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Content-Based Image Retrieval and Precision-Recall graphs using Color Histograms in MATLAB
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Closed 7 years ago.
I want to implement the following Matlab function:
function hist = binnedRgbHist(im, numChannelBins)
Given an image im and a number between 1 and 256 numChannelBins, it should create a histogram sized (numChannelBins)^3.
For example, if numChannelBins is 2, it should produce the following 8-sized histogram:
Number of pixels with R < 128, G < 128, B < 128
Number of pixels with R < 128, G < 128, B >= 128
Number of pixels with R < 128, G >= 128, B < 128
Number of pixels with R < 128, G >= 128, B >= 128
Number of pixels with R > 128, G < 128, B < 128
Number of pixels with R > 128, G < 128, B >= 128
Number of pixels with R > 128, G >= 128, B < 128
Number of pixels with R > 128, G >= 128, B >= 128
It is like creating a cube where each axis represents one of (R,G and B), where each axis is divided into 2 bins => Finally there are 8 bins in the cube.
My questions:
It there a built-in function for it?
If not, how is it better to implement it in manners of runtinme using the GPU? Should I better iterate over the pixels once and create the histogram manually, or should I better iterate over the bins and each time count the number of pixels which satisfy the bin's conditions?
accumarray is very suited for this. Let
im: input image;
N: number of bins per color component.
Then
result = accumarray(reshape(permute(ceil(im/255*N), [3 1 2]), 3, []).', 1, [N N N]);
How it works
ceil(im/255*N) quantizes each color vaue to 1, 2, ..., N.
reshape(permute(..., [3 1 2]), 3, []).' transforms the quantized image into a three-column matrix where each row is a pixel and each column is a (quantized) color component.
accumarray(..., 1, [N N N]) considers each row of that matrix as 3D index, and counts how many times each index appears, giving filling indices that don't appear with a 0.
Example 1
Data:
>> N = 2;
>> im = randi(256,4,5,3)
im(:,:,1) =
113 152 157 65 229
138 71 215 39 41
13 108 230 160 153
142 128 125 220 214
im(:,:,2) =
208 215 182 27 230
205 161 8 95 180
225 53 73 129 31
103 97 160 83 255
im(:,:,3) =
242 29 185 89 55
202 225 156 174 96
160 197 35 87 113
244 176 146 85 120
Result:
result(:,:,1) =
1 1
3 4
result(:,:,2) =
2 4
3 2
It can be checked for example that there is only 1 pixel with all R,G,B less than 128.
Example 2
Data:
>> im = repmat(150,20,30,3);
>> N = 4;
Result:
result(:,:,1) =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
result(:,:,2) =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
result(:,:,3) =
0 0 0 0
0 0 0 0
0 0 600 0
0 0 0 0
result(:,:,4) =
0 0 0 0
0 0 0 0
0 0 0 0
0 0 0 0
In this case all pixels belong to the same 3D-bin:
I see #Luis Mendo gave a great one-liner solution as I was writing this. In case it provides some deeper intuition, my solution makes use of histcounts and accumarray:
im = randi([1 255],[10,5,3]); %// A random 10-by-5 "image"
numChannelBins = 2;
[~,~,binR]=histcounts(im(:,:,1),[1 ceil((1:numChannelBins)*(255/numChannelBins))]);
[~,~,binG]=histcounts(im(:,:,2),[1 ceil((1:numChannelBins)*(255/numChannelBins))]);
[~,~,binB]=histcounts(im(:,:,3),[1 ceil((1:numChannelBins)*(255/numChannelBins))]);
hist=accumarray([binR(:) binG(:) binB(:)],1,[numChannelBins,numChannelBins,numChannelBins])
Explanation:
the three calls to histcounts bin the red, green, blue pixels separately -- the third output [~,~,binX] of histcounts gives the bin index for each pixel
accumarray accumulates all the unique index triplets
I need to make a handwritten image to be tested with a neural network in Matlab. When I see the data contained in the training images from the MNIST I see that it is an array of different gray scales like:
Columns 343 through 351
0 -0.0240 0.4002 0.6555 0.0235 -0.0062 0 0 0
Columns 352 through 360
0 0 0 0 0 0 0 0 0
Columns 361 through 369
0 0 0 -0.0079 0.1266 0.3272 -0.0233 0.0005
corresponding to a 20x20 image, unrolled into a 1*400 dimensional array.
