2d histogram on OpenCV using one grayscale image (one plane)?

I might haven't completely understood histograms... but i think i could get a 2-dimension of a grayscale image, right?

The one dimension is fine:

from cv import *
import os, glob, sys


original = LoadImage('test.jpg')
gray  = CreateImage(GetSize(original), IPL_DEPTH_8U, 1)
canny = CreateImage(GetSize(original), IPL_DEPTH_8U, 1)
NamedWindow('Circles', 1)


CvtColor(original, gray, CV_BGR2GRAY)

bins = 30
scale = 10
hist = CreateHist([bins], CV_HIST_ARRAY, [[0,256]], 1)
CalcHist([gray], hist)


hist开发者_StackOverflow中文版_img = CreateImage([bins*scale,50], 8, 1)
Rectangle(hist_img, (0,0), (bins*scale,50), CV_RGB(255,255,255), -1)


(_, max_value, _, _) = GetMinMaxHistValue(hist)

for i in range(0,bins):
  bin_val = QueryHistValue_1D(hist, i)
  #print bin_val
  norm = Round((bin_val/max_value)*50)
  Rectangle(hist_img, (i*scale, 50), (i*scale+scale-1,50-norm), CV_RGB(0, 0, 0), CV_FILLED)             


ShowImage('Circles', hist_img)
WaitKey(0)

But the 2d one, when i call CalcHist, says he needs two planes, or images:

from cv import *
import os, glob, sys


original = LoadImage('test.jpg')
gray  = CreateImage(GetSize(original), IPL_DEPTH_8U, 1)
NamedWindow('Circles', 1)


CvtColor(original, gray, CV_BGR2GRAY)

bins = 30
scale = 3

hist = CreateHist([bins,bins], CV_HIST_ARRAY, [[0,255], [0,255]], 1)
CalcHist([gray], hist)


hist_img = CreateImage([bins*scale,bins*scale], 8, 1)
#Rectangle(hist_img, (0,0), (bins*scale,50), CV_RGB(255,255,255), -1)
Zero(hist_img)

(_, max_value, _, _) = GetMinMaxHistValue(hist)

for h in range(0,bins):
  for s in range(0,bins):
    bin_val = QueryHistValue_2D(hist, h, s)
    inte = Round(bin_val*255/max_value)
    Rectangle(hist_img, (h*scale, s*scale), ((h+1)*scale-1,(s+1)*scale-1), CV_RGB(inte, inte, inte), CV_FILLED)             


ShowImage('Circles', hist_img)
WaitKey(0)

This error:

OpenCV Error: Bad argument (Unknown array type) in cvarrToMat, file /opt/local/var/macports/build/_opt_local_var_macports_sources_rsync.macports.org_release_ports_graphics_opencv/work/OpenCV-2.2.0/modules/core/src/matrix.cpp, line 641
Traceback (most recent call last):
  File "hist2d.py", line 16, in <module>
    CalcHist([gray], hist, 0)
cv.error: Unknown array type

If i use:

CalcHist([gray, gray], hist, 0)

it works, but i get a screwed up histogram (diagonal colored and the rest is black)

So... can someone enlighten me?

---

A grayscale image already is a two-dimensional histogram: the intensity of pixel (a, b) is value of the bin defined by a along the x-dimension and b along the y-dimension. Typically when one speaks of histograms in computer vision, one is speaking of a histogram over intensity values. For a grayscale image, this is a one-dimensional histogram where each bin corresponds to a range of intensity values and has a count corresponding to the number of pixels whose intensity falls in that bin.

Higher-dimensional histograms only make sense if the image is of multiple channels. For example, one might compute the three-dimensional histogram of RGB values over a color image. Calling CalcHist([gray, gray], hist, 0) results in a diagonal line because every pixel in the first image (gray) has the same value as the corresponding pixel in the second image (gray). This fills all of the bins along the diagonal in the output histogram.

Also, note that a multi-dimension histogram is very different from three one-dimensional histogram.

---

bins = 10 # specify the number of bins
ranges = (10,255) % specify the top and bottom range of the bins. This truncates the image
hist = cv.CreateHist([bins], cv.CV_HIST_ARRAY, [ranges], 1) # create histograms
cv.CalcHist([gr], hist) # calculate the histograms for the image
(min_value, max_value, min_idx, max_idx) = cv.GetMinMaxHistValue(hist) # get the min and max values for the histogram

---

Higher-dim. hists do not only make sence in RGB-image-analysis - those are only intensity hists - but also in feature-extraction like in the GLCM (gray-level co-occurrence matrix, 2D), shape context (dim. depends on the algorithm) etc.