32 nWord]).*repmat(2.^(31:(-1):0)',[1 nWord ]))
To refine what was suggested by Jacob in his answer and mtrw in his comment, here's the most succinct version I can come up with (given a 1-by-N variable vec containing the values 0 through 3):
value = uint32(vec(1:16)*4.^(0:15)');
This treats the first element in the array as the least-significant bit in the result. To treat the first element as the most-significant bit, use the following:
value = uint32(vec(16:-1:1)*4.^(0:15)');
EDIT: This addresses the new revision of the question...
If the number of elements in your vector isn't a multiple of 16, then the last series of numbers you extract from it will have less than 16 values. You will likely want to pad the higher bits of the series with zeroes to make it a 16-element vector. Depending on whether the first element in the series is the least-significant bit (LSB) or most-significant bit (MSB), you will end up padding the series differently:
v = [2 3 1 1 3 1 2 2]; % A sample 8-element vector
v = [v zeros(1,8)]; % If v(1) is the LSB, set the higher bits to zero
% or...
v = [zeros(1,8) v]; % If v(1) is the MSB, again set the higher bits to zero
If you want to process the entire vector all at once, here is how you would do it (with any necessary zero-padding included) for the case when vec(1) is the LSB:
nValues = numel(vec);
nRem = rem(nValues,16);
vec = [vec(:) zeros(1,nRem)]; % Pad with zeroes
vec = reshape(vec,16,[])'; % Reshape to an N-by-16 matrix
values = uint32(vec*4.^(0:15)');
and when vec(1) is the MSB:
nValues = numel(vec);
nRem = rem(nValues,16);
vec = [vec(1:(nValues-nRem)) zeros(1,nRem) ...
vec((nValues-nRem+1):nValues)]; % Pad with zeroes
vec = reshape(vec,16,[])'; % Reshape to an N-by-16 matrix
values = uint32(fliplr(vec)*4.^(0:15)');
I think you should have a look at bitget and bitshift. It should be possible to be something like this (pseudo-matlab code as I haven't worked with Matlab for a long time):
result = 0;
for i = 1:16 do
result += bitshift(bitget(vector(i), 2:-1:1), 2);
Note that this will give you the last bits of the first vector in the highest bits, so you might want to descend i from 16 to 1 instead
Related
I would like to generate an array which contains all ordered samples of length k taken from a set of n elements {a_1,...,a_n}, that is all the k-tuples (x_1,...,x_k) where each x_j can be any of the a_i (repetition of elements is allowed), and whose total number is n^k.
Is there a built-in function in Matlab to obtain it?
I have tried to write a code that iteratively uses the datasample function, but I couldn't get what desired so far.
An alternative way to get all the tuples is based on k-base integer representation.
If you take the k-base representation of all integers from 0 to n^k - 1, it gives you all possible set of k indexes, knowing that these indexes start at 0.
Now, implementing this idea is quite straightforward. You can use dec2base if k is lower than 10:
X = A(dec2base(0:(n^k-1), k)-'0'+1));
For k between 10 and 36, you can still use dec2base but you must take care of letters as there is a gap in ordinal codes between '9' and 'A':
X = A(dec2base(0:(n^k-1), k)-'0'+1));
X(X>=17) = X(X>=17)-7;
Above 36, you must use a custom made code for retrieving the representation of the integer, like this one. But IMO you may not need this as 2^36 is quite huge.
What you are looking for is ndgrid: it generates the grid elements in any dimension.
In the case k is fixed at the moment of coding, get all indexes of all elements a this way:
[X_1, ..., X_k] = ndgrid(1:n);
Then build the matrix X from vector A:
X = [A(X_1(:)), ..., A(X_k(:))];
If k is a parameter, my advice would be to look at the code of ndgrid and adapt it in a new function so that the output is a matrix of values instead of storing them in varargout.
