I have matrix (a) with (1:10),<10 x 1> double. I would like to copy the values and rearrange them column wise into another matrix var. (b). See example below. Also, what method would be most efficient at this task?
matrix a matrix b
1 1
2 2 2
3 3 3 3
4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6 6
7 7 7 7 7 7 7 7
8 8 8 8 8 8 8 8 8
9 9 9 9 9 9 9 9 9 9
10 10 10 10 10 10 10 10 10 10 10
update:
Hi once again Amro. How about if I wanted to define which values to copy. See below example:
matrix a matrix b
column: 1 2 3 4 5 6 7
1 1
2 2 2
3 3 3
4 4
5 5
6 6
7 7
8 8
9 9
10 10 10
Try:
>> a = (1:10)'
a =
1
2
3
4
5
6
7
8
9
10
>> b = tril(repmat(a,1,10))
b =
1 0 0 0 0 0 0 0 0 0
2 2 0 0 0 0 0 0 0 0
3 3 3 0 0 0 0 0 0 0
4 4 4 4 0 0 0 0 0 0
5 5 5 5 5 0 0 0 0 0
6 6 6 6 6 6 0 0 0 0
7 7 7 7 7 7 7 0 0 0
8 8 8 8 8 8 8 8 0 0
9 9 9 9 9 9 9 9 9 0
10 10 10 10 10 10 10 10 10 10
I think in the second matrix you specified you made an error. I'm assuming you wanted to do something like this:
b =
1 0 0 0 0 0
2 2 0 0 0 0
0 3 3 0 0 0
0 0 4 4 0 0
0 0 0 5 5 0
0 0 0 0 6 6
this is simple to do:
%define vector of arbitrary length
a=1:6;
%generate b with shifted diagonal matrices
b=diag(a)+diag(a(2:end),-1);
the second argument of diag just shifts the resulting diagonal.
Related
Assuming I'm having a vectors of numbers A, for example: A=[1 3 5 3 9 6](A's length >= 2) and an Integer X=6. Need to find how many pairs (A[i],A[j]) where i<j exist in the vector which answer this condition: A[i]+A[j]=X. The number of pairs is printed.
Not allowed to use for/while. Allowed only ceil,floor,mod,repmat,reshape,size,length,transpose,sort,isempty,all,any,find ,sum,max,min.
With repmat, length and sum -
integer1 = 6; %// One of the paramters
A_ind = 1:length(A) %// Get the indices array
%// Expand A_ind into rows and A_ind' into columns, to form a meshgrid structure
A_ind_mat1 = repmat(A_ind,[length(A) 1])
A_ind_mat2 = repmat(A_ind',[1 length(A)]) %//'
%// Expand A into rows and A' into columns, to form a meshgrid structure
A_mat1 = repmat(A,[length(A) 1])
A_mat2 = repmat(A',[1 length(A)]) %//'
%// Form the binary matrix of -> (A[i],A[j]) where i<j
cond1 = A_ind_mat1 < A_ind_mat2
%// Use the binary matrix as a logical mask to select elements from the two
%// matrices and see which element pairs satisfy -> A[i]+A[j]=X and get a
%// count of those pairs with SUM
pairs_count = sum((A_mat1(cond1) + A_mat2(cond1))==integer1)
Outputs from code run to make it clearer -
A =
1 3 5 3 9 6
A_ind =
1 2 3 4 5 6
A_ind_mat1 =
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
1 2 3 4 5 6
A_ind_mat2 =
1 1 1 1 1 1
2 2 2 2 2 2
3 3 3 3 3 3
4 4 4 4 4 4
5 5 5 5 5 5
6 6 6 6 6 6
A_mat1 =
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
1 3 5 3 9 6
A_mat2 =
1 1 1 1 1 1
3 3 3 3 3 3
5 5 5 5 5 5
3 3 3 3 3 3
9 9 9 9 9 9
6 6 6 6 6 6
cond1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 1 0 0 0 0
1 1 1 0 0 0
1 1 1 1 0 0
1 1 1 1 1 0
pairs_count =
2
A bit more explanation -
Taking few more steps to clarify why pairs_count must be 2 here -
Set all values in A_mat1 and A_mat2 to be zeros that do not satisfy the less than criteria
>> A_mat1(~cond1)=0
A_mat1 =
0 0 0 0 0 0
1 0 0 0 0 0
1 3 0 0 0 0
1 3 5 0 0 0
1 3 5 3 0 0
1 3 5 3 9 0
>> A_mat2(~cond1)=0
A_mat2 =
0 0 0 0 0 0
3 0 0 0 0 0
5 5 0 0 0 0
3 3 3 0 0 0
9 9 9 9 0 0
6 6 6 6 6 0
Now, add A_mat1 and A_mat2 and see how many 6's you got -
>> A_mat1 + A_mat2
ans =
0 0 0 0 0 0
4 0 0 0 0 0
6 8 0 0 0 0
4 6 8 0 0 0
10 12 14 12 0 0
7 9 11 9 15 0
As you can see there are two 6's and thus our result is verified.