I have downloaded an image in jpeg format and did the following:
im=imread('image.jpg');
gi=rgb2gray(im);
gi=gi(:);
gi=gi';
that generates me an array gi that says <1*400 uint8>, the last part of uint8 does not appear in the MNIST samples when I put it in Matlab. When I check it up my array it appear the following values:
Columns 289 through 306
58 105 128 133 142 131 76 21 1 0 3 0 2 4 17 12 7 0
Columns 307 through 324
1 15 42 75 97 105 98 73 31 4 1 0 0 0 0 2 4 3
Columns 325 through 342
0 0 1 4 21 37 55 59 46 26 9 0 0 0 0 0 0 0
Columns 343 through 360
1 1 0 0 0 1 7 14 21 21 14 5 0 0 0 0 0 0
Columns 361 through 378
0 0 0 0 0 0 0 0 0 1 2 1 0 0 0 2 0 0
when I visualize them all is fine, but when I want to run my program the following message appears:
??? Error using ==> mtimes
MTIMES is not fully supported for integer classes. At least one input must be scalar.
Error in ==> predict at 15
h1 = sigmoid([ones(m, 1) X] * Theta1');
Error in ==> ex4 at 241
pred = predict(Theta1, Theta2, gi);
situation that does not occur when I test my program even with one random sample ofc the MNIST data; any help?
You could try something like this:
imfile = 'image.jpg';
im = double(rgb2gray(imread(imfile))); % double and convert to grayscale
im = imresize(im,[20,20]); % change to 20 by 20 dimension
im = im(:); % unroll matrix to vector
im = im./max(im);
Note the MNIST dataset is intended to be a good dataset to require minimal preprocessing and the images were actually originally black and white (bilevel) whereas you are using color image. Also they do normalisation and other preprocessing to make nice 28 by 28 image dataset, my brief snippet of code above is unlikely to be anywhere near as good as MNIST dataset and is just intended to attempt to fix your error.
Your specific error is likely because you don't use double().
You may also get further errors because your code needs right dimensions, which can be achieved using imresize.
More information on MNIST dataset here:
http://yann.lecun.com/exdb/mnist/
I have written a routine to give the column index position for the furthers right cell containing a 1, marking the right edge of a polygon in a mask array.
But when I print the index arrays (I realize the row position array is redundant), all rows are reporting a position (other than 0), which I know shouldn't be the case. I can't seem to find where the following could be incorrect.
Example mask array:
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 0 0 0
0 0 0 0 0 0 1 1 1 0 0 0 0 0
0 0 0 0 0 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 0 0 0
0 0 0 1 1 1 1 1 1 1 1 0 0 0
0 0 0 0 0 0 1 1 1 1 0 0 0 0
0 0 0 0 0 0 0 0 0 1 0 0 0 0
0 0 0 0 0 0 0 0 0 0 0 0 0 0
Desired output:
[0,10,10,11,11,11,9,8,0]
Is this a Fortran thing, or is my logic just off (
Routine:
subroutine get_right_idx(mask_array, idx_x, idx_y)
integer, parameter :: x = 169 ! num columns
integer, parameter :: y = 124 ! num rows
integer i ! iterator for rows
integer j ! iterator for columns
integer row_x ! x position for furthest south cell in row
integer row_y ! y position for furthest south cell in row
integer :: idx_x(y) ! index positions for lowest lat in model grid - x
integer :: idx_y(y) ! index positions for lowest lat in model grid - y
real mask_array(y,x) ! mask array of zeros, containing polygon of ones
do j=1,y
row_x = 0
row_y = 0
do i=1,x
if (mask_array(j,i).eq.1) then
row_x = i
row_y = j
endif
enddo
idx_x(j)=row_x
idx_y(j)=row_y
enddo
endsubroutine get_right_idx
Actual mask (zoomed out in Open Office):
Below is the mask that I am trying to evaluate. My resulting array has a non-zero value in for all elements, where there should be zero elements at the start and end of the array, no matter which direction it is evaluated from.
