14
$\begingroup$

How can I extract data from this picture of a graph?

Graph

i = Import["http://i.stack.imgur.com/Ac8m0.png"];

The caption of the picture reads: " Two-dimensional histogram values measured . Tick labels on the color bar are bin counts (17 force bins, 32 lifetime bins, and n = 803 observations)"

Plan of Attack

I divided the image into two parts: 1st the main graph and 2nd the legend. I am trying to extract average pixel in the each block but I am not able to do so becuase the edges between the boxes are not well defined... so, I am trying to define the edges, which is kinda hard to do accurately.

I am trying to use ComponentMeasurements and MorphologicalComponents. Before using those two tools, I need to have well-defined boxes.

Do you guys have better algorithm to extract data?

UPDATE

@SimonWoods did a great job of extracting all these data. Now, I am trying to make sense out of this data. Using the data from legend, I made a list:

m = {{0, keycols[[1]]}, {1, keycols[[2]]}, {2, keycols[[3]]}, {3, keycols[[4]]}, {4, keycols[[5]]}, {5, keycols[[6]]}, {6, keycols[[7]]}, {7, keycols[[8]]}, {8, keycols[[9]]}, {9, keycols[[10]]}, {10, keycols[[11]]}, {11, keycols[[12]]}, {12, keycols[[13]]}, {13, keycols[[14]]}, {14, keycols[[15]]}, {35, keycols[[19]]}, {96, keycols[[24]]}}

Then, I tried to interpolate:

ip = Interpolation[m, Method -> "Spline", InterpolationOrder -> 1] (I don't know if assuming it's linear is correct or not but if the total count is correct then I can be sure if not then I can try different interpolation order)

and I got this:

enter image description here

However, what I want is a function that takes in the RGB and gives count. So, I want inverse of this interpolation, which I couldn't get.

$\endgroup$
  • $\begingroup$ Why do you need to go to the edges of the blocks? The sampling is uniform - can't you only consider pixels in the middle of the blocks? $\endgroup$ – Simon Woods Jun 29 '16 at 21:00
  • $\begingroup$ @SimonWoods I think I can do that but since I am new to this and doing image processing first time, I was trying to follow: ComponentMeasurements, where you can see that well-defined edges helps a lot. $\endgroup$ – psimeson Jun 29 '16 at 21:03
  • $\begingroup$ This looks like a MATLAB plot, shown with the (old) default colormap. If so, I think that the scale may be linear in Hue. $\endgroup$ – mikado Jun 29 '16 at 21:23
  • $\begingroup$ @mikado I do not know how they made this graph. I would also guess it as a colormap. I grabbed it from a Science paper and I want to extract data from this graph to do some analysis. I really do not understand their scaling. They do not do a good job describing it in the paper $\endgroup$ – psimeson Jun 29 '16 at 21:27
  • 1
    $\begingroup$ Have you considered asking the authors for their data? $\endgroup$ – Eric Towers Jun 30 '16 at 6:01
20
$\begingroup$

This is not a complete solution, but might get you on the way.

With a little bit of trial and error you can identify the coordinates of the blocks and the key in the image:

img = Import["http://i.stack.imgur.com/Ac8m0.png"];

pts = Table[{x, y}, {x, 65, 598, 33}, {y, 64, 810, 24}];    
key = Table[{697, y}, {y, 60, 810, 30}];

HighlightImage[img, {Flatten[pts, 1], ImageMarker[key, "Circle"]}]

enter image description here

Sample the image at the "key" coordinates to get the key colours:

keycols = ImageValue[img, key];

Row[RGBColor /@ keycols]

enter image description here

You can now sample the colours of the blocks and map the results into the key index using Nearest:

nf = Nearest[keycols -> Automatic];

data = Map[First[nf@ImageValue[img, #]] &, pts, {2}];

data is an array of values from 1 (white) to 26 (dark red). To check it, we can reconstruct the original blocks from the key colours:

Graphics @ Raster[Map[keycols[[#]] &, Transpose[data], {2}]]

enter image description here

What remains is to convert the key colour indices (1 to 26) into actual values (0 to 100?) taking into account the non-linear scale.

$\endgroup$
  • 1
    $\begingroup$ For reference: here is a Mathematica implementation of the jet() colormap. $\endgroup$ – J. M. will be back soon Jun 30 '16 at 0:37
  • $\begingroup$ According to the OP the figure caption reads " ... n = 803 observations". Since the figure is a DensityHistogram, this total number of observations can be used to calibrate the non-linear color scaling. $\endgroup$ – Romke Bontekoe Jul 2 '16 at 6:46
1
$\begingroup$

For the legend at the bottom, the traditional method of identifying color boundaries is the Fast Fourier Transform (FFT): the resulting frequencies will be large where the color is changing quickly -- the boundaries. Obviously, FFT will not work on the far right-hand side, but then there are no observations there either. After applying the FFT, choose a frequency cutoff value that selects the number of boundaries most closely approximating the real number.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.