Can someone check did I guess correct number of neurons in input/hidden/output layer and overall params please.
My idea of this ANN:
Input neurons : 784 (28x28)
Hidden Layers : 1
Size of hidden layer(s) : 25
Activation function : Log-sigmoid
Training method : gradient descent
Data size : 400 + 200
There are 400 bmp images used for training of it, and 200 for checking (however only 1-50 get guessed with 100% rate and others 0% rate...)
clear all;
clc
for kk=1:400
pl=ones(28,28); %³õʼ»¯28*28¶þֵͼÏñΪȫ°×
m=strcat('b',int2str(kk),'.bmp'); %Á¬½Ó×Ö·ûµÃµ½Ñù±¾ÎļþÃû
x=imread(m,'bmp'); %¶ÁÈëÑб¾ÎļþͼÏñ
pl=im2bw(x,0.5); %°ÑÑù±¾Í¼Ïñת»¯Îª¶þֵͼ
for m=0:27 %ÐγÉÉñ¾ÍøÂçÊäÈëÏòÁ¿
p(m*28+1:(m+1)*28,kk)=pl(1:28,m+1);
end
end
%ÊÖдÌåÑù±¾¶ÔÓ¦µÄÊý×Ö£¨´Ób1.bmpµ½b400.bmp ¹²400¸ö£©£º
t=[5 0 4 1 9 2 1 3 1 4 3 6 3 6 1 7 2 8 6 9 4 0 9 1 1 2 4 3 2 7 8 8 6 9 0 5 6 0 7......
6 1 8 7 9 3 9 8 5 9 3 3 0 7 4 9 8 0 9 4 1 4 4 6 0 4 5 6 1 0 0 1 7 1 6 3 0 2 1......
1 7 8 0 2 6 7 8 3 9 0 4 6 7 4 6 8 0 7 8 3 1 5 7 1 7 1 1 6 3 0 2 9 3 1 1 0 4 9......
2 0 0 2 0 2 7 1 8 6 4 1 6 3 4 1 9 1 3 3 9 5 4 7 7 4 2 8 5 8 6 0 3 4 6 1 9 9 6......
0 3 7 2 8 2 9 4 4 6 4 9 7 0 9 2 7 5 1 5 9 1 2 3 1 3 5 9 1 7 6 2 8 2 2 6 0 7 4......
9 7 8 3 2 1 1 8 3 6 1 0 3 1 0 0 1 1 2 7 3 0 4 6 5 2 6 4 7 1 8 9 9 3 0 7 1 0 2......
0 3 5 4 6 5 8 6 3 7 5 8 0 9 1 0 3 1 2 2 3 3 6 4 7 5 0 6 2 7 9 8 5 9 2 1 1 4 4......
5 6 4 1 2 5 3 9 3 9 0 5 9 6 5 7 4 1 3 4 0 4 8 0 4 3 6 8 7 6 0 9 7 5 7 2 1 1 6......
8 9 4 1 5 2 2 9 0 3 9 6 7 2 0 3 5 4 3 6 5 8 9 5 4 7 4 2 7 3 4 8 9 1 9 2 1 7 9......
1 8 7 4 1 3 1 1 0 2 3 9 4 9 2 1 6 8 4 7 7 4 4 9 2 5 7 2 4 4 2 1 9 2 2 8 7 6 9......