Output:
125 104 104 104 104 104 104 114 114 114 114 103 103 103 108 108 103 103 103 103 97 97 97 107 107 107 107 107 107 107 107 107 97 101 101 101 101 101 111 111 111 111 111 111 101 101 100 105 105 105 105 105 105 100 115 115 104 104 104 104 104 104 104 104 104 104 98 98 98 98 108 108 108 108 108 108 108 108 98 102 102 102 102 102 112 112 112 112 112 112 101 101 101 106 106 106 101 101 101 95 95 105 105 105 105 105 105 105 105 105 105 99 99 99 99 99 109 109 109 109 109 109 109 99 99
I made a quick template of your code, but with my suggestion. Since you want to find the largest index j that has a 1 in it, you can simply test this by starting your j-loop with the maximum (in the example, 14) and working downwards but exiting the inner do-loop if mask_array==1.
You could generalize this by replacing 14 and 9 with the maximum values for those dimensions.
program get_right
implicit none
integer, dimension(9,14) :: mask_array
integer :: i,j,mask_out(9),j_tmp
mask_array(1,:)=[0,0,0,0,0,0,0,0,0,0,0,0,0,0]
mask_array(2,:)=[0,0,0,0,0,0,0,1,0,0,0,0,0,0]
mask_array(3,:)=[0,0,0,0,0,0,1,1,1,0,0,0,0,0]
mask_array(4,:)=[0,0,0,0,0,1,1,1,1,1,1,0,0,0]
mask_array(5,:)=[0,0,0,1,1,1,1,1,1,1,1,0,0,0]
mask_array(6,:)=[0,0,0,1,1,1,1,1,1,1,1,0,0,0]
mask_array(7,:)=[0,0,0,0,0,0,1,1,1,1,0,0,0,0]
mask_array(8,:)=[0,0,0,0,0,0,0,0,0,1,0,0,0,0]
mask_array(9,:)=[0,0,0,0,0,0,0,0,0,0,0,0,0,0]
mask_out=0
do i=1,9
j_tmp=0
print '(14(i0,2x))',mask_array(i,:)
do j=14,1,-1
if(mask_array(i,j)==1) then
j_tmp=j
exit
endif
enddo
mask_out(i)=j
enddo
print *,""
print '(9(i0,2x))',mask_out
end program get_right
I got 0,8,9,11,11,11,10,10,0 as the answer (which is what you got but backwards).
Or, if typing makes you tired you could evaluate this
maxval(mask_array*spread([(ix,ix=lbound(mask_array,dim=2),&
ubound(mask_array,dim=2))],dim=1, ncopies=size(mask_array,dim=1)),dim=2)
As you can see this makes a temporary array (which may be undesirable if your masks are large) using spread and an implied-do loop, each element of this temporary contains its own column index number. Then it multiplies the temporary with the mask_array, performing element-wise multiplication not matrix multiplication. Finally take the maxval of each row of the result. This returns the vector in the same order as Kyle's code does.
I've edited the code to use lbound and ubound rather than 1 and size in case you want to use the code on arrays with lower bounds other than 1.
Bob's yer uncle, but don't ask whether this is faster than Kyle's code. If you are interested in execution speed test and measure.
Incidentally, since this returns just one vector and doesn't modify its arguments or have any other side effects, I'd package it as a function rather than as a subroutine.
The following code works for all positive inputs correctly but, negative values are bogus. It would work if negative values are set to 0 initially, because they do add. Example at the end.
