I would like your help to make more efficient (maybe, by vectorising) the Matlab code below. The code below does essentially the following: take a row vector A; consider the maximum elements of such a row vector and let, for example, be i and j their positions; construct two columns vectors, the first with all zeros but a 1 positioned at i, the second with all zeros but a 1 positioned at j.
This is my attempt with loops, but it looks more complicated than needed.
clear
rng default
A=[3 2 3];
max_idx=ismember(A,max(A));
vertex=cell(size(A,2),1);
for j=1:size(max_idx,2)
if max_idx(j)>0
position=find(max_idx(j));
vertex_temp=zeros(size(A,2),1);
vertex_temp(position)=1;
vertex{j}=vertex_temp;
else
vertex{j}=[];
end
end
vertex=vertex(~cellfun('isempty',vertex));
Still using a for loop, but more readable:
A = [3 2 3];
% find max indices
max_idx = find(A == max(A));
vertex = cell(numel(max_idx),1);
for k = 1:numel(max_idx)
vertex{k} = zeros(size(A,2),1); % init zeros
vertex{k}(max_idx(k)) = 1; % set value in vector to 1
end
If you really wanted to avoid a for loop, you could probably also use something like this:
A=[3 2 3];
max_idx = find(A==max(A));
outMat = zeros(numel(A), numel(max_idx));
outMat((0:(numel(max_idx)-1)) * numel(A) + max_idx) = 1;
then optionally, if you want them in separate cells rather than columns of a matrix:
outCell = mat2cell(outMat, numel(A), ones(1,numel(max_idx)))';
However, I think this might be less simple and readable than the existing answers.
Is there a specific reason you want a cell array rather than a matrix?
If you can have it all in one vector:
A = [3 2 3]
B_rowvec = A == max(A)
B_colvec = B_rowvec'
If you need them separated into separate vectors:
A = [3 2 3]
Nmaxval = sum(A==max(A))
outmat = zeros(length(A),Nmaxval)
for i = 1:Nmaxval
outmat(find(A==max(A),i),i)=1;
end
outvec1 = outmat(:,1)
outvec2 = outmat(:,2)
Basically, the second input for find will specify which satisfactory instance of the first input you want.
so therefore
example = [ 1 2 3 1 2 3 1 2 3 ]
first = find(example == 1, 1) % returns 1
second = find(example == 1, 2) % returns 4
third = find(example == 1, 3) % returns 7
I was looking at the docs of tensorflow about tf.nn.conv2d here. But I can't understand what it does or what it is trying to achieve. It says on the docs,
#1 : Flattens the filter to a 2-D matrix with shape
[filter_height * filter_width * in_channels, output_channels].
Now what does that do? Is that element-wise multiplication or just plain matrix multiplication? I also could not understand the other two points mentioned in the docs. I have written them below :
# 2: Extracts image patches from the the input tensor to form a virtual tensor of shape
[batch, out_height, out_width, filter_height * filter_width * in_channels].
# 3: For each patch, right-multiplies the filter matrix and the image patch vector.
It would be really helpful if anyone could give an example, a piece of code (extremely helpful) maybe and explain what is going on there and why the operation is like this.
I've tried coding a small portion and printing out the shape of the operation. Still, I can't understand.
I tried something like this:
op = tf.shape(tf.nn.conv2d(tf.random_normal([1,10,10,10]),
tf.random_normal([2,10,10,10]),
strides=[1, 2, 2, 1], padding='SAME'))
with tf.Session() as sess:
result = sess.run(op)
print(result)
I understand bits and pieces of convolutional neural networks. I studied them here. But the implementation on tensorflow is not what I expected. So it raised the question.
