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How to fill OpenCV image with one solid color?
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I am new in OpenCV. I have a image what I want is to change the color of each and every pixel of image with any single color.
I found that code but when I run this part of code then a exception is generated.
for (i = 0;i < img1.rows;i++) {
for (j = 0;j < img1.cols;j++) {
img1.at<Vec3b>(i, j) = Vec3b(255, 105, 240);
}
}
Can anyone please tell me the solution.
Or what I found is that this take a lot of time for the conversion So if their is any other approach then please tell me.
// Make a 3 channel image
cv::Mat img(480,640,CV_8UC3);
// Fill entire image with cyan (blue and green)
img = cv::Scalar(255,255,0);
You can use Mat::operator() to select the roi and then assign a value to it.
void main()
{
cv::Mat img = cv::Mat::ones(5, 5, CV_8UC3);
img({2, 4}, {2, 4}) = cv::Vec3b(7, 8, 9);
cout << img;
}
This will print
[ 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0;
1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0;
1, 0, 0, 1, 0, 0, 7, 8, 9, 7, 8, 9, 1, 0, 0;
1, 0, 0, 1, 0, 0, 7, 8, 9, 7, 8, 9, 1, 0, 0;
1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0, 1, 0, 0]
To fill image with single color, use rectangle with CV_FILLED argument.
(there might be some reasons for exception - image was not created, wrong pixel format etc - it is hard to diagnose a problem with given information)
I came across this:
cv::Mat Mat_out;
cv::Mat Mat2(openFingerCentroids.size(), CV_8UC1, cv::Scalar(2)); imshow("Mat2", Mat2);
cv::Mat Mat3(openFingerCentroids.size(), CV_8UC1, cv::Scalar(3)); imshow("Mat3", Mat3);
cv::bitwise_and(Mat2, Mat3, Mat_out); imshow("Mat_out", Mat_out);
Why does Mat_out contain all 2? Bit-wise operation of a matrix of all 2s and 3s should give me 0, right? Since 2 is not equal to 3?
Anyway, this is the simple thing I tried to implement: (like find function of MATLAB)
Mat_A = {1, 1, 0, 9, 0, 5;
5, 0, 0, 0, 9, 0;
1, 2, 0, 0, 0, 0};
Output expected, if I'm searching for all 5s:
Mat_out = {0, 0, 0, 0, 0, 5;
5, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0 };
How can I do this in OpenCV using C++??
What is the safest way to add integer into array value without having to call the array value ? In this case I would rather not to call the array value because the array is inside a nested loop and the loop itself can repeat for thousands times.
For example dots[1] = I want to add value of this array with 3. Here's my sample code :
void box(const Mat &dots, Mat &newdots, int rows, int cols)
{
for (int i = 0; i < dots.rows; i++) {
for (int j = 0; j < dots.cols; j++) {
newdots.at<Vec3b>(i, j)[0] = dots.at<Vec3b>(i, j)[0];
newdots.at<Vec3b>(i, j)[1] = dots.at<Vec3b>(i, j)[1]; //add this with 3
newdots.at<Vec3b>(i, j)[2] = dots.at<Vec3b>(i, j)[2]; //add this with 5
}
}
Is it possible ? Any suggestion how to do it ? Thanks.
This simplest way is to use cv::add, which overloads the + operator for the Mat class:
// Create a Mat of all 0's
cv::Mat dots = cv::Mat(5, 4, CV_8UC3, cv::Scalar(0,0,0));
std::cout << "dots:\n" << dots << std::endl;
// Add 0 to the B channel, 3 to the G channel, and 5 to R
cv::Mat newdots = dots + cv::Scalar(0, 3, 5);
std::cout << "newdots:\n" << newdots << std::endl;
Result:
dots:
[ 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, 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]
newdots:
[ 0, 3, 5, 0, 3, 5, 0, 3, 5, 0, 3, 5;
0, 3, 5, 0, 3, 5, 0, 3, 5, 0, 3, 5;
0, 3, 5, 0, 3, 5, 0, 3, 5, 0, 3, 5;
0, 3, 5, 0, 3, 5, 0, 3, 5, 0, 3, 5;
0, 3, 5, 0, 3, 5, 0, 3, 5, 0, 3, 5]
Note that dots += Scalar(0,3,5) also works if you just want to modify the original Mat.
Given a number, n, I need to efficiently find how many times this number is a multiple of all powers of 4 less than the given number.
For examples:
16 is a multiple of 4, and 16, so the result would be 2.
64 is a multiple of 4, 16, and 64, so the result would be 3.
256 is a multiple of 4, 16, 64, and 256, so the result would be 4.
14 is not a multiple of any power of 4, so the result would be 0.
35 is not a multiple of any power of 4, so the result would be 0.
Bitwise operations are preferred, and this is in a very tight loop so it is inside of a bottleneck that needs to be efficient. My code at the moment is the obvious answer, but I have to believe there is something more mathematical that can figure out the result in less steps:
power = 4;
while (power < n) {
result += !(n & (power - 1));
power *= 4;
}
You could use logarithms. A quick Google search for "fast log2 c++" brought up a pretty long list of ideas. Then your answer is log2(x)/2, and you'd have to find some way to make sure that your result is a whole number if you only want an answer for exact powers of 4.