#include <stdio.h> /* Necessary header */
#define MAX_RESPONDENTS 6
#define MIN_RESPONSE_VALUE -100 /* Abbreviated MiRV */
#define MAX_RESPONSE_VALUE 86 /* Abbreviated MaRV */
#define RESPONSE_VALUE 187 /* Equals |(MiRV)| + or - |(MaRV)| + 1 */
#define STOP 3
#define BREAK 1
int main(void)
{
CountRating();
return 0;
}
CountRating()
{
int ratingCounters, rating[RESPONSE_VALUE] = {0}, Counter, response;
for (ratingCounters = 0; ratingCounters < MAX_RESPONDENTS;)
{
int response = ratingCounters, stop = STOP;
printf("Enter a integer rating between %d and %d for the product: ", MIN_RESPONSE_VALUE, MAX_RESPONSE_VALUE);
scanf("%d", &response);
if (response <= MAX_RESPONSE_VALUE && response >= MIN_RESPONSE_VALUE)
stop = STOP;
else
{
int stopElse = stop;
if (stopElse < BREAK)
break;
else
{
do
{
printf("\nNot within range. Try again.\nYou have %d more attempts before program outputs total:", stop);
scanf("%d", &response);
printf("\n");
--stop;
if (stop < BREAK)
break;
} while (response > MAX_RESPONSE_VALUE || response < MIN_RESPONSE_VALUE);
} if (stop < BREAK)
break;
}
++rating[response];
++ratingCounters;
}
printf("\nRating Number of Responses\n");
printf("------ -------------------");
for (Counter = MAX_RESPONSE_VALUE; Counter >= MIN_RESPONSE_VALUE; --Counter)
{
printf("\n%3d%24d", Counter, rating[Counter]);
}
}
Example Output:
Enter a integer rating between -100 and 86 for the product: 66
Enter a integer rating between -100 and 86 for the product: 66
Enter a integer rating between -100 and 86 for the product: 66
Enter a integer rating between -100 and 86 for the product: 66
Enter a integer rating between -100 and 86 for the product: 66
Enter a integer rating between -100 and 86 for the product: 55
Rating Number of Responses
------ -------------------
86 0
85 0
84 0
83 0
82 0
81 0
80 0
79 0
78 0
77 0
76 0
75 0
74 0
73 0
72 0
71 0
70 0
69 0
68 0
67 0
66 5
65 0
64 0
63 0
62 0
61 0
60 0
59 0
58 0
57 0
56 0
55 1
54 0
53 0
52 0
51 0
50 0
49 0
48 0
47 0
46 0
45 0
44 0
43 0
42 0
41 0
40 0
39 0
38 0
37 0
36 0
35 0
34 0
33 0
32 0
31 0
30 0
29 0
28 0
27 0
26 0
25 0
24 0
23 0
22 0
21 0
20 0
19 0
18 0
17 0
16 0
15 0
14 0
13 0
12 0
11 0
10 0
9 0
8 0
7 0
6 0
5 0
4 0
3 0
2 0
1 0
0 0
-1 55
-2 2686244
-3 2686244
-4 -3
-5 4206858
-6 4199517
-7 1965606432
-8 -2
-9 -533773290
-10 1965657301
-11 2686916
-12 1965975817
-13 2685900
-14 28
-15 2685972
-16 0
-17 2686720
-18 -1792050746
-19 1965606432
-20 1966156064
-21 1
-22 1965606483
-23 2685944
-24 17
-25 1965552124
-26 2685880
-27 1966146752
-28 1965532226
-29 2685868
-30 1966156064
-31 1
-32 0
-33 1966156064
-34 0
-35 28
-36 6105760
-37 1
-38 1965551966
-39 1965572357
-40 -2
-41 -533740746
-42 1965657301
-43 2685928
-44 1966156064
-45 2685768
-46 28
-47 1965572357
-48 0
-49 1966156064
-50 28
-51 -1792050874
-52 1965572357
-53 1
-54 1965572401
-55 2685816
-56 6097440
-57 1965552169
-58 2685752
-59 36
-60 1966146816
-61 0
-62 1966146816
-63 36
-64 0
-65 -1792050942
-66 1965552271
-67 1965552263
-68 2685756
-69 6105754
-70 4104
-71 4096
-72 6097428
-73 36
-74 -2
-75 358724962
-76 1999401429
-77 2685928
-78 -1
-79 2685472
-80 16777216
-81 6094848
-82 1
-83 0
-84 6105755
-85 6105760
-86 6095044
-87 8
-88 2
-89 1999167428
-90 1
-91 6095044
-92 -218103565
-93 2685640
-94 -254549010
-95 0
-96 1136989510
-97 -503245737
-98 2752512
-99 6098688
-100 0
Process returned 0 (0x0) execution time : 9.734 s
Press any key to continue.
