I have a training set that has input and outputs in this way:
Input:
0.832 64.643
0.818 78.843
1.776 45.049
0.597 88.302
1.412 63.458
1.468 49.535
1.985 33.387
2.073 30.279
1.431 55.231
1.116 68.521
1.617 44.362
2.159 66.512
Output:
0 0 1
0 0 1
0 1 0
0 0 1
0 0 1
1 0 0
0 0 1
1 0 0
1 0 0
0 0 1
0 0 1
0 1 0
1 0 0
1 0 0
0 1 0
0 1 0
I need to implement one linear layer neural network that can represent the data set best in MATLAB. What would be the algorithm to do it in MATLAB?
The target output is "1 for a particular class that the corresponding input belongs to and "0 for the remaining 2 outputs.
Consider this example of training a feed-forward ANN of one hidden layer (with 3 nodes).
Since your data seems to have more output points than input, I'm using a demo dataset, but the idea is the same:
%# load sample data
laod simpleclass_dataset
input = simpleclassInputs; %# 2x1000, 2-dimensional points
output = simpleclassTargets; %# 4x1000, 4 classes
%# split data into training/testing sets
trainInd = 1:500;
testInd = 501:1000;
%# create ANN and initialize network weights
net = newpr(input, output, 3);
net = init(net);
net.trainParam.epochs = 25; %# max number of iterations
%# learn net weights from training data
net = train(net, input(:,trainInd), output(:,trainInd));
%# predict output of net on testing data
pred = sim(net, input(:,testInd));
%# classification confusion matrix
[err,cm] = confusion(output(:,testInd), pred);
The output is:
err =
0.075075
cm =
81 0 0 0
0 82 0 0
9 0 52 16
0 0 0 93
Obviously you will need access to the Neural Network Toolbox.
Related
I basically want to use the function ismember, but for a range. For example, I want to know what data points in array1 are within n distance to array2, for each element in array2.
I have the following:
array1 = [1,2,3,4,5]
array2 = [2,2,3,10,20,40,50]
I want to know what values in array2 are <= 2 away from array1:
indices(1,:) (where array1(1) = 1) = [1 1 1 0 0 0 0]
indices(2,:) (where array1(2) = 2) = [1 1 1 0 0 0 0]
indices(3,:) (where array1(3) = 3) = [1 1 1 0 0 0 0]
indices(4,:) (where array1(4) = 4) = [1 1 1 0 0 0 0]
indices(5,:) (where array1(5) = 5) = [0 0 1 0 0 0 0]
Drawbacks:
My array1 is 496736 elements, my array2 is 9268 elements, so aI would rather not use a loop.
Looping is a valid option here. Intialise an array output to the size of array1 X array2, then loop over all elements in array1 an subtract array2 from that, then check whether the absolute value is less than or equal to 2:
array1 = [1,2,3,4,5];
array2 = [2,2,3,10,20,40,50];
output = zeros(numel(array1), numel(array2),'logical');
for ii = 1:numel(array1)
output(ii,:) = abs(array1(ii)-array2)<=2;
end
output =
1 1 1 0 0 0 0
1 1 1 0 0 0 0
1 1 1 0 0 0 0
1 1 1 0 0 0 0
0 0 1 0 0 0 0
i.e. loops are not the problem.
Thanks to Rahnema1's suggestion, you can initialise output directly as a logical matrix:
output = zeros(numel(array1),numel(array2),'logical');
whose size is just 4.3GB.
On timings: Hans' code runs in a matter of seconds for array1 = 5*rand(496736,1); array2 = 25*rand(9286,1);, the looped solution takes about 15 times longer. Both solutions are equal to one another. obcahrdon's ismembertol solution is somewhere in between on my machine.
On RAM usage:
Both implicit expansion, as per Hans' answer, as well as the loop suggested in mine work with just 4.3GB RAM on your expanded problem size (496736*9286)
pdist2 as per Luis' answer and bsxfun as per Hans' on the other hand try to create an intermediate double matrix of 34GB (which doesn't even fit in my RAM, so I cannot compare timings).
obchardon's ismembertol solution outputs a different form of the solution, and takes ~5.04GB (highly dependent on the amount of matches found, the more, the larger this number will be).
In general this leads me to the conclusion that implicit expansion should be your option of choice, but if you have R2016a or earlier, ismembertol or a loop is the way to go.
