These aerial images, taken at different times, represent overlapping views of the same terrain in Concord, Massachusetts. This example demonstrates that input imagesAandRefcan be of different sizes and image types.

Load an RGB image and a reference grayscale image.

A = imread('concordaerial.png');Ref = imread ('concordorthophoto.png');

Get the size ofA.

size(A)

ans =1×32036 3060 3

Get the size ofRef.

size(Ref)

ans =1×22215 2956

Note that imageAandRefare different in size and type. ImageAis a truecolor RGB image, while imageRefis a grayscale image. Both images are of data typeuint8.

Generate the histogram matched output image. The example matches each channel ofAagainst the single histogram ofRef. Output imageBtakes on the characteristics of imageA- it is an RGB image whose size and data type is the same as imageA. The number of distinct levels present in each RGB channel of imageBis the same as the number of bins in the histogram built from grayscale imageRef. In this example, the histogram ofRefandBhave the default number of bins, 64.

B = imhistmatch(A,Ref);

Display the RGB imageAthe reference imageRef, and the histogram matched RGB imageB. The images are resized before display.

imshow(A) title('RGB Image with Color Cast')

imshow(Ref) title('Reference Grayscale Image')

imshow(B) title('Histogram Matched RGB Image')

Read a color image and a reference image. To demonstrate the polynomial method, assign the reference image to be the darker of the two images.

I = imread('office_4.jpg');ref = imread('office_2.jpg');montage({I,ref}) title('Input Image (Left) vs Reference Image (Right)');

Use the polynomial method to adjust the intensity of imageIso that it matches the histogram of reference imageref. For comparison, also adjust the intensity of imageIusing the uniform method.

J = imhistmatch(I,ref,'method','polynomial');K = imhistmatch(I,ref,'method','uniform');montage({J,K}) title('Histogram-Matched Image Using Polynomial Method (Left) vs Uniform Method (Right)');

The histogram-matched image using the uniform method introduces false colors in the sky and road. The histogram-matched image using the polynomial method does not exhibit this artifact.

This example shows how you can vary the number of bins in the target histogram to improve histogram equalization.

加载两个图像的数据类型uint8into the workspace. The images were taken with a digital camera and represent two different exposures of the same scene.Ais an underexposed image and appears dark.refis a reference image with good exposure and brightness.

A = imread('office_2.jpg');ref = imread('office_4.jpg');

Display the images in a montage.

montage({A,ref}) title('Dark Image (Left) and Reference Image (Right)')

Display the histogram of each color channel using 256 bins. You can use the helper function,displayHistogramChannelsthat is included with the example.

displayHistogramChannels(A,ref)

ImageA, being the darker image, has most of its pixels in the lower bins. The reference image, ref, fully populates all 256 bins values in all three RGB channels.

Count the number of unique 8-bit level values for each color channel of the dark and reference image. You can use the helper function,countUniqueValuesthat is included with the example.

numVals = countUniqueValues(A,ref); table(numVals(:,1),numVals(:,2),numVals(:,3),...'VariableNames',["Red""Green""Blue"],...'RowNames',["A""ref"])

ans=2×3 table红绿蓝___ _____ ____ A 205 193 224 ref 256 256 256

Equalize the histogram of the dark image using three different values ofnbins: 64, 128 and 256. 64 is the default number of bins and 256 is the maximum number of bins foruint8pixel data.

[B64,hgram64] = imhistmatch(A,ref,64); [B128,hgram128] = imhistmatch(A,ref,128); [B256,hgram256] = imhistmatch(A,ref,256); figure montage({B64,B128,B256},'Size',[1 3]) title('Output Image B64 | Output Image B128 | Output Image B256')

Display the histogram of each color channel using 256 bins. You can use the helper function,displayThreeHistogramChannelsthat is included with the example.

displayThreeHistogramChannels(B64,B128,B256)

Count the number of unique 8-bit level values for each color channel of the three histogram equalized images. Asnbinsincreases, the number of levels in each RGB channel of output imageBalso increases.

numVals = countUniqueValues(B64,B128,B256); table(numVals(:,1),numVals(:,2),numVals(:,3),...'VariableNames',["Red""Green""Blue"],...'RowNames',["B64""B128""B256"])

ans=3×3 table红绿蓝___ _____ ____ B64 57 60 58 B128 101 104 104 B256 134 135 136

This example shows how to perform histogram matching with different numbers of bins.

Load a 16-bit DICOM image of a knee imaged via MRI.

K = dicomread('knee1.dcm');% read in original 16-bit imageLevelsK = unique(K(:));% determine number of unique code valuesdisp(['image K: ',num2str(length(LevelsK)),' distinct levels']);

image K: 448 distinct levels

disp(['max level = 'num2str( max(LevelsK) )]);

max level = 473

disp(['min level = 'num2str( min(LevelsK) )]);

min level = 0

All 448 discrete values are at low code values, which causes the image to look dark. To rectify this, scale the image data to span the entire 16-bit range of [0, 65535].

Kdouble = double(K);% cast uint16 to doublekmult = 65535/(max(max(Kdouble(:))));% full range multiplierRef = uint16 (kmult * Kdouble);% full range 16-bit reference image

Darken the reference imageRefto create an imageAthat can be used in the histogram matching operation.

%Build concave bow-shaped curve for darkening |Ref|.ramp = [0:65535]/65535; ppconcave = spline([0 .1 .50 .72 .87 1],[0 .025 .25 .5 .75 1]); Ybuf = ppval( ppconcave, ramp); Lut16bit = uint16( round( 65535*Ybuf ) );% Pass image |Ref| through a lookup table (LUT) to darken the image.A = intlut(Ref,Lut16bit);

View the reference imageRefand the darkened imageA. Note that they have the same number of discrete code values, but differ in overall brightness.

subplot(1,2,1) imshow(Ref) title('Ref: Reference Image') subplot(1,2,2) imshow(A) title('A: Darkened Image');

Generate histogram-matched output images using histograms with different number of bins. First use the default number of bins, 64. Then use the number of values present in imageA, 448 bins.

B16bit64 = imhistmatch(A(:,:,1),Ref(:,:,1));% default: 64 binsN = length(LevelsK);% number of unique 16-bit code values in image A.B16bitUniq = imhistmatch(A(:,:,1),Ref(:,:,1),N);

View the results of the two histogram matching operations.

figure subplot(1,2,1) imshow(B16bit64) title('B16bit64: 64 bins') subplot(1,2,2) imshow(Ref) title(['B16bitUniq: ',num2str(N),' bins'])