How to resize without interpolation (zero-padding) in OpenCV?

Question

Is there an efficient way to resize an image in OpenCV without using any interpolation? Instead of the conventional "resize" I would like my image to remap the pixels into a larger image but pad everything else with 0.

e.g. to scale up img1 below 2x to img2:

img1 = [ 1, 2, 3,
         4, 5, 6,
         7, 8, 9 ]

cv::resize(img1, img2, cv::Size(6, 6));

img2 = [ 1, 0, 2, 0, 3, 0,
         0, 0, 0, 0, 0, 0,
         4, 0, 5, 0, 6, 0,
         0, 0, 0, 0, 0, 0,
         7, 0, 8, 0, 9, 0,
         0, 0, 0, 0, 0, 0 ]

I know the obvious way is to just use a for loop, but I'm wondering if there is a more efficient way using an OpenCV call?

Answer 1

One option that comes to mind would be to use cv::resize with INTER_NEAREST and then mask out the unwanted pixels.

Example:

#include <opencv2/opencv.hpp>

#include <cstdint>
#include <iostream>

int main()
{
    cv::Mat m1((cv::Mat_<uint8_t>(3, 3) << 1, 2, 3, 4, 5, 6, 7, 8, 9));
    std::cout << "Input:\n" << m1 << "\n\n";

    cv::Mat mask((cv::Mat_<uint8_t>(2, 2) << 255, 0, 0, 0));
    mask = cv::repeat(mask, m1.rows, m1.cols);
    std::cout << "Mask:\n" << mask << "\n\n";

    cv::Mat m2;
    cv::resize(m1, m2, cv::Size(), 2, 2, cv::INTER_NEAREST);
    std::cout << "Resized:\n" << m2 << "\n\n";

    cv::bitwise_and(m2, mask, m2);
    std::cout << "Masked:\n" << m2 << "\n\n";
}

Console output:

Input:
[  1,   2,   3;
   4,   5,   6;
   7,   8,   9]

Mask:
[255,   0, 255,   0, 255,   0;
   0,   0,   0,   0,   0,   0;
 255,   0, 255,   0, 255,   0;
   0,   0,   0,   0,   0,   0;
 255,   0, 255,   0, 255,   0;
   0,   0,   0,   0,   0,   0]

Resized:
[  1,   1,   2,   2,   3,   3;
   1,   1,   2,   2,   3,   3;
   4,   4,   5,   5,   6,   6;
   4,   4,   5,   5,   6,   6;
   7,   7,   8,   8,   9,   9;
   7,   7,   8,   8,   9,   9]

Masked:
[  1,   0,   2,   0,   3,   0;
   0,   0,   0,   0,   0,   0;
   4,   0,   5,   0,   6,   0;
   0,   0,   0,   0,   0,   0;
   7,   0,   8,   0,   9,   0;
   0,   0,   0,   0,   0,   0]

---

Update

If we eliminate parts of Miki's code that are unnecessary for our specific scenario, we pretty much reduce it to a simple loop.

Doing some quick comparisons, this turns out to be somewhat faster.

#include <opencv2/opencv.hpp>

#include <chrono>
#include <cstdint>
#include <iostream>

cv::Mat resize_1(cv::Mat image)
{
    cv::Mat result(cv::Mat::zeros(image.rows * 2, image.cols * 2, CV_8UC1));

    for (int ra(0); ra < image.rows; ++ra) {
        for (int ca = 0; ca < image.cols; ++ca) {
            result.at<uint8_t>(ra * 2, ca * 2) = image.at<uint8_t>(ra, ca);
        }
    }

    return result;
}

cv::Mat resize_2(cv::Mat image)
{
    cv::Mat mask((cv::Mat_<uint8_t>(2, 2) << 255, 0, 0, 0));
    mask = cv::repeat(mask, image.rows, image.cols);

    cv::Mat result;
    cv::resize(image, result, cv::Size(), 2, 2, cv::INTER_NEAREST);
    cv::bitwise_and(result, mask, result);