Using implicit expansion, introduced in MATLAB R2016b, you can simply write:
abs(array1.' - array2) <= 2
ans =
1 1 1 0 0 0 0
1 1 1 0 0 0 0
1 1 1 0 0 0 0
1 1 1 0 0 0 0
0 0 1 0 0 0 0
For earlier MATLAB versions, you can get this using the bsxfun function:
abs(bsxfun(#minus, array1.', array2)) <= 2
ans =
1 1 1 0 0 0 0
1 1 1 0 0 0 0
1 1 1 0 0 0 0
1 1 1 0 0 0 0
0 0 1 0 0 0 0
Hope that helps!
P.S. On the "MATLAB is slow for loops" myth, please have a look at that blog post for example.
EDIT: Please read Adriaan's answer on the RAM consumption using this and/or his approach!
If you have the Statistics Toolbox, you can use pdist2 to compute the distance matrix:
result = pdist2(array1(:), array2(:)) <= 2;
As noted by #Adriaan, this is not efficient for large input sizes. In that case a better approach is a loop with preallocation of the logical matrix output, as in his answer.
You can also use ismembertol with some specific option:
A = [1,2,3,4,5];
B = [2,2,3,10,20,40,5000];
tol = 2;
[~,ind] = ismembertol([A-tol;A+tol].',[B-tol;B+tol].',tol, 'ByRows', true, ...
'OutputAllIndices', true, 'DataScale', [1,Inf])
It will create a 5x1 cell array containing the corresponding linear indice
ind{1} = [1,2,3]
ind{2} = [1,2,3]
...
ind{5} = [3]
In this case using linear indices instead of logical indices will greatly reduce the memory consumption.
I am trying to produce a 3D array from a 2D data frame in R and could really use some help. So far, I haven't found a solution to this problem from similar questions that have been posted previously.
My input data are available here: https://www.dropbox.com/s/7f8td34mpzgpvgh/example_data.csv?dl=0, and they resemble the following basic structure: I have 14 sites (i.e. Field) with 6 replicates each (i.e. Replicate) in which a subset of 32 subjects (i.e. columns of species codes: AMGO, BASW, etc.) were counted when present in a survey.
A subset of the input data looks like so:
example_data[1:5, 1:5]
Field Replicate AMGO BASW BHCO
1 Brinkman 1 2 0 0
72 Brinkman 2 10 0 0
190 Brinkman 3 6 0 0
283 Brinkman 4 0 0 0
342 Brinkman 5 2 1 0
I'd like to reformat these input data to resemble a 3D array (i.e. 14 sites x 6 replicates x 32 subjects), as exemplified below with the AMGO species:
, , = AMGO
1 2 3 4 5 6
Brinkman 0 0 0 0 0 0
Clara 0 0 0 0 0 0
Esckelson 0 0 0 0 0 0
GarnerEast 0 0 0 0 0 0
GarnerWest 0 0 0 0 0 0
KHess 0 0 0 0 0 0
Lounsbury 0 0 0 0 0 0
McCallum 0 0 0 0 0 0
Pomeroy 0 0 0 0 0 0
Sattelberg 0 0 0 0 0 0
THess 0 0 0 0 0 0
Turner 0 0 0 0 0 0
VollmarEast 0 0 0 0 0 0
VollmarWest 0 0 0 0 0 0
...
Note that, in the solution, many of the zeroes above would likely be replaced by non-zero counts when AMGO (and other species) was (were) actually encountered during a survey.
Please let me know if there's anything I need to clarify, and thanks in advance!
Here's a solution using the reshape() function from base R. I'm applying the function to each of the subject columns and creating a list of reshaped dataframes.
df <- read.csv("C:\\Users\\Shrivatav\\Downloads\\example_data.csv", encoding = "UTF-8")
# Extract subject columns
list.of.cols <- colnames(df)[3:34]
# Function for reshaping
func.for.reshaping <- function(column){
# Subset the data, keep only Field, replicate and the column input in the
# function
to.keep <- c("Field", "Replicate", column)
subset.df <- df[to.keep]
# reshape from long to wide
reshaped.df <- reshape(subset.df, idvar = "Field", timevar = "Replicate", direction = "wide")
return(reshaped.df)
}
# Apply the function over all subject columns, reulting
# in a list of dataframes
list.of.reshaped.dfs <- lapply(list.of.cols, func.for.reshaping)
# Name the list for easy access
names(list.of.reshaped.dfs) <- list.of.cols
You can access the elements of the list like: list.of.reshaped.dfs$AMGO and so forth.