EDIT:
So, I implemented a much simpler code. But I can't figure out what's going on. I mean how the results are like this. It would be extremely helpful if anyone could tell me what process yields this output.
input = tf.Variable(tf.random_normal([1,2,2,1]))
filter = tf.Variable(tf.random_normal([1,1,1,1]))
op = tf.nn.conv2d(input, filter, strides=[1, 1, 1, 1], padding='SAME')
init = tf.initialize_all_variables()
with tf.Session() as sess:
sess.run(init)
print("input")
print(input.eval())
print("filter")
print(filter.eval())
print("result")
result = sess.run(op)
print(result)
output
input
[[[[ 1.60314465]
[-0.55022103]]
[[ 0.00595062]
[-0.69889867]]]]
filter
[[[[-0.59594476]]]]
result
[[[[-0.95538563]
[ 0.32790133]]
[[-0.00354624]
[ 0.41650501]]]]
Ok I think this is about the simplest way to explain it all.
Your example is 1 image, size 2x2, with 1 channel. You have 1 filter, with size 1x1, and 1 channel (size is height x width x channels x number of filters).
For this simple case the resulting 2x2, 1 channel image (size 1x2x2x1, number of images x height x width x x channels) is the result of multiplying the filter value by each pixel of the image.
Now let's try more channels:
input = tf.Variable(tf.random_normal([1,3,3,5]))
filter = tf.Variable(tf.random_normal([1,1,5,1]))
op = tf.nn.conv2d(input, filter, strides=[1, 1, 1, 1], padding='VALID')
Here the 3x3 image and the 1x1 filter each have 5 channels. The resulting image will be 3x3 with 1 channel (size 1x3x3x1), where the value of each pixel is the dot product across channels of the filter with the corresponding pixel in the input image.
Now with a 3x3 filter
input = tf.Variable(tf.random_normal([1,3,3,5]))
filter = tf.Variable(tf.random_normal([3,3,5,1]))
op = tf.nn.conv2d(input, filter, strides=[1, 1, 1, 1], padding='VALID')
Here we get a 1x1 image, with 1 channel (size 1x1x1x1). The value is the sum of the 9, 5-element dot products. But you could just call this a 45-element dot product.
Now with a bigger image
input = tf.Variable(tf.random_normal([1,5,5,5]))
filter = tf.Variable(tf.random_normal([3,3,5,1]))
op = tf.nn.conv2d(input, filter, strides=[1, 1, 1, 1], padding='VALID')
The output is a 3x3 1-channel image (size 1x3x3x1).
Each of these values is a sum of 9, 5-element dot products.
Each output is made by centering the filter on one of the 9 center pixels of the input image, so that none of the filter sticks out. The xs below represent the filter centers for each output pixel.
.....
.xxx.
.xxx.
.xxx.
.....
Now with "SAME" padding:
input = tf.Variable(tf.random_normal([1,5,5,5]))
filter = tf.Variable(tf.random_normal([3,3,5,1]))
op = tf.nn.conv2d(input, filter, strides=[1, 1, 1, 1], padding='SAME')
This gives a 5x5 output image (size 1x5x5x1). This is done by centering the filter at each position on the image.
Any of the 5-element dot products where the filter sticks out past the edge of the image get a value of zero.
So the corners are only sums of 4, 5-element dot products.
Now with multiple filters.
input = tf.Variable(tf.random_normal([1,5,5,5]))
filter = tf.Variable(tf.random_normal([3,3,5,7]))
op = tf.nn.conv2d(input, filter, strides=[1, 1, 1, 1], padding='SAME')
This still gives a 5x5 output image, but with 7 channels (size 1x5x5x7). Where each channel is produced by one of the filters in the set.
Now with strides 2,2:
input = tf.Variable(tf.random_normal([1,5,5,5]))
filter = tf.Variable(tf.random_normal([3,3,5,7]))
op = tf.nn.conv2d(input, filter, strides=[1, 2, 2, 1], padding='SAME')
Now the result still has 7 channels, but is only 3x3 (size 1x3x3x7).