8 2 3 8 1 6 5 1 1 0];
%´´½¨BPÍøÂç
pr(1:784,1)=0;
pr(1:784,2)=1;
t1=clock; %¼Æʱ¿ªÊ¼
%ÉèÖÃѵÁ·²ÎÊý
net=newff(pr,[25 1],{'logsig','purelin'},'traingdx','learngdm');
net.trainParam.epochs=5000; %ÉèÖÃѵÁ·´ÎÊý
net.trainParam.goal=0.05; %ÉèÖÃÐÔÄܺ¯Êý
net.trainParam.show=10; %ÿ10ÏÔʾ
net.trainParam.Ir=0.05; %ÉèÖÃѧϰËÙÂÊ
net=train(net,p,t); %ѵÁ·BPÍøÂç
datat=etime(clock,t1) %¼ÆËãÉè¼ÆÍøÂçµÄʱ¼äΪ66.417s
%Éú³É²âÊÔÑù±¾
pt(1:784,1)=1;
pl=ones(28,28); %³õʼ»¯28*28¶þֵͼÏñÏñËØ
for kk=401:600
pl=ones(28,28); %³õʼ»¯28*28¶þֵͼÏñΪȫ°×
m=strcat('b',int2str(kk),'.bmp'); %Á¬½Ó×Ö·ûµÃµ½Ñù±¾ÎļþÃû
x=imread(m,'bmp'); %¶ÁÈëÑб¾ÎļþͼÏñ
pl=im2bw(x,0.5); %°ÑÑù±¾Í¼Ïñת»¯Îª¶þֵͼ
for m=0:27 %ÐγÉÉñ¾ÍøÂçÊäÈëÏòÁ¿
pt(m*28+1:(m+1)*28,kk-400)=pl(1:28,m+1);
end
end
[a,Pf,Af]=sim(net,pt); %ÍøÂç·ÂÕæ
a=round(a) %Êä³öʶ±ð½á¹û
%²âÊÔÑù±¾¶ÔÓ¦µÄÊý×Ö£¨´Ób401.bmpµ½b600.bmp ¹²200¸ö£©£º
tl=[2 6 4 5 8 3 1 5 1 9 2 7 4 4 4 8 1 5 8 9 5 6 7 9 9 3 7 0 9......
0 6 6 2 3 9 0 7 5 4 8 0 9 4 1 1 8 7 1 2 6 1 0 3 0 1 1 8 2 0 3 9 4 0 5 0 6 1 7......
7 8 1 9 2 0 5 1 2 2 7 3 5 4 4 7 1 8 3 9 6 0 3 1 1 2 0 3 5 7 6 8 2 9 5 8 5 7 4......
1 1 3 1 7 5 5 5 2 5 8 2 0 9 7 7 5 0 9 0 0 8 9 2 4 8 1 6 1 6 5 1 8 3 4 0 5 5 8......
3 4 2 3 9 2 1 1 5 2 1 3 2 8 7 3 7 2 4 6 9 7 2 4 2 8 1 1 3 8 4 0 6 5 9 3 0 9 2......
4 7 1 1 9 4 2 6 1 8 9 0 6 6 7];
k=0;
for i=1:200
if a(i)==tl(i)
k=k+1;
end
end
rate=1.00*k/200; %¼ÆËã×îºóÕýÈ·ÂÊΪ0.495
I might be wrong, since you don't specify the number of output neurons and the number of patterns per class in your dataset. However, it seems that you have created only one output neuron for your network. In this case, the network assings ALL patterns to the same class, and the classification accuracy you get is equal to the a priori probability. If, for example, the first 50 patterns of your dataset belong to the same class, and the rest to different classes, a classifier with one output will assign all patterns to the first class, so you will get the first 50 right.
If this is the case, you should create a classifier with N outputs, where N is the number of classes in your dataset. In this case, the classifier will vote for each class, and the pattern will be assigned to the class with the maximum output. If for example you have 3 classes, and the output for a specific pattern is [0.2, 0.83, 0.6], the pattern will be assigned to the second class.
Moreover, converting to image to black-and-white is probably not the best way. It would be better to convert the image to grayscale (to preserve the histogram to some extent), and use some normalization to compensate for differences in lighting.
Finally, keep in mind that neural networks essentially detect similarity between input vectors. So, if you need to classify pictures, you need to find a representation such that similar images produce similar input vectors. Feeding the values of the pixels into the classifier is not a such representation. For example, if you turn the image upside down, the input vector changes completely, even though it still shows the same object. You don't want that. You want features that depend on the object shown, and not on lighting/angle etc. However, extracting such features is a different matter altogether (for example see some examples for image preprocessing and feature extraction from the OpenCV framework, the standard image processing and computer vision tool in C++/python)
If you are interested in neural networks and not image processing, it would be better to start with some standard classification problems from the UCI repository (eg. iris flower, wisconsin breast cancer) and practice with them until you produce good results and feel comfortable with the tools you are using.
