I have recorded data containing a vector of bit-sequences, which I would like to re-arrange efficiently. One value in the vector of data could look like this:
bit0, bit1, bit2, ... bit7
I would like to re-arrange this bit-sequence into this order:
bit0, bit7, bit1, bit6, bit2, bit5, bit3, bit4
If I had only one value this would work nicely via:
sum(uint32(bitset(0,1:8,bitget(uint32(X), [1 8 2 7 3 6 4 5]))))
Unfortunately bitset and bitget are not capable of handling vectors of bit-sequences. Since I have a fairly large dataset I am interested in efficient solutions.
Any help would be appreciated, thanks!
dec2bin and bin2dec can process vectors, you can input all numbers at once and permute the matrix:
input=1:23;
pattern = [1 8 2 7 3 6 4 5];
bit=dec2bin(input(:),numel(pattern));
if size(bit,2)>numel(pattern)
warning('input numbers to large for pattern, leading bits will be cut off')
end
output=bin2dec(bit(:,pattern));
if available, I would use de2bi and bi2de instead.
I don't know if I may get the question wrong, but isn't it just solvable by indexing wrapped into cellfun?
%// example data
BIN{1} = dec2bin(84,8)
BIN{2} = dec2bin(42,8)
%// pattern and reordering
pattern = [1 8 2 7 3 6 4 5];
output = cellfun(#(x) x(pattern), BIN, 'uni', 0)
Or what is the format of you input and desired output?
BIN =
'01010100' '00101010'
output =
'00100110' '00011001'
The most efficient way is probably to use bitget and bitset as you did in your question, although you only need an 8-bit integer. Suppose you have a uint8 array X which describes your recorded data (for the example below, X = uint8([169;5]), for no particular reason. We can inspect the bits by creating a useful anonymous function:
>> dispbits = #(W) arrayfun(#(X) disp(bitget(X,1:8)),W)
>> dispbits =
#(W)arrayfun(#(X)disp(bitget(X,1:8)),W)
>> dispbits(X)
1 0 0 1 0 1 0 1
1 0 1 0 0 0 0 0
and suppose you have some pattern pattern according to which you want to reorder the bits stored in this vector of integers:
>> pattern
pattern =
1 8 2 7 3 6 4 5
You can use arrayfun and find to reorder the bits according to pattern:
Y = arrayfun(#(X) uint8(sum(bitset(uint8(0),find(bitget(X,pattern))))), X)
Y =
99
17
We get the desired answer stored efficiently in a vector of 8 bit integers:
>> class(Y)
ans =
uint8
>> dispbits(Y)
1 1 0 0 0 1 1 0
1 0 0 0 1 0 0 0
Related
If there is a vector like this,
T = [1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16]
(the size of vector T can be flexible)
How can I get a array of 'sum of divisions'?
For example,
fn(T, 5) = [ (1+2+3+4+5) , (6+7+8+9+10), (11+12+13+14+15) , 16]
One option, which doesn't require the padding of zeros on the original array, is the use of accumarray and ceil:
div = 5;
out = accumarray(ceil((1:numel(T))/div).',T(:))
Another option using cumsum and diff instead:
div = 5;
T(ceil(numel(T)/div)*div) = 0;
cs = cumsum(T)
out = diff( [0 cs(div:div:end) ] )
Edit: once the padding is done, cumsum and diff are a little overkill and one should proceed as in Bentoy's answer.
Another way, close to the 2nd option of thewaywewalk:
div = 5;
T(ceil(numel(T)/div)*div) = 0;
out = sum(reshape(T,div,[])).'; % transpose if you really want a column vector
Also, one one-liner solution (I prefer this one):
out = blockproc(T,[1 5], #(blk) sum(blk.data), 'PadPartialBlocks',true);
Don't forget to set the parameter 'PadPartialBlocks', this is the key of avoiding explicit padding.
There is an in-built function vec2mat in Communications System Toolbox to convert a vector into a 2D matrix that cuts off after every N elements and puts into separate rows, padding the leftover places at the end with zeros to maintain 2D size . So, after using vec2mat, summing all the rows would be enough to give you the desired output. Here's the implementation -
sum(vec2mat(T,5),2)
Sample run -
>> T = 1:16;
>> vec2mat(T,5)
ans =
1 2 3 4 5
6 7 8 9 10
11 12 13 14 15
16 0 0 0 0
>> sum(vec2mat(T,5),2)
ans =
15
40
65
16
How can I find all the cells that have the same values in a multi-dimensional array?