I have an array A of 1s and 0s and want to see if the larger array of bits B contains those bits in that exact order?
Example: A= [0 1 1 0 0 0 0 1]
B= [0 1 0 0 1 1 0 0 0 0 1 0 1 0 1]
would be true as A is contained in B
Most solutions I have found only determine if a value IS contained in another matrix, this is no good here as it is already certain that both matrices will be 1s and 0s
Thanks
One (albeit unusual) option, since you're dealing with integer values, is to convert A and B to character arrays and use the contains function:
isWithin = contains(char(B), char(A));
There are some obtuse vectorized ways to to do this, but by far the easiest, and likely just as efficient, is to use a loop with a sliding window,
A = [0 1 1 0 0 0 0 1];
B = [0 1 0 0 1 1 0 0 0 0 1 0 1 0 1];
vec = 0:(numel(A)-1);
for idx = 1:(numel(B)-numel(A)-1)
if all(A==B(idx+vec))
fprintf('A is contained in B\n');
break; % exit the loop as soon as 1 match is found
end
end
Or if you want to know the location(s) in B (of potentially multiple matches) then,
A = [0 1 1 0 0 0 0 1];
B = [0 1 0 0 1 1 0 0 0 0 1 0 1 0 1];
C = false(1,numel(B)-numel(A)-1);
vec = 0:(numel(A)-1);
for idx = 1:numel(C)
C(idx) = all(A==B(idx+vec));
end
if any(C)
fprintf('A is contained in B\n');
end
In this case
>> C
C =
1×6 logical array
0 0 0 1 0 0
You can use the cross-correlation between two signals for this, as a measure of local similarity.
For achieving good results, you need to shift A and B so that you don't have the value 0 any more. Then compute the correlation between the two of them with conv (keeping in mind that the convolution is the cross-correlation with one signal flipped), and normalize with the energy of A so that you get a perfect match whenever you get the value 1:
conv(B-0.5, flip(A)-0.5, 'valid')/sum((A-0.5).^2)
In the normalization term, flipping is removed as it does not change the value.
It gives:
[0 -0.5 0.25 1 0 0 -0.25 0]
4th element is 1, so starting from index equal to 4 you get a perfect match.
So I have this matrix A, which is made of 1 and zeros, I have about 10 to 14 white spots of many pixels, but I want only 1 white pixel/centers coordinate for every cluster of white, how do I calculate how many cluster there are and their centers.
Try to imagine the matrix A as the night sky with white starts in black sky and how to I count the stars and the stars centers, plus the star are made of cluster of white pixels.
also the clusters are not all exactly the same size.
Here is some code using bwlabel and/or regioprops, which are used to identify connected components in a matrix and a buch of other properties, respectively. I think it suits your problem quite well; however you might want to adapt my code a bit as its more of a starting point.
clear
clc
%// Create dummy matrix.
BW = logical ([ 1 1 1 0 1 1 1 0
1 1 1 0 1 1 1 0
1 1 1 0 1 1 1 0
0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 0
1 1 1 1 0 1 1 0
1 1 1 1 0 1 1 0
1 1 1 1 0 0 0 0]);
%// Identify clusters.
L = bwlabel(BW,4)
Matrix L looks like this:
L =
1 1 1 0 3 3 3 0
1 1 1 0 3 3 3 0
1 1 1 0 3 3 3 0
0 0 0 0 0 0 0 0
0 0 0 0 0 4 4 0
2 2 2 2 0 4 4 0
2 2 2 2 0 4 4 0
2 2 2 2 0 0 0 0
Here you have many ways to locate the center of the clusters. The first one uses the output of bwlabel to find each cluster and calculate the coordinates in a loop. It works and its didactic but it's a bit long and not so efficient. The 2nd method, as mentioned by #nkjt, uses regionprops which does exactly what you want using the 'Centroid' property. So here are the 2 methods:
Method 1: a bit complicated
So bwlabel identified 4 clusters, which makes sense. Now we need to identify the center of each of those clusters. My method could probably be simplified; but I'm a bit out of time so fell free to modify it as you see fit.
%// Get number of clusters
NumClusters = numel(unique(L)) -1;
Centers = zeros(NumClusters,2);
CenterLinIdices = zeros(NumClusters,1);
for k = 1:NumClusters
%// Find indices for elements forming each cluster.