Rating is an array of size 187. You're indexing negative values, which means C is accessing memory before the beginning of the array and printing whatever garbage values are stored there. You can only access values between 0 and 186 legally, so you should just add 100 to the index whenever you access and store rating. That way, the number -100 corresponds to index 0, -99 to index 1, etc.
The index of an array always starts from zero. You should not pass a negative index for an array. If you want to have negative indexes, add a constant to the index.
In your example code, you can add -MIN_RESPONSE_VALUE to the index, so accessing rating[k - MIN_RESPONSE] if you want to access index k.
Your code is assigning to negative indices of an array. This is generally not good - you have only reserved memory from 0:RESPONSE_VALUE-1, writing to anything outside that range will have unpredictable consequences. You have two general options. One is to map your range MIN_RESPONSE_VALUE:MAX_RESPONSE_VALUE to 0:(MAX_RESPONSE_VALUE-MIN_RESPONSE_VALUE). I.e. subtract MIN_RESPONSE_VALUE whenever you index rating:
++rating[response-MIN_RESPONSE_VALUE];
The other is to repoint response, giving it a new zero point. This will save you from having to subtract MIN_RESPONSE_VALUE all the time, but may be confusing to readers:
rating -= MIN_RESPONSE_VALUE;
Do this before the for loop. After doing this, rating[MIN_RESPONSE_VALUE] is a valid index, even though it is negative.
Actually the negative index is totally acceptable. The problem is that you index goes into areas outside memory dedicated for array on the stack. Changing rating[-1] you changing your stack outside the array and thus corrupting it. No wonder the values there are not initialized to zero with rating[RESPONSE_VALUE] = {0}, again they are outside the array and show you some data on your stack.
Took me a while but I got it. Thanks.
#include <stdio.h> /* Necessary header */
#define MAX_RESPONDENTS 6
#define MIN_RESPONSE_VALUE -100 /* Abbreviated MiRV */
#define MAX_RESPONSE_VALUE 86 /* Abbreviated MaRV */
#define RESPONSE_VALUE 187 /* Equals |(MiRV)| + or - |(MaRV)| + 1 */
#define STOP 3
#define BREAK 1
#define DOUBLE_RESPONSE_VALUE 374
int main(void)
{
CountRating();
return 0;
}
CountRating()
{
int ratingCounters, rating[DOUBLE_RESPONSE_VALUE] = {0}, Counter, response;
for (ratingCounters = 0; ratingCounters < MAX_RESPONDENTS;)
{
int response = ratingCounters, stop = STOP;
printf("Enter a integer rating between %d and %d for the product: ", MIN_RESPONSE_VALUE, MAX_RESPONSE_VALUE);
scanf("%d", &response);
if (response <= MAX_RESPONSE_VALUE && response >= MIN_RESPONSE_VALUE)
stop = STOP;
else
{
int stopElse = stop;
if (stopElse < BREAK)
break;
else
{
do
{
printf("\nNot within range. Try again.\nYou have %d more attempts before program outputs total:", stop);
scanf("%d", &response);
printf("\n");
--stop;
if (stop < BREAK)
break;
} while (response > MAX_RESPONSE_VALUE || response < MIN_RESPONSE_VALUE);
} if (stop < BREAK)
break;
}
if (response < 0)
++rating[response + abs(RESPONSE_VALUE)];
else
++rating[response + abs(RESPONSE_VALUE)];
++ratingCounters;
}
printf("\nRating Number of Responses\n");
printf("------ -------------------");
for (Counter = MAX_RESPONSE_VALUE; Counter >= MIN_RESPONSE_VALUE; --Counter)
{
printf("\n%3d%24d", Counter, rating[Counter + (RESPONSE_VALUE)]);
}
}
I am hesitant to post my code because this is for a school assignment, but basically I need to iterate through a uint64_t that can be very large.