What about this solution, I don't know if it's as fast as yours, but do you think is correct?
function Y = ordsampwithrep(X,K)
%ordsampwithrep Ordered samples with replacement
% Generates an array Y containing in its rows all ordered samples with
% replacement of length K with elements of vector X
X = X(:);
nX = length(X);
Y = zeros(nX^K,K);
Y(1,:) = datasample(X,K)';
k = 2;
while k < nX^K +1
temprow = datasample(X,K)';
%checknew = find (temprow == Y(1:k-1,:));
if not(ismember(temprow,Y(1:k-1,:),'rows'))
Y(k,:) = temprow;
k = k+1;
end
end
end
I have a 3D matrix called mat. Every column may or may not comprise a variable number of leading zeros. I need to replace them with NaNs. It's important to recognize that there might follow even more zeros in any column after the occurence of the first non-zero elements. That is, just indexing ALL zeros in the matrix and replacing them with NaN won't lead to the correct result.
I do have a working solution. However, it contains two for-loops. I am wondering whether it's possible to vectorize and get rid of the loop. In reality, mat could be very big, something like 10000x15x10000. Therefore, I am quite sensitive to execution speed.
Here's my toy example:
% Create test matrix
mat = randi(100,20,5,2);
mat(1:5,1,1) = 0;
mat(1:7,2,1) = 0;
mat(1:3,4,1) = 0;
mat(1:10,5,1) = 0;
mat(1:2,1,2) = 0;
mat(1:3,3,2) = 0;
mat(1:7,4,2) = 0;
mat(1:4,5,2) = 0;
% Find first non-zero element in every column
[~, firstNonZero] = max( mat ~= 0 );
% Replace leading zeros with NaN
% How to vectorize this part???
[nRows, nCols, nPlanes] = size(mat);
for j = 1 : nPlanes
for i = 1 : nCols
mat(1:firstNonZero(1, i, j)-1, i, j) = NaN;
end
end
You could use cumsum to create a cumulative sum down each column, then all leading zeros have a cumulative sum of zero whilst all intermediate zeros have a cumulative sum greater than zero...
mat( cumsum(mat,1) == 0 ) = NaN;
As suggested in the comments, if your mat has negative values then there's a chance the cumulative sum will be 0 later on... use the sum of absolute values instead
mat( cumsum(abs(mat),1) == 0 ) = NaN;
Note that by default, cumsum operates along the first non-singleton dimension, you can use the optional dim argument to specify the dimension. I've used dim=1 to enforce column-wise operation in case your mat could be of height 1, but this is the default for any matrix with height greater than 1.
Note this uses == for comparison, you may want to read Why is 24.0000 not equal to 24.0000 in MATLAB? and use a threshold for your equality comparison.
im new to matlab and i've run into a slight problem. I want to check my matrix that generated random number if they are divisible by 12. Then i want to list number of digit divisible by 12 and the total sum of those.
clc
clear
format compact
a=4
b=0
N=50+a
R=randi([100+a,159+b], 1, N) % generate random no. from 100+a to 159 on a matrix 1xN
s1=0
N1=0
for i = 1
for j= 1:N
if rem(R,12)==0
N1=N1+1;
s1=s1+R(i,j);
else
N1=N1+0;
s1=s1+0;
end
end
end
numberof1=N1
sum1=s1
Your code isn't working because your are calling rem(R, 12) (remainder of all elements) as opposed to the remainder of the specific element (rem(R(i,j), 12)).
The better approach though would be to remove the for loop and generate a logical matrix the size of R that is true when that number is divisible by 12 and false otherwise by passing the entire matrix to rem.
is_divisible_by_12 = rem(R, 12) == 0;
Then we can use this to compute the sum of these by using this logical array as an index into R
subset = R(is_divisible_by_12);
number = numel(subset);
s1 = sum(subset);
I need to calculate the frequency of each value in another vector in MATLAB.
I can use something like
for i=1:length(pdata)
gt(i)=length(find(pf_test(:,1)==pdata(i,1)));
end
But I prefer not to use loop because my dataset is quite large. Is there anything like histc (which is used to find the frequency of values in one vector) to find the frequency of one vector value in another vector?