I can get it partially to work with result=A(:,:,1)==A(:,:,2) but I'm not sure how to also include A(:,:,3)
I tried result=A(:,:,1)==A(:,:,2)==A(:,:,3) but the results come back as all 0 when there should be 1 correct answer
which is where the number 8 is located in the same cell on all the pages of the array. Note: this is just a test the repeating number could be found multiple times and as different numbers.
PS: I'm using octave 3.8.1 which is like matlab
See code below:
clear all, tic
%graphics_toolkit gnuplot %use this for now it's older but allows zoom
A(:,:,1)=[1 2 3; 4 5 6; 7 9 8]; A(:,:,2)=[9 1 7; 6 5 4; 7 2 8]; A(:,:,3)=[2 4 6; 8 9 1; 3 5 8]
[i j k]=size(A)
for ii=1:k
maxamp(ii)=max(max(A(:,:,ii)))
Ainv(:,:,ii)=abs(A(:,:,ii)-maxamp(ii));%the extra max will get the max value of all values in array
end
%result=A(:,:,1)==A(:,:,2)==A(:,:,3)
result=A(:,:,1)==A(:,:,2)
result=double(result); %turns logical index into double to do find
[row col page] = find(result) %gives me the col, row, page
This is the output it gives me:
>>>A =
ans(:,:,1) =
1 2 3
4 5 6
7 9 8
ans(:,:,2) =
9 1 7
6 5 4
7 2 8
ans(:,:,3) =
2 4 6
8 9 1
3 5 8
i = 3
j = 3
k = 3
maxamp = 9
maxamp =
9 9
maxamp =
9 9 9
result =
0 0 0
0 1 0
1 0 1
row =
3
2
3
col =
1
2
3
page =
1
1
1
Use bsxfun(MATLAB doc, Octave doc) and check to see if broadcasting the first slice is equal across all slices with a call to all(MATLAB doc, Octave doc):
B = bsxfun(#eq, A, A(:,:,1));
result = all(B, 3);
If we're playing code golf, a one liner could be:
result = all(bsxfun(#eq, A, A(:,:,1)), 3);
The beauty of the above approach is that you can have as many slices as you want in the third dimension, other than just three.
Example
%// Your data
A(:,:,1)=[1 2 3; 4 5 6; 7 9 8];
A(:,:,2)=[9 1 7; 6 5 4; 7 2 8];
A(:,:,3)=[2 4 6; 8 9 1; 3 5 8];
B = bsxfun(#eq, A, A(:,:,1));
result = all(B, 3);
... gives us:
>> result
result =
0 0 0
0 0 0
0 0 1
The above makes sense since the third row and third column for all slices is the only value where every slice shares this same value (i.e. 8).
Here's another approach: compute differences along third dimension and detect when all those differences are zero:
result = ~any(diff(A,[],3),3);
You can do
result = A(:,:,1) == A(:,:,2) & A(:,:,1) == A(:,:,3);
sum the elements along the third dimension and divide it with the number of dimensions. We get back the original value if the values are the same in all dimension. Otherwise a different (e.g. a decimal) value. Then find the location where A and the summation are equal over the third dimension.
all( A == sum(A,3)./size(A,3),3)
ans =
0 0 0
0 0 0
0 0 1
or
You could also do
all(A==repmat(sum(A,3)./size(A,3),[1 1 size(A,3)]),3)
where repmat(sum(A,3)./size(A,3),[1 1 size(A,3)]) would highlight the implicit broadcasting of this when compared with A.
or
you skip the broadcasting altogether and just compare it with the first slice of A
A(:,:,1) == sum(A,3)./size(A,3)
Explanation
3 represents the third dimension .
sum(A,3) means that we are taking the sum over the third dimension.
Then we divide that sum by the number of dimensions.
It's basically the average value for that position in the third dimension.
If you add three values and then divide it by three then you get the original value back.
For example, A(3,3,:) is [8 8 8]. (8+8+8)/3 = 8.
If you take another example, i.e. the value above, A(2,3,:) = [6 4 1].