    return result;
}

template<typename T>
void timeit(T f)
{
    using std::chrono::high_resolution_clock;
    using std::chrono::duration_cast;
    using std::chrono::microseconds;

    cv::Mat m1((cv::Mat_<uint8_t>(3, 3) << 1, 2, 3, 4, 5, 6, 7, 8, 9));
    m1 = cv::repeat(m1, 1024, 1024);

    high_resolution_clock::time_point t1 = high_resolution_clock::now();
    for (uint32_t i(0); i < 256; ++i) {
        cv::Mat result = f(m1);
    }
    high_resolution_clock::time_point t2 = high_resolution_clock::now();

    auto duration = duration_cast<microseconds>(t2 - t1).count();
    double t_ms(static_cast<double>(duration) / 1000.0);
    std::cout
        << "Total = " << t_ms << " ms\n"
        << "Iteration = " << (t_ms / 256) << " ms\n"
        << "FPS = " << (256 / t_ms * 1000.0) << "\n";
}

int main()
{
    timeit(&resize_1);
    timeit(&resize_2);
}

Timing:

resize_1

Total = 6344.86 ms
Iteration = 24.7846 ms
FPS = 40.3476

resize_2

Total = 7271.31 ms
Iteration = 28.4036 ms
FPS = 35.2068

Update 2

Parallelized version:

class ResizeInvoker : public cv::ParallelLoopBody
{
public:
    ResizeInvoker(cv::Mat const& src, cv::Mat& dst)
        : image(src)
        , result(dst)
    {
    }

    void operator()(const cv::Range& range) const
    {
        for (int y(range.start); y < (range.end); ++y) {
            for (int x(0); x < image.cols; ++x) {
                result.at<uint8_t>(y * 2, x * 2) = image.at<uint8_t>(y, x);
            }
        }
    }

    cv::Mat const& image;
    cv::Mat& result;
};

cv::Mat resize_3(cv::Mat image)
{
    cv::Mat result(cv::Mat::zeros(image.rows * 2, image.cols * 2, CV_8UC1));

    ResizeInvoker loop_body(image, result);
    cv::parallel_for_(cv::Range(0, image.rows)
        , loop_body
        , result.total() / (double)(1 << 16));

    return result;
}

Timing:

resize_3

Total = 3876.63 ms
Iteration = 15.1431 ms
FPS = 66.0367

Update 3

We can do a little better if we use raw pointers in the invoker:

void operator()(const cv::Range& range) const
{
    for (int y(range.start); y < (range.end); ++y) {
        uint8_t* D = result.data + result.step * y * 2;
        uint8_t const* S = image.data + image.step * y;
        for (int x(0); x < image.cols; ++x) {
            D[x * 2] = S[x];
        }
    }
}

Timing:

Total = 3387.87 ms
Iteration = 13.2339 ms
FPS = 75.5636

Answer 2

You can use the Kronecker product of your image and a pattern like:

1, 0
0, 0

The result is:

Input:
[1, 2, 3;
 4, 5, 6;
 7, 8, 9]

Output:
[1, 0, 2, 0, 3, 0;
 0, 0, 0, 0, 0, 0;
 4, 0, 5, 0, 6, 0;
 0, 0, 0, 0, 0, 0;
 7, 0, 8, 0, 9, 0;
 0, 0, 0, 0, 0, 0]

Code:

#include <opencv2/opencv.hpp>
#include <iostream>

using namespace std;
using namespace cv;

Mat1b kron(const Mat1b& A, const Mat1b& B)
{
    Mat1b K(A.rows * B.rows, A.cols * B.cols, uchar(0));
    for (int ra = 0; ra < A.rows; ++ra)
    {
        for (int ca = 0; ca < A.cols; ++ca)
        {
            K(Range(ra*B.rows, (ra + 1)*B.rows), Range(ca*B.cols, (ca + 1)*B.cols)) = B.mul(A(ra, ca));
        }
    }
    return K;
}

int main()
{
    Mat1b img = (Mat1b(3, 3) << 1, 2, 3, 4, 5, 6, 7, 8, 9);
    std::cout << "Input:\n" << img << "\n\n";