Is there a way to find the maximum flow between each pair of vertices in matlab?
c = sparse([1 1 2 2 3 4 4 5 5 6 7 8 9 9],[2 3 3 4 5 6 7 6 7 8 9 10 8 10],[15 10 3 8 9 7 5 6 2 12 10 6 10 8],10,10)
a = [2 3 4 5 6 7 8 9 10]
b = arrayfun(#(x)max_flow(c,1,x),a)
OR
b = arrayfun(#(x)graphmaxflow(c,1,x),a)
b =
15 13 8 9 13 7 16 7 13
So, I can take a sparse matrix and get the maximum flow from one vertex to all others. Is there a way to continue this to obtain the max flow for all of the pairs?
I'd eventually like to be able to find the all-pair max flow for a directed, weighted graph. . .
Got it to work:
c = sparse([1 1 2 2 3 4 4 5 5 6 7 8 9 9],[2 3 3 4 5 6 7 6 7 8 9 10 8 10],[15 10 3 8 9 7 5 6 2 12 10 6 10 8],10,10)
for a=1:10
for b=1:10
if a==b
continue
end
t(b,a)=graphmaxflow(c,a,b);
p=t(:);
end
end
I couldn't figure out a way to use arrayfun to do this.
Each maximum flow value:
t =
0 0 0 0 0 0 0 0 0 0
15 0 0 0 0 0 0 0 0 0
13 3 0 0 0 0 0 0 0 0
8 8 0 0 0 0 0 0 0 0
9 3 9 0 0 0 0 0 0 0
13 10 6 7 6 0 0 0 0 0
7 7 2 5 2 0 0 0 0 0
16 11 8 12 8 12 10 0 10 0
7 7 2 5 2 0 10 0 0 0
13 11 8 11 8 6 10 6 14 0
p =
0
15
13
8
9
13
7
...
I have 2 matrices:
T3(:,:,1) =
0 0 0 0 1 0 0 0 0
0 0 0 0 2 0 0 0 0
0 0 0 0 3 0 0 0 0
0 1 0 1 4 2 0 4 0
0 3 0 2 6 3 0 5 0
2 4 2 5 7 5 4 6 5
4 5 5 7 8 8 5 7 6
5 6 6 8 9 9 8 9 8
T3(:,:,2) =
2 1 1 1 1 1 1 1 1
3 3 2 2 2 2 2 2 2
4 4 4 3 3 3 3 3 3
5 5 5 5 4 4 4 4 4
6 6 6 6 6 5 5 5 5
7 7 7 7 7 7 6 6 6
8 8 8 8 8 8 8 7 7
9 9 9 9 9 9 9 9 8
How do I make values present in T3(:,:,1) turn to zero in T3(:,:,2)?
e.g. in the first column of T3(:,:,1) the values are 2,4,5. I'd like the first column of T3(:,:,2) to have the the values 2,4,5 as zero.
T3(:,:,2) =
0 0 1 0 0 1 1 1 1
3 0 0 0 0 0 2 2 2
0 0 4 3 0 0 3 3 3
0 0 0 0 0 4 0 0 4
6 0 0 6 0 0 0 0 0
7 7 7 0 0 7 6 0 0
8 8 8 0 0 0 0 0 7
9 9 9 9 0 0 9 0 0
I wonder if there is a way to do this using setdiff or unique.
for y=1:H-1
for z=1:H-1
for h=1:H
for d=1:D-1
if T3(y,h,d+1) == T3(z,h,d)
T3(y,h,d+1)=0;
end
end
end
end
end
I can do it as a loop where H=number of columns (9) and D= number of dimensions (2). There must be a better way :)?
Many thanks guys.
How can I create the following matrix?
[0 0 0
8 0 0
9 8 0
6 9 8
5 6 9
4 5 6]
How about the toeplitz function?
c=[0 8 9 6 5 4 3]
r=[0 0 0]
t=toeplitz(c,r)
(Disclaimer: untested!)
T should be:
0 0 0
8 0 0
9 8 0
6 9 8
5 6 9
4 5 6
3 4 5