Then (6+4+1)/3=3.667. This is not equal to A(2,3,:).
sum(A,3)./size(A,3)
ans =
4.0000 2.3333 5.3333
6.0000 6.3333 3.6667
5.6667 5.3333 8.0000
Therefore, we know that the elements are not the same
throughout the third dimension. This is just a trick I use
to determine that. You also have to remember that
sum(A,3)./size(A,3) is originally a 3x3x1 matrix
that will be automatically expanded (i.e. broadcasted) to a
3x3x3 matrix when we do the comparison with A (A == sum(A,3)./size(A,3)).
The result of that comparison will be a logical array with 1 for the positions that are the same throughout the third dimension.
A == sum(A,3)./size(A,3)
ans =
ans(:,:,1) =
0 0 0
0 0 0
0 0 1
ans(:,:,2) =
0 0 0
1 0 0
0 0 1
ans(:,:,3) =
0 0 0
0 0 0
0 0 1
Then use all(....,3) to get those. The result is a 3x3x1
matrix where a 1 indicates that the value is the same in the
third dimension.
all( A == sum(A,3)./size(A,3),3)
ans =
0 0 0
0 0 0
0 0 1
I'm trying to create a script to solve my problem, but I got stuck in one place.
So I have imported .txt file with 4x1 sized matrix (simplified to give an example in my case it might be 1209x1 matrix) which contains some coordinate X. And it's look like this:
0
1
2
3
That's coordinates for one period, and I need to get one column for different number of periods N . Each period is the same and lenght=L
So you can do it manually by this script, for example for N=3 periods:
X=[X; X+L; X+2*L];
so for example if L=3
then i will get
0
1
2
3
3
4
5
6
6
7
8
9
it works well but it's not efficient in case if I need to work with number of periods let's say N=1000 or if I need to change their number quickly. Any solution to parameterize this operation so I can just put number for N and get X for N periods?
Thanks and Regards
I don't have MATLAB on this machine so I can't test, but the most straightforward implementation would be something like:
n = 1000;
L = 3;
nvalues = length(X); % Assuming X is your initial vector
newx = zeros(n*nvalues, 1); % Preallocate new array
for ii = 0:(n-1)
startidx = (nvalues*ii) + 1;
endidx = nvalues*(ii+1);
newx(startidx:endidx) = X + ii*L
end
You can use bsxfun to create X, X+L, X+2*L, ... and then reshape it to a vector
>> F=bsxfun(#plus, X, (0:(N-1))*L)
F =
0 3 6
1 4 7
2 5 8
3 6 9
>> X=F(:)
X =
0
1
2
3
3
4
5
6
6
7
8
9
or in a more concise form:
>> X=reshape(bsxfun(#plus, X, (0:(N-1))*L), [], 1)
X =
0
1
2
3
3
4
5
6
6
7
8
9
In MATLAB, say I have a set of square matrices, say A, with trace(A)=0 as follows:
For example,
A = [0 1 2; 3 0 4; 5 6 0]
How can I remove the zeros and then vertically collapse the matrix to become as follow:
A_reduced = [1 2; 3 4; 5 6]
More generally, what if the zeroes can appear anywhere in the column (i.e., not necessarily at the long diagonal)? Assuming, of course, that the total number of zeros for all columns are the same.
The matrix can be quite big (hundreds x hundreds in dimension). So, an efficient way will be appreciated.
