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MathGroup Archive 2004

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Re: importing image and getting numbers from the gray intensity

  • To: mathgroup at smc.vnet.net
  • Subject: [mg49405] Re: importing image and getting numbers from the gray intensity
  • From: sean_incali at yahoo.com (sean kim)
  • Date: Sat, 17 Jul 2004 06:40:10 -0400 (EDT)
  • References: <200407151100.HAA11123@smc.vnet.net> <cd8agm$ohr$1@smc.vnet.net>
  • Sender: owner-wri-mathgroup at wolfram.com

hi Yas

Thanks for the reply. 

I was kinda puzzled in an amazed way...

you use.. 

lane1 = ListPlot[Reverse[Transpose[d][[26]]], PlotStyle -> Gold, 
      Frame ->  True, PlotJoined -> True, ImageSize -> 600, 
      DisplayFunction -> Identity]; 
lane2 = ListPlot[Reverse[Transpose[d][[65]]], PlotStyle -> Blue, 
      Frame -> True, PlotRange -> All, PlotJoined -> True, ImageSize
-> 600,
      DisplayFunction -> Identity]; 
lane3 = ListPlot[Reverse[Transpose[d][[110]]], PlotStyle -> Red, 
      Frame -> True, PlotRange -> All, PlotJoined -> True, ImageSize
-> 600,
      DisplayFunction -> Identity]; 

and 

Length[d]

reveals that imported and pixleated image is a list of 289 items... 
How did you know that part 26, 65, 110 belong to the lanes?  I'm truly
baffled.

Thanks for any thoughts in advance 

sean 


Yasvir Tesiram <yat at omrf.ouhsc.edu> wrote in message news:<cd8agm$ohr$1 at smc.vnet.net>...
> Hi,
> It is possible. But you may have to consider the accuracy of the 
> scanned image with respect to the concentration / intensity 
> relationship. Nevertheless here are some quick notes on the same image 
> from the web page. Also,
> 
> http://www.wolfram.com/products/applications/digitalimage/
> 
> but I've never used it, don't have it and so I can't say anymore.
> 
> (* get the image *)
> 
> g = Import["~/Desktop/SDS-PAGE.jpg"];
> Show[g]
> 
> (*extract the numbers*)
> 
> d = g[[1, 1]];
> 
> (* uncomment below to get dimensions of image*)
> (* Dimensions[d] *)
> 
> (* here is a display of the image *)
> ListDensityPlot[d, Mesh -> False, AspectRatio -> Automatic, ImageSize 
> -> 600]
> 
> (* to get a trace of the data you can get one column *)
> (* I've reversed it so that the top blob in lane 1 is on the left *)
> (* but really I shouldn't have *)
> (*Reverse[Transpose[d][[26]]]*)
> 
> (* let's look at this data as if it were several chromatograms *)
> 
> << Graphics`Colors`
> lane1 = ListPlot[Reverse[Transpose[d][[26]]], PlotStyle -> Gold, Frame 
> -> \
> True, PlotJoined -> True, ImageSize -> 600, DisplayFunction -> 
> Identity];
> lane2 = ListPlot[Reverse[Transpose[d][[65]]], PlotStyle -> Blue, Frame 
> ->
>          True, PlotRange -> All, PlotJoined -> True, ImageSize -> 600, \
> DisplayFunction -> Identity];
> lane3 = ListPlot[Reverse[Transpose[d][[110]]], PlotStyle -> Red, Frame 
> ->
>        True, PlotRange -> All, PlotJoined ->
>        True, ImageSize -> 600, DisplayFunction -> Identity];
> 
> Show[{lane1, lane2, lane3}, PlotRange ->
>       All, Background -> IvoryBlack, FrameTicks -> {Automatic, 
> Automatic, None,
>       None}, ImageSize -> 700, DisplayFunction -> $DisplayFunction]
> 
> I hope this gives you some idea of how to get at the image data from a 
> jpeg image and of course you can now integrate peaks from each lane 
> etc.
> Hint, to get at the lane number, select the graphic, hold down the ALT 
> or CMND (Mac) key and navigate around the graphic. On the bottom left 
> hand side of the frontend you will see the coordinates 
> (http://forums.wolfram.com/mathgroup/archive/2003/Feb/msg00429.html)
> 
> Cheers
> Yas
> 
> 
> 
> On Jul 15, 2004, at 6:00 AM, sean kim wrote:
> 
> > let's say you have an image of gray bands of varying intensity.
> >
> > such as the one seen on the website,
> >
> > http://encyclopedia.thefreedictionary.com/SDS-PAGE
> >
> >
> > Is there anyway to change the picture or do something and be able to
> > get numerical values for those bands?
> >
> > maybe define a vertical line and then go to the center of the each
> > bands and measure how intense the gray is.
> >
> > or maybe define a rectangle and get the gray intensity in that...
> >
> > is there a packages or codes for doing that?  NIH image program which
> > is freely avaialble can do this. But I wanted to do it in Mathematica.
> >
> > Thanks for any and all insights.
> >
> > sean


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