So, I have a value that is basically going to be maybe 10 characters long. It originally was an int, and the code partially works (but doesn't finish creating the output it's supposed to) but then I'd get the error "warning: integer constant is so large that it is unsigned".
I'd then change that int to uint64_t (it's being multiplied by 10), and it would give me a floating point exception.
Any insight without the code?
EDIT: Adding little bits of code...
uint64_t number = 1U;
...
for (int x = 0; x < y; x++) {
number *= 10;
}
^ this gives me the floating point exception.
Not sure if this is enough to be helpful.
By default, integer literals are signed. If you use the U suffix on the literal, that tells the compiler to make it unsigned, e.g. 123456789U. That should clear up the warning about the constant being too large. Alternatively, you could use the UINT64_C() macro from <stdint.h>.
It's hard to know what's causing your FPE without seeing some code.
Have you checked how many times of the for loop executed? In your for loop, the number increased with a exponential ratio.I wrote a small program to show the situation.
1 #include <stdio.h>
2 #include <stdlib.h>
3 #include <stdint.h>
4
5 #define MAXLOOP 100
6
7 int
8 main(void)
9 {
10 uint64_t number = 1U;
11 int i;
12
13 for(i = 0; i < MAXLOOP; ++i){
14 number *= 10;
15 printf("%d\t%lu\n", i, number);
16 }
17
18 return 0;
19 }
20
The output is:
0 10
1 100
2 1000
3 10000
4 100000
5 1000000
6 10000000
7 100000000
8 1000000000
9 10000000000
10 100000000000
11 1000000000000
12 10000000000000
13 100000000000000
14 1000000000000000
15 10000000000000000
16 100000000000000000
17 1000000000000000000
18 10000000000000000000
19 7766279631452241920 <--- overflow, uint64_t can store 18,446,744,073,709,551,616 as its max value
20 3875820019684212736
21 1864712049423024128
22 200376420520689664
23 2003764205206896640
24 1590897978359414784
25 15908979783594147840
26 11515845246265065472
27 4477988020393345024
28 7886392056514347008
29 5076944270305263616
30 13875954555633532928
31 9632337040368467968
32 4089650035136921600
33 4003012203950112768
34 3136633892082024448
35 12919594847110692864
36 68739955140067328
37 687399551400673280
38 6873995514006732800
39 13399722918938673152
40 4870020673419870208
41 11806718586779598848
42 7386721425538678784
43 80237960548581376
44 802379605485813760
45 8023796054858137600
46 6450984253743169536
47 9169610316303040512
48 17909126868192198656
49 13070572018536022016
50 1578511669393358848
51 15785116693933588480
52 10277214349659471872
53 10538423128046960640
54 13150510911921848320
55 2377900603251621888
56 5332261958806667264
57 16429131440647569408
58 16717361816799281152
59 1152921504606846976
60 11529215046068469760
61 4611686018427387904
62 9223372036854775808
63 0
64 0
65 0
66 0
67 0
68 0
69 0
70 0
71 0
72 0
73 0
74 0
75 0
76 0
77 0
78 0
79 0
80 0
81 0
82 0
83 0
84 0
85 0
86 0
87 0
88 0
89 0
90 0
91 0
92 0
93 0
94 0
95 0
96 0
97 0
98 0
99 0
I think it is easy for you to understand.:->