    // Define the pattern
    Mat1b pattern = (Mat1b(2, 2) << 1, 0, 
                                    0, 0);

    Mat1b out = kron(img, pattern);
    std::cout << "Output:\n" << out << "\n\n";

    return 0;
}

OpenCV doesn't have the Kronecker product implemented, so you need to write a custom function. For a more general implementation that works with all data types (1 channel) have a look here.

---

I found that @Dan Masek approach is faster. This is because my kron implementation is not optimized. I expect this approach to work pretty well with a good implementation.

Answer 3

Thought of sharing the following approach as it is a bit different. I don't know how efficient this would be compared to other methods. At least you can use opencv calls without any looping and easily use arbitrary scale factors for x and y.

First convert your image to floating point type, then scale it using warpAffine (use linear interpolation). Resize the same image using nearest-neighbor method. Compare the two resulting images element-wise to get a mask. Use this mask to copy the relevant elements from any of the result images.

Here's the code and some results I get:

uchar data[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9 };
Mat im(3, 3, CV_8U, data);
im.convertTo(im, CV_32F);
// x and y scale
int xscale = 2, yscale = 2;
Size size(im.cols * xscale, im.rows * yscale);
float tr[] = {xscale, 0, 0, 0, yscale, 0};
Mat m(2, 3, CV_32F, tr);    // transformation matrix

Mat resized1, resized2;
warpAffine(im, resized1, m, size);  // affine scaling with linear interpolation
resize(im, resized2, size, 0, 0, INTER_NEAREST);    // resize with nearest neighbor
// get the mask
Mat resized = resized1 == resized2;
// copy the pixels
resized1.copyTo(resized, resized);

cout << "image:\n" << im << endl;
cout << "M:\n" << m << endl;
cout << "affine(scaled):\n" << resized1 << endl;
cout << "resized:\n" << resized2 << endl;
cout << "mask:\n" << resized << endl;
cout << "output:\n" << resized << endl;

For xscale = 2, yscale = 2

image:
[1, 2, 3;
  4, 5, 6;
  7, 8, 9]
M:
[2, 0, 0;
  0, 2, 0]
affine(scaled):
[1, 1.5, 2, 2.5, 3, 1.5;
  2.5, 3, 3.5, 4, 4.5, 2.25;
  4, 4.5, 5, 5.5, 6, 3;
  5.5, 6, 6.5, 7, 7.5, 3.75;
  7, 7.5, 8, 8.5, 9, 4.5;
  3.5, 3.75, 4, 4.25, 4.5, 2.25]
resized:
[1, 1, 2, 2, 3, 3;
  1, 1, 2, 2, 3, 3;
  4, 4, 5, 5, 6, 6;
  4, 4, 5, 5, 6, 6;
  7, 7, 8, 8, 9, 9;
  7, 7, 8, 8, 9, 9]
mask:
[1, 0, 2, 0, 3, 0;
  0, 0, 0, 0, 0, 0;
  4, 0, 5, 0, 6, 0;
  0, 0, 0, 0, 0, 0;
  7, 0, 8, 0, 9, 0;
  0, 0, 0, 0, 0, 0]
output:
[1, 0, 2, 0, 3, 0;
  0, 0, 0, 0, 0, 0;
  4, 0, 5, 0, 6, 0;
  0, 0, 0, 0, 0, 0;
  7, 0, 8, 0, 9, 0;
  0, 0, 0, 0, 0, 0]

For xscale = 4, yscale = 3

output:
[1, 0, 0, 0, 2, 0, 0, 0, 3, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  4, 0, 0, 0, 5, 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;
  7, 0, 0, 0, 8, 0, 0, 0, 9, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0;
  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]