If your values are only integers, you could do the following:
range = min(pf_test):max(pf_test);
count = histc(pf_test,range);
gt = count(ismember(range,a));
gt(~ismember(unique(a),b)) = 0;
If you can't guarantee that the values are integers, it's a bit more complicated. One possible method of it would be the following:
%restrict yourself to values that appear in the second vector
filter = ismember(pf_test,pdata);
% sort your first vector (ignore this if it is already sorted)
spf_test = sort(pf_test);
% Find the first and last occurrence of each element
[~,last] = unique(spf_test(filter));
[~,first] = unique(spf_test(filter),'first');
% Initialise gt
gt = zeros(length(pf_test));
% Fill gt
gt(filter) = (last-first)+1;
EDIT: Note that I may have got the vectors the wrong way around - if this doesn't work as expected, switch pf_test and pdata. It wasn't immediately clear to me which was which.
You mention histc. Why are you not using it (in its version with two input parameters)?
>> pdata = [1 1 3 2 3 1 4 4 5];
>> pf_test = 1:6;
>> histc(pdata,pf_test)
ans =
3 1 2 2 1 0
I have two vectors, idx1 and idx2, and I want to obtain the values between them. If idx1 and idx2 were numbers and not vectors, I could do that the following way:
idx1=1;
idx2=5;
values=idx1:idx2
% Result
% values =
%
% 1 2 3 4 5
But in my case, idx1 and idx2 are vectors of variable length. For example, for length=2:
idx1=[5,9];
idx2=[9 11];
Can I use the colon operator to directly obtain the values in between? This is, something similar to the following:
values = [5 6 7 8 9 9 10 11]
I know I can do idx1(1):idx2(1) and idx1(2):idx2(2), this is, extract the values for each column separately, so if there is no other solution, I can do this with a for-loop, but maybe Matlab can do this more easily.
Your sample output is not legal. A matrix cannot have rows of different length. What you can do is create a cell array using arrayfun:
values = arrayfun(#colon, idx1, idx2, 'Uniform', false)
To convert the resulting cell array into a vector, you can use cell2mat:
values = cell2mat(values);
Alternatively, if all vectors in the resulting cell array have the same length, you can construct an output matrix as follows:
values = vertcat(values{:});
Try taking the union of the sets. Given the values of idx1 and idx2 you supplied, run
values = union(idx1(1):idx1(2), idx2(1):idx2(2));
Which will yield a vector with the values [5 6 7 8 9 10 11], as desired.
I couldn't get #Eitan's solution to work, apparently you need to specify parameters to colon. The small modification that follows got it working on my R2010b version:
step = 1;
idx1 = [5, 9];
idx2 = [9, 11];
values = arrayfun(#(x,y)colon(x, step, y), idx1, idx2, 'UniformOutput', false);
values=vertcat(cell2mat(values));
Note that step = 1 is actually the default value in colon, and Uniform can be used in place of UniformOutput, but I've included these for the sake of completeness.
There is a great blog post by Loren called Vectorizing the Notion of Colon (:). It includes an answer that is about 5 times faster (for large arrays) than using arrayfun or a for-loop and is similar to run-length-decoding:
The idea is to expand the colon sequences out. I know the lengths of
each sequence so I know the starting points in the output array. Fill
the values after the start values with 1s. Then I figure out how much
to jump from the end of one sequence to the beginning of the next one.
If there are repeated start values, the jumps might be negative. Once
this array is filled, the output is simply the cumulative sum or
cumsum of the sequence.
function x = coloncatrld(start, stop)
% COLONCAT Concatenate colon expressions
% X = COLONCAT(START,STOP) returns a vector containing the values
% [START(1):STOP(1) START(2):STOP(2) START(END):STOP(END)].
% Based on Peter Acklam's code for run length decoding.
len = stop - start + 1;
% keep only sequences whose length is positive
pos = len > 0;
start = start(pos);
stop = stop(pos);
len = len(pos);
if isempty(len)
x = [];
return;
end
% expand out the colon expressions
endlocs = cumsum(len);
incr = ones(1, endlocs(end));
jumps = start(2:end) - stop(1:end-1);
incr(endlocs(1:end-1)+1) = jumps;
incr(1) = start(1);
x = cumsum(incr);