If you are programming for an x86 processor, you can use BitScanForward & BitScanReverse to find the set bit, and use it to compute log2. The following code works in Visual Studio, for GCC or others, there are other ways to do inline assembly.
uint32_t exact_power_of_4_scan(uint32_t num)
{
unsigned long reverse;
unsigned long forward;
if (!_BitScanReverse(&reverse, num)) return 0;
_BitScanForward(&forward, num);
if (reverse != forward) return 0; // makes sure only a single bit is set
if (reverse & 0x1) return 0; // only want every other power of 2
return reverse / 2;
}
If you need a portable solution, table lookup might be the way to go, but is more complicated.
uint8_t not_single_bit[256] = {
1, 0, 0, 1, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
};
uint8_t log2_table[256] = {
0, 0, 1, 0, 2, 0, 0, 0, 3, 0, 0, 0, 0, 0, 0, 0,
4, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
5, 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,
6, 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, 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,
7, 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, 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, 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, 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
};
uint32_t exact_power_of_2(uint32_t num)
{
auto a = not_single_bit[num & 0xff];
auto b = not_single_bit[(num >> 8) & 0xff];
auto c = not_single_bit[(num >> 16) & 0xff];
auto d = not_single_bit[(num >> 24) & 0xff];
if (a + b + c + d != 3) {
return 0;
}
if (!a) {
return log2_table[num & 0xff];
}
if (!b) {
return log2_table[(num >> 8) & 0xff] + 8;
}
if (!c) {
return log2_table[(num >> 16) & 0xff] + 16;
}
return log2_table[(num >> 24) & 0xff] + 24;
}
uint32_t exact_power_of_4(uint32_t num)
{
auto ret = exact_power_of_2(num);
if (ret & 0x1) return 0;
return ret / 2;
}
Both are linear algorithms. The first will probably beat out looping for almost any value of num, but I haven't tested it. The second is probably only good for largish nums.
The mathematics would be to keep dividing by 4 until the result is no longer divisible by 4.
If you really want to do it with bitwise operations, techniques here can be used to count the number of trailing zero bits (i.e. the number of times a value is divisible by 2). Those can be adjusted to count pairs of trailing bits (i.e. divisibility by a power of 4 rather than 2).
Note that you will need to work with unsigned values to avoid certain cases of undefined or unspecified behaviours.
I would dispute your assertion that bitwise operations will make for a more efficient solution. It is not a given without testing, particularly with modern compilers.
It seems to me that using Matrix with Ranges as an L-value (assignment target) should work or not (and if not a compiler error would be nice) but not both depending on the particulars of a legitimate r-value.
cout << "hi mom" << endl;
Mat Img0=Mat::zeros(7,7,CV_8UC1);
Mat Img1=Mat::ones(7,7,CV_8UC1);
cout << Img0 << endl;
cout << Img1 << endl;
Img0(Range::all(), Range::all()) = Img1;
cout << Img0 << endl;
Img0(Range::all(), Range::all()) = 1;
cout << Img0 << endl;
Below is the output from the above. The first two matrix print outs are of Img0 and Img1 as initialized by Mat::zeros and Mat::ones respectively.
The third matrix print out is Img0 again but after
Img0(Range::all(), Range::all()) = Img1;
which I expected would set Img0 to Img1; i.e. all ones; but it's not. It's still all zeros.
The fourth/last matrix print out is the result of
Img0(Range::all(), Range::all()) = 1;
Which has the same L value as the third assignment but it works when a scalar is the Rvalue (unlike the third which as a matrix as the RValue).
Is there some sense in this that I'm missing? Should this r-value distinction behavior be allowed? It seems inconsistent to me.
[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, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 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, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0]
[1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1;
1, 1, 1, 1, 1, 1, 1]
No, this is not a bug.
This line Img0(Range::all(), Range::all()) = Img1; doesn't work as expected because Img0(Range::all(), Range::all()) forms a temporary header that is further assigned to another header, which is Img1. Remember that each of these operations is O(1), that is, no data is copied. Thus, no real assignment happens.
You can realize this effect more clearly by doing this:
(Img0(Range::all(), Range::all()) = Img1) = 2;
cout << Img0 << endl;
cout << Img1 << endl;
If you have understood what I described above, you should be aware of that the code will only change the value of Img1. And the output is:
[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, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0;
0, 0, 0, 0, 0, 0, 0]
[2, 2, 2, 2, 2, 2, 2;
2, 2, 2, 2, 2, 2, 2;
2, 2, 2, 2, 2, 2, 2;
2, 2, 2, 2, 2, 2, 2;
2, 2, 2, 2, 2, 2, 2;
2, 2, 2, 2, 2, 2, 2;
2, 2, 2, 2, 2, 2, 2]
Further reading: check out similar effect happened to Mat::row().