To compress the matrix vertically (assuming every column has the same number of zeros):
A_reduced_v = reshape(nonzeros(A), nnz(A(:,1)), []);
To compress the matrix horizontally (assuming every row has the same number of zeros):
A_reduced_h = reshape(nonzeros(A.'), nnz(A(1,:)), []).';
Case #1
Assuming that A has equal number of zeros across all rows, you can compress it horizontally (i.e. per row) with this -
At = A' %//'# transpose input array
out = reshape(At(At~=0),size(A,2)-sum(A(1,:)==0),[]).' %//'# final output
Sample code run -
>> A
A =
0 3 0 2
3 0 0 1
7 0 6 0
1 0 6 0
0 16 0 9
>> out
out =
3 2
3 1
7 6
1 6
16 9
Case #2
If A has equal number of zeros across all columns, you can compress it vertically (i.e. per column) with something like this -
out = reshape(A(A~=0),size(A,1)-sum(A(:,1)==0),[]) %//'# final output
Sample code run -
>> A
A =
0 3 7 1 0
3 0 0 0 16
0 0 6 6 0
2 1 0 0 9
>> out
out =
3 3 7 1 16
2 1 6 6 9
This seems to work, quite fiddly to get the behaviour right with transposing:
>> B = A';
>> C = B(:);
>> reshape(C(~C==0), size(A) - [1, 0])'
ans =
1 2
3 4
5 6
As your zeros are always in the main diagonal you can do the following:
l = tril(A, -1);
u = triu(A, 1);
out = l(:, 1:end-1) + u(:, 2:end)
A correct and very simple way to do what you want is:
A = [0 1 2; 3 0 4; 5 6 0]
A =
0 1 2
3 0 4
5 6 0
A = sort((A(find(A))))
A =
1
2
3
4
5
6
A = reshape(A, 2, 3)
A =
1 3 5
2 4 6
I came up with almost the same solution as Mr E's though with another reshape command. This solution is more universal, as it uses the number of rows in A to create the final matrix, instead of counting the number of zeros or assuming a fixed number of zeros..
B = A.';
B = B(:);
C = reshape(B(B~=0),[],size(A,1)).'
This question already has answers here:
Repeat copies of array elements: Run-length decoding in MATLAB
(5 answers)
Closed 8 years ago.
My question is similar to this one, but I would like to replicate each element according to a count specified in a second array of the same size.
An example of this, say I had an array v = [3 1 9 4], I want to use rep = [2 3 1 5] to replicate the first element 2 times, the second three times, and so on to get [3 3 1 1 1 9 4 4 4 4 4].
So far I'm using a simple loop to get the job done. This is what I started with:
vv = [];
for i=1:numel(v)
vv = [vv repmat(v(i),1,rep(i))];
end
I managed to improve by preallocating space:
vv = zeros(1,sum(rep));
c = cumsum([1 rep]);
for i=1:numel(v)
vv(c(i):c(i)+rep(i)-1) = repmat(v(i),1,rep(i));
end
However I still feel there has to be a more clever way to do this... Thanks
Here's one way I like to accomplish this:
>> index = zeros(1,sum(rep));
>> index(cumsum([1 rep(1:end-1)])) = 1;
index =
1 0 1 0 0 1 1 0 0 0 0
>> index = cumsum(index)
index =
1 1 2 2 2 3 4 4 4 4 4
>> vv = v(index)
vv =
3 3 1 1 1 9 4 4 4 4 4
This works by first creating an index vector of zeroes the same length as the final count of all the values. By performing a cumulative sum of the rep vector with the last element removed and a 1 placed at the start, I get a vector of indices into index showing where the groups of replicated values will begin. These points are marked with ones. When a cumulative sum is performed on index, I get a final index vector that I can use to index into v to create the vector of heterogeneously-replicated values.
To add to the list of possible solutions, consider this one:
vv = cellfun(#(a,b)repmat(a,1,b), num2cell(v), num2cell(rep), 'UniformOutput',0);
vv = [vv{:}];
This is much slower than the one by gnovice..
What you are trying to do is to run-length decode. A high level reliable/vectorized utility is the FEX submission rude():
% example inputs
counts = [2, 3, 1];
values = [24,3,30];
the result
rude(counts, values)
ans =
24 24 3 3 3 30
Note that this function performs the opposite operation as well, i.e. run-length encodes a vector or in other words returns values and the corresponding counts.
accumarray function can be used to make the code work if zeros exit in rep array
function vv = repeatElements(v, rep)
index = accumarray(cumsum(rep)'+1, 1);
vv = v(cumsum(index(1:end-1))+1);
end
This works similar to solution of gnovice, except that indices are accumulated instead being assigned to 1. This allows to skip some indices (3 and 6 in the example below) and remove corresponding elements from the output.
>> v = [3 1 42 9 4 42];
>> rep = [2 3 0 1 5 0];
>> index = accumarray(cumsum(rep)'+1, 1)'
index =
0 0 1 0 0 2 1 0 0 0 0 2
>> cumsum(index(1:end-1))+1
ans =
1 1 2 2 2 4 5 5 5 5 5
>> vv = v(cumsum(index(1:end-1))+1)
vv =
3 3 1 1 1 9 4 4 4 4 4