I have a matrix having rows with repeated numbers.
A= [ 2 3 6;
4 7 4;
8 7 2;
1 3 1;
7 8 2 ]
The codes below find those rows and replace them with a Dummy_row [1 2 3]
new_A=[ 2 3 6;
1 2 3;
8 7 2;
1 2 3;
7 8 2 ]
This are the codes:
CODE NUMBER 1 (#Bruno)
Dummy_row = [1 2 3];
b = any(~diff(sort(A,2),1,2),2);
A(b,:) = repmat(Dummy_row,sum(b),1)
CODE NUMBER 2 (#Kamtal)
Dummy_row = [1 2 3];
b = diff(sort(A,2),1,2);
b = sum(b == 0,2);
b = b > 0;
c = repmat(Dummy_row,sum(b),1);
b = b' .* (1:length(b));
b = b(b > 0);
newA = A;
newA(b,:) = c
Note: both codes Number 1 and 2 perform the task efficiently.
Question
How can this code(either code num 1 or num 2) be modified such that it also replaces any rows having at least one zero with the Dummy_row?
Code 1
b = any(~diff(sort(A,2),1,2),2) | any(A==0,2); % <-- Only change
A(b,:) = repmat(Dummy_row,sum(b),1);
Code 2
b = diff(sort(A,2),1,2);
b = sum(b == 0,2);
b = (b > 0) | any(A==0,2); % <-- Only change
c = repmat(Dummy_row,sum(b),1);
b = b' .* (1:length(b));
b = b(b > 0);
newA = A;
newA(b,:) = c;
By the way: Code1 basically does the same thing that Code2 does, just that it uses logical-indexing instead of doing the unnecessary conversion from logical indexes to index positions.
Related
I'm trying to generate an n x n matrix like
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
where n = 5 or n 50. I'm at an impasse and can only generate a portion of the matrix. It is Problem 2.14 from Numerical Methods using MATLAB 3rd Edition by Penny and Lindfield. This is the best I have so far:
n = 5;
m = n;
A = zeros(m,n);
for i = 1:m
for j = 1:n
A(i,j) = m;
end
m = m - 1;
end
Any feedback is appreciated.
That was a nice brain-teaser, here’s my solution:
[x,y] = meshgrid(5:-1:1);
out = min(x,y)
Output:
ans =
5 4 3 2 1
4 4 3 2 1
3 3 3 2 1
2 2 2 2 1
1 1 1 1 1
Here's one loop-based approach:
n = 5;
m = n;
A = zeros(m, n);
for r = 1:m
for c = 1:n
A(r, c) = n+1-max(r, c);
end
end
And here's a vectorized approach (probably not faster, just for fun):
n = 5;
A = repmat(n:-1:1, n, 1);
A = min(A, A.');
That's one of the matrices in Matlab's gallery, except that it needs a 180-degree rotation, which you can achieve with rot90:
n = 5;
A = rot90(gallery('minij', n), 2);
Can someone help me to provide an efficient way or help me to perform the provide code to do make same results in minimal possible steps. I shall be grateful to you.
I have an Original Array:
A = [1 1 1 4.3 4.5 4 4.3 3 1 0 0 2 6.2 6.3 6 6.2 7.4 8 7.2 2 2 3 3 2];
Output Looks like:
A = [1 1 1 4 4 4 4 3 1 0 0 2 6 6 6 6 6 7 7 2 2 3 3 2];
I apply some restrictions and removed some values from array of local maxima’s after that I received some new arrays.
Yposlocfiltered = [6 15 23];
idx = [4 6 3];
Yneglocfiltered = [2 9 20];
idx_neg = [1 1 2];
Where I will find idx(local maxima value) I will check if values behind and ahead are greater make a window.
Example
If I will find 4 and 4.5, 4.3 is greater than 4 include it in backward window
4.3 is greater than 4 include it in forward window.
I need to find range of values behind local maxima and ahead local maxima.
I have tried to write a code that’s works fine but issue is that it’s too long.
Can someone please provide me an idea to perform this action in minimal steps and in faster ways?
I have only provided code for positive local maxima’s as for negative local maxima code Is just replica of this.