This is because instead of centering the filters at every point on the image, the filters are centered at every other point on the image, taking steps (strides) of width 2. The x's below represent the filter center for each output pixel, on the input image.
x.x.x
.....
x.x.x
.....
x.x.x
And of course the first dimension of the input is the number of images so you can apply it over a batch of 10 images, for example:
input = tf.Variable(tf.random_normal([10,5,5,5]))
filter = tf.Variable(tf.random_normal([3,3,5,7]))
op = tf.nn.conv2d(input, filter, strides=[1, 2, 2, 1], padding='SAME')
This performs the same operation, for each image independently, giving a stack of 10 images as the result (size 10x3x3x7)
2D convolution is computed in a similar way one would calculate 1D convolution: you slide your kernel over the input, calculate the element-wise multiplications and sum them up. But instead of your kernel/input being an array, here they are matrices.
In the most basic example there is no padding and stride=1. Let's assume your input and kernel are:
When you use your kernel you will receive the following output: , which is calculated in the following way:
14 = 4 * 1 + 3 * 0 + 1 * 1 + 2 * 2 + 1 * 1 + 0 * 0 + 1 * 0 + 2 * 0 + 4 * 1
6 = 3 * 1 + 1 * 0 + 0 * 1 + 1 * 2 + 0 * 1 + 1 * 0 + 2 * 0 + 4 * 0 + 1 * 1
6 = 2 * 1 + 1 * 0 + 0 * 1 + 1 * 2 + 2 * 1 + 4 * 0 + 3 * 0 + 1 * 0 + 0 * 1
12 = 1 * 1 + 0 * 0 + 1 * 1 + 2 * 2 + 4 * 1 + 1 * 0 + 1 * 0 + 0 * 0 + 2 * 1
TF's conv2d function calculates convolutions in batches and uses a slightly different format. For an input it is [batch, in_height, in_width, in_channels] for the kernel it is [filter_height, filter_width, in_channels, out_channels]. So we need to provide the data in the correct format:
import tensorflow as tf
k = tf.constant([
[1, 0, 1],
[2, 1, 0],
[0, 0, 1]
], dtype=tf.float32, name='k')
i = tf.constant([
[4, 3, 1, 0],
[2, 1, 0, 1],
[1, 2, 4, 1],
[3, 1, 0, 2]
], dtype=tf.float32, name='i')
kernel = tf.reshape(k, [3, 3, 1, 1], name='kernel')
image = tf.reshape(i, [1, 4, 4, 1], name='image')
Afterwards the convolution is computed with:
res = tf.squeeze(tf.nn.conv2d(image, kernel, [1, 1, 1, 1], "VALID"))
# VALID means no padding
with tf.Session() as sess:
print sess.run(res)
And will be equivalent to the one we calculated by hand.
For examples with padding/strides, take a look here.
Just to add to the other answers, you should think of the parameters in
filter = tf.Variable(tf.random_normal([3,3,5,7]))
as '5' corresponding to the number of channels in each filter. Each filter is a 3d cube, with a depth of 5. Your filter depth must correspond to your input image's depth. The last parameter, 7, should be thought of as the number of filters in the batch. Just forget about this being 4D, and instead imagine that you have a set or a batch of 7 filters. What you do is create 7 filter cubes with dimensions (3,3,5).
It is a lot easier to visualize in the Fourier domain since convolution becomes point-wise multiplication. For an input image of dimensions (100,100,3) you can rewrite the filter dimensions as
filter = tf.Variable(tf.random_normal([100,100,3,7]))
In order to obtain one of the 7 output feature maps, we simply perform the point-wise multiplication of the filter cube with the image cube, then we sum the results across the channels/depth dimension (here it's 3), collapsing to a 2d (100,100) feature map. Do this with each filter cube, and you get 7 2D feature maps.