[r, c] = find(L==k);
%// Sort the elements to know hot many rows and columns the cluster is spanning.
[~,y] = sort(r);
c = c(y);
r = r(y);
NumRow = numel(unique(r));
NumCol = numel(unique(c));
%// Calculate the approximate center of the cluster.
CenterCoord = [r(1)+floor(NumRow/2) c(1)+floor(NumCol/2)];
%// Actually this array is not used here but you might want to keep it for future reference.
Centers(k,:) = [CenterCoord(1) CenterCoord(2)];
%// Convert the subscripts indices to linear indices for easy reference.
CenterLinIdices(k) = sub2ind(size(BW),CenterCoord(1),CenterCoord(2));
end
%// Create output matrix full of 0s, except at the center of the clusters.
BW2 = false(size(BW));
BW2(CenterLinIdices) = 1
BW2 =
0 0 0 0 0 0 0 0
0 1 0 0 0 1 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 0 0
0 0 0 0 0 0 1 0
0 0 1 0 0 0 0 0
0 0 0 0 0 0 0 0
Method 2 Using regionprops and the 'Centroid' property.
Once you have matrix L, apply regionprops and concatenate the output to get an array containing the coordinates directly. Much simpler!
%// Create dummy matrix.
BW = logical ([ 1 1 1 0 1 1 1 0
1 1 1 0 1 1 1 0
1 1 1 0 1 1 1 0
0 0 0 0 0 0 0 0
0 0 0 0 0 1 1 0
1 1 1 1 0 1 1 0
1 1 1 1 0 1 1 0
1 1 1 1 0 0 0 0]);
%// Identify clusters.
L = bwlabel(BW,4)
s = regionprops(L,'Centroid');
CentroidCoord = vertcat(s.Centroid)
which gives this:
CentroidCoord =
2.0000 2.0000
2.5000 7.0000
6.0000 2.0000
6.5000 6.0000
Which is much simpler and gives the same output once you use floor.
Hope that helps!
I have a large matrix of 1's and 0's, and am looking for a way to "fill" up areas that are locally dense with 1's.
I first did this task for an array, and counted the number of 1's within a certain radius of the element in questions. If the radius was 5, for example, and my threshold was 4, then a point that had 4 elements marked "1" within 5 elements to the left or right would be changed to a 1.
Basically I would like to generalized this to a two - dimensional array and have a resulting matrix that has "smooth" and "connected" regions of 1's and no "patchy" spots.
As an example, the matrix
1 0 0 1 0 0 0
0 0 1 0 1 0 0
0 1 0 1 0 0 0
0 0 1 1 1 0 0
would ideally be changed to
1 0 0 1 1 0 0
0 0 1 1 1 0 0
0 1 1 1 1 0 0
0 0 1 1 1 0 0
or something similar
For binary images, the morphologial operations that are implemented in MATLAB are perfect for manipulating the shape and size of connected regions. Specifically, the process of image closing is designed to fill holes in connected regions. In MATLAB, the function is imclose, which takes the image and a structuring element, similar to a filter kernel, for how neighboring pixels effect the filling of holes and gaps. A simple invocation of imclose is,
IM2 = imclose(IM,strel(ones(3)));
Larger gaps can be filled by increasing the area of the influence of of neighboring pixes, via larger structuring elements. For example, we an use a disk of radius 10 pixels:
IM2 = imclose(IM,strel('disk',10));
While, imclose supports grayscale and binary (0 and 1) images, the function bwmorph is designed for operation on binary images only but provides a generic interface to all of the morphological operations and various neat combinations of operations (e.g. 'bothat', 'tophat', etc.). The syntax for closing is simplified with bwmorph:
BW2 = bwmorph(BW,'close');
Here the structuring element is the standard ones(3).
A simple filter such as the following might do the trick:
h = [ 0 1 0
1 0 1
0 1 0];
img2=(imfilter(img,h)>2) | img;
For instance:
img =
1 0 0 1 0 0 0
0 0 1 0 1 0 0
0 1 0 1 0 0 0
0 0 1 1 1 0 0
img2 =
1 0 0 1 0 0 0
0 0 1 1 1 0 0
0 1 1 1 1 0 0
0 0 1 1 1 0 0
You can try different filters to modify the output img2.
This uses the image processing toolbox. If you don't have that, you may want to look up equivalent routines from the matlab exchange.