Code:only for positive local maximas
clc
clear all
A = [1 0 1 4.3 4.5 5 4.3 3 0 0 0 2 6.2 6.3 7 6.2 7.4 8 7.2 1 2 3 4 2];
Yposlocfiltered = [ 6 15 23];
idx = [4 6 3];
Yneglocfiltered = [2 9 20];
idx_neg = [1 1 2];
for b = 1: numel(idx)
for c = 1:numel(A)
f = Yposlocfiltered(1,b)-c ;
g = Yposlocfiltered(1,b)+c ;
% if (f > 0 && g <= numel(X))
if(f > 0)
if (A(Yposlocfiltered(1,b)-c))> idx(1,b)
else
d= f+1;
z(b)= d;
backward_window = z;
break
end
end
end
end
for r = 1: numel(idx)
for s = 1:numel(A)
u = Yposlocfiltered(1,r)-s ;
v = Yposlocfiltered(1,r)+s ;
% if (f > 0 && g <= numel(X))
if(v <=numel(A))
if (A(Yposlocfiltered(1,r)+s))> idx(1,r)
else
w= v-1;
y(r)= w;
forward_window = y;
break
end
end
end
end
n=4
for i=1:length(backward_window)
range = backward_window(i): forward_window(i);
p = range
if n <= numel(p)
p = range(1:n)
A( p) = idx(i);
else
% if (size(range)<= 3)
A( p) = idx(i);
end
end
From the first look at your code, I believe you can combine your first two for loops into one.
sA = numel(A);
sI = numel(idx);
for i = 1:sI
f = Yposlocfiltered(i) - [1:sA];
g = Yposlocfiltered(i) + [1:sA];
f(f<1) = [];
g(g>sA) = [];
backward_window(i) = f(find(A(f) <= idx(i), 1)) + 1;
forward_window(i) = g(find(A(g) <= idx(i), 1)) - 1;
end
Here, you can use find to locate the element of an array matching the specified condition, i.e. g <= numel(X) or A(f) <= idx(i).
Your last loop which modifies A can also be integrated into the same loop, so you can have:
sA = numel(A);
sI = numel(idx);
n=4;
for i = 1:sI
f = Yposlocfiltered(i) - [1:sA];
g = Yposlocfiltered(i) + [1:sA];
f(f<1) = [];
g(g>sA) = [];
backward_window(i) = f(find(A(f) <= idx(i), 1)) + 1;
forward_window(i) = g(find(A(g) <= idx(i), 1)) - 1;
range = backward_window(i) : forward_window(i);
if n <= numel(range)
A(range(1:n)) = idx(i);
else
A(range) = idx(i);
end
end
I have a matrix having rows with repeated numbers. I want to find those rows and replace them with a dummy row so as to keep the number of rows of the matrix constant.
Dummy_row = [1 2 3]
(5x3) Matrix A
A = [2 3 6;
4 7 4;
8 7 2;
1 3 1;
7 8 2]
(5x3) Matrix new_A
new_A = [2 3 6;
1 2 3;
8 7 2;
1 2 3;
7 8 2]
I tried the following which deleted the rows having repeated numbers.
y = [1 2 3]
w = sort(A,2)
v = all(diff(t,1,2)~=0|w(:,1:2)==0,2) % When v is zero, the row has repeated numbers
z = A(w,:)
Can you please help?