I tried to implement conv2d (for my studying). Well, I wrote that:
def conv(ix, w):
# filter shape: [filter_height, filter_width, in_channels, out_channels]
# flatten filters
filter_height = int(w.shape[0])
filter_width = int(w.shape[1])
in_channels = int(w.shape[2])
out_channels = int(w.shape[3])
ix_height = int(ix.shape[1])
ix_width = int(ix.shape[2])
ix_channels = int(ix.shape[3])
filter_shape = [filter_height, filter_width, in_channels, out_channels]
flat_w = tf.reshape(w, [filter_height * filter_width * in_channels, out_channels])
patches = tf.extract_image_patches(
ix,
ksizes=[1, filter_height, filter_width, 1],
strides=[1, 1, 1, 1],
rates=[1, 1, 1, 1],
padding='SAME'
)
patches_reshaped = tf.reshape(patches, [-1, ix_height, ix_width, filter_height * filter_width * ix_channels])
feature_maps = []
for i in range(out_channels):
feature_map = tf.reduce_sum(tf.multiply(flat_w[:, i], patches_reshaped), axis=3, keep_dims=True)
feature_maps.append(feature_map)
features = tf.concat(feature_maps, axis=3)
return features
Hope I did it properly. Checked on MNIST, had very close results (but this implementation is slower). I hope this helps you.
In addition to other answers, conv2d operation is operating in c++ (cpu) or cuda for gpu machines that requires to flatten and reshape data in certain way and use gemmBLAS or cuBLAS(cuda) matrix multiplication.
It's performing convulition throught the picture when you are trying for example image classifation thuis function has all the parameters need to do that.
When you are basically can chose the filter dimension. Strides. Padding. Before to used its need to undestant the concepts of convolution
this explanation complements:
Keras Conv2d own filters
I had some doubts about the filter parameters in keras.conv2d because when I learned I was supposed to set my own filter design. But this parameters tells how many filters to test and keras itself will try to find the best filters weights.
I've written a simple function that takes a vector vec, iterates through it, performing an operation whose result is stored in another vector vecRes of same size at same index, and returns vecRes upon completing the loop. Below is function code:
function [ vecRes ] = squareTerms( vec )
vecSize = size(vec);
vecRes = zeros(vecSize);
for i = 1:vecSize
vecRes(i) = vec(i)^2;
end
end
Problem is that it seems to exit too early, after only one iteration in fact as the output appears as:
vecRes = 1 0 0 0 0 0 0 0 0 0
For input:
vec = 1 2 3 4 5 6 7 8 9 10
I can't figure out why it does so. Any help is greatly appreciated.
Size returns 2 values, rows and columns. Probably you are a having a 1xN vector. So size returns [1 N] and your loop runs 1 time.
>>> size ([1 2 3])
>
> ans =
>
> 1 3
>
>>> 1:size ([1 2 3])
>
> ans =
>
> 1
Others have pointed out the problem. My preferred solution in this sort of case is to use numel, i.e.
vecRes = zeros(size(vec));
for i = 1:numel(vec)
vecRes(i) = vec(i) ^ 2;
end
Of course, in this case, vectorisation is better still:
vecRes = vec .^ 2;
Replace
for i = 1:vecSize
with
for i = 1:vecSize(2)
vecSize is an array of numbers, not just a single value. For example, if vec is a 1 by 8 vector, then size(vec) will return [1, 8].
Therefore, your for-loop-statement,
for i = 1:vecSize
, is actually equivalent to something like:
for i = 1:[1, 8]
This doesn't make much sense. There are a number of ways to fix the problem. You could write:
for i = 1:length(vec)
or
for i = 1:numel(vec) % "numel" stands for "number of elements"
If the vector is 1xn instead of nx1, you can write:
for i = 1:size(vec, 2)
Yet another alternative is:
for i = 1:max(vecSize)
However, the most sensible option is not to write the squareTerms function at all and simply write
vecRes = vec.^2;
Note the dot before the caret. vec^2 and vec . ^2 are not the same thing.
If you put a dot before an operator sign, the operation will be performed element-wise. For example,
C = A * B
performs matrix multiplication, but
C = A .* B
will cause the first element of A to by multiplied by the first element of B, and the result will be assigned to the first element of C. Then, the product of the second elements of A and B will be taken, and the result will be stuck in the second element of C, and so on.
vecRes = vec.^2;