bsxfun based solution -
%// Create a row mask of the elements that are to be edited
mask = any(sum(bsxfun(#eq,A,permute(A,[1 3 2])),2)>1,3);
%// Setup output variable and set to-be-edited rows as copies of [1 2 3]
new_A = A;
new_A(mask,:) = repmat(Dummy_row,sum(mask),1)
Code run -
A =
2 3 6
4 7 4
8 7 2
1 3 1
7 8 2
new_A =
2 3 6
1 2 3
8 7 2
1 2 3
7 8 2
You could use the following:
hasRepeatingNums = any(diff(sort(A, 2), 1, 2)==0, 2);
A(hasRepeatingNums,:) = repmat(Dummy_row, nnz(hasRepeatingNums), 1);
See if this works for you,
A= [ 2 3 6;
4 7 4;
8 7 2;
5 5 5;
1 8 8;
1 3 1;
7 8 2 ];
Dummy_row = [1 2 3];
b = diff(sort(A,2),1,2);
b = sum(b == 0,2);
b = b > 0;
c = repmat(Dummy_row,sum(b),1);
b = b' .* (1:length(b));
b = b(b > 0);
newA = A;
newA(b,:) = c;
gives,
newA =
2 3 6
1 2 3
8 7 2
1 2 3
1 2 3
1 2 3
7 8 2
Edit
Not much change is needed, try this,
Dummy_row = [1 2 3];
b = sum(A == 0,2);
b = b > 0;
c = repmat(Dummy_row,sum(b),1);
b = b' .* (1:length(b));
b = b(b > 0);
newA = A;
newA(b,:) = c;
I am trying to match 1st column of A with 1st to 3rd columns of B, and append corresponding 4th column of B to A.
For example,
A=
1 2
3 4
B=
1 2 4 5 4
1 2 3 5 3
1 1 1 1 2
3 4 5 6 5
I compare A(:,1) and B(:, 1:3)
1 and 3 are in A(:,1)
1 is in the 1st, 2nd, 3rd rows of B(:, 1:3), so append B([1 2 3], 4:end)' to A's 1st row.
3 is in the 2nd and 4th rows of B(:,1:3), so append B([2 4], 4:end)' to A's 2nd row.
So that it becomes:
1 2 5 4 5 3 1 2
3 4 5 3 6 5 0 0
I could code this using only for and if.
clearvars AA A B mem mem2 mem3
A = [1 2 ; 3 4]
B = [1 2 4 5 4; 1 2 3 5 3; 1 1 1 1 2; 3 4 5 6 5]
for n=1:1:size(A,1)
mem = ismember(B(:,[1:3]), A(n,1));
mem2 = mem(:,1) + mem(:,2) + mem(:,3);
mem3 = find(mem2>0);
AA{n,:} = horzcat( A(n,:), reshape(B(mem3,[4,5])',1,[]) ); %'
end
maxLength = max(cellfun(#(x)numel(x),AA));
out = cell2mat(cellfun(#(x)cat(2,x,zeros(1,maxLength-length(x))),AA,'UniformOutput',false))
I am trying to make this code efficient, by not using for and if, but couldn't find an answer.
Try this
a = A(:,1);
b = B(:,1:3);
z = size(b);
b = repmat(b,[1,1,numel(a)]);
ab = repmat(permute(a,[2,3,1]),z);
row2 = mat2cell(permute(sum(ab==b,2),[3,1,2]),ones(1,numel(a)));
AA = cellfun(#(x)(reshape(B(x>0,4:end)',1,numel(B(x>0,4:end)))),row2,'UniformOutput',0);
maxLength = max(cellfun(#(x)numel(x),AA));
out = cat(2,A,cell2mat(cellfun(#(x)cat(2,x,zeros(1,maxLength-length(x))),AA,'UniformOutput',false)))
UPDATE Below code runs in almost same time as the iterative code
a = A(:,1);
b = B(:,1:3);
z = size(b);
b = repmat(b,[1,1,numel(a)]);
ab = repmat(permute(a,[2,3,1]),z);
df = permute(sum(ab==b,2),[3,1,2])';
AA = arrayfun(#(x)(B(df(:,x)>0,4:end)),1:size(df,2),'UniformOutput',0);
AA = arrayfun(#(x)(reshape(AA{1,x}',1,numel(AA{1,x}))),1:size(AA,2),'UniformOutput',0);
maxLength = max(arrayfun(#(x)(numel(AA{1,x})),1:size(AA,2)));
out2 = cell2mat(arrayfun(#(x,i)((cat(2,A(i,:),AA{1,x},zeros(1,maxLength-length(AA{1,x}))))),1:numel(AA),1:size(A,1),'UniformOutput',0));
How about this:
%# example data
A = [1 2
3 4];
B = [1 2 4 5 4
1 2 3 5 3
1 1 1 1 2
3 4 5 6 5];
%# rename for clarity & reshape for algorithm's convenience
needle = permute(A(:,1), [2 3 1]);
haystack = B(:,1:3);
data = B(:,4:end).';
%# Get the relevant rows of 'haystack' for each entry in 'needle'
inds = any(bsxfun(#eq, haystack, needle), 2);
%# Create data that should be appended to A
%# All data and functionality in this loop is local and static, so speed
%# should be optimal.
append = zeros( size(A,1), numel(data) );
for ii = 1:size(inds,3)
newrow = data(:,inds(:,:,ii));
append(ii,1:numel(newrow)) = newrow(:);
end
%# Now append to A, stripping unneeded zeros
A = [A append(:, ~all(append==0,1))]
I need to generate (I prefere MATLAB) all "unique" integer tuples k = (k_1, k_2, ..., k_r) and
its corresponding multiplicities, satisfying two additional conditions:
1. sum(k) = n
2. 0<=k_i<=w_i, where vector w = (w_1,w_2, ..., w_r) contains predefined limits w_i.
"Unique" tuples means, that it contains unique unordered set of elements
(k_1,k_2, ..., k_r)
[t,m] = func(n,w)
t ... matrix of tuples, m .. vector of tuples multiplicities
Typical problem dimensions are about:
n ~ 30, n <= sum(w) <= n+10, 5 <= r <= n
(I hope that exist any polynomial time algorithm!!!)
Example:
n = 8, w = (2,2,2,2,2), r = length(w)
[t,m] = func(n,w)
t =
2 2 2 2 0
2 2 2 1 1
m =
5
10
in this case exist only two "unique" tuples:
(2,2,2,2,0) with multiplicity 5
there are 5 "identical" tuples with same set of elements
0 2 2 2 2
2 0 2 2 2
2 2 0 2 2
2 2 2 0 2
2 2 2 2 0
and
(2,2,2,1,1) with multiplicity 10
there are 10 "identical" tuples with same set of elements
1 1 2 2 2
1 2 1 2 2
1 2 2 1 2
1 2 2 2 1
2 1 1 2 2
2 1 2 1 2
2 1 2 2 1
2 2 1 1 2
2 2 1 2 1
2 2 2 1 1
Thanks in advance for any help.
Very rough (extremely ineffective) solution. FOR cycle over 2^nvec-1 (nvec = r*maxw) test samples and storage of variable res are really terrible things!!!
This solution is based on tho following question.
Is there any more effective way?
function [tup,mul] = tupmul(n,w)
r = length(w);
maxw = max(w);
w = repmat(w,1,maxw+1);
vec = 0:maxw;
vec = repmat(vec',1,r);
vec = reshape(vec',1,r*(maxw+1));
nvec = length(vec);
res = [];
for i = 1:(2^nvec - 1)
ndx = dec2bin(i,nvec) == '1';
if sum(vec(ndx)) == n && all(vec(ndx)<=w(ndx)) && length(vec(ndx))==r
res = [res; vec(ndx)];
end
end
tup = unique(res,'rows');
ntup = size(tup,1);
mul = zeros(ntup,1);
for i=1:ntup
mul(i) = size(unique(perms(tup(i,:)),'rows'),1);
end
end
Example:
> [tup mul] = tupmul(8,[2 2 2 2 2])
tup =
0 2 2 2 2
1 1 2 2 2
mul =
5
10
Or same case but with changed limits for first two positions:
>> [tup mul] = tupmul(8,[1 1 2 2 2])
tup =
1 1 2 2 2
mul =
10
This is far more better algorithm, created by Bruno Luong (phenomenal MATLAB programmer):
function [t, m, v] = tupmul(n, w)
v = tmr(length(w), n, w);
t = sort(v,2);
[t,~,J] = unique(t,'rows');
m = accumarray(J(:),1);
end % tupmul
function v = tmr(p, n, w, head)
if p==1
if n <= w(end)
v = n;
else
v = zeros(0,1);
end
else
jmax = min(n,w(end-p+1));
v = cell2mat(arrayfun(#(j) tmr(p-1, n-j, w, j), (0:jmax)', ...
'UniformOutput', false));
end
if nargin>=4 % add a head column
v = [head+zeros(size(v,1),1,class(head)) v];
end
end %tmr