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Image transform

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Manuel Urrutia Avisrror

Image transform

What would be the code to transform a RGB image to 8-bit indexed color
or to create a 8-bit indexed color image ?

Thanks for the help

Manuel Urrutia Avisrror MD, PhD
[hidden email]

Liu, Dongfang (NIH/NIAID) [F]

Re: Image transform

Dear friends,

Could you tell me what is the unit of "center of mass" in the ImageJ? The unit of "Center of Mass" is pixels? When I try to track a dot, we can get the "center of mass", such as XM and YM.
I don't know the unit of "center of mass" .
Thank you very much.

Best regards

> ----------
> From: Manuel Urrutia Avisrror
> Reply To: Manuel Urrutia Avisrror
> Sent: Friday, March 3, 2006 7:21 AM
> To: List IMAGEJ
> Subject: Image transform
>
> What would be the code to transform a RGB image to 8-bit indexed color
> or to create a 8-bit indexed color image ?
>
> Thanks for the help
>
> Manuel Urrutia Avisrror MD, PhD
> [hidden email]
>
>

Gabriel Landini

Re: Image transform

On Friday 03 March 2006 14:40, Liu, Dongfang (NIH/NIAID) [E] wrote:
> Could you tell me what is the unit of "center of mass" in the ImageJ? The
> unit of "Center of Mass" is pixels?

The coordinates in the image.

How can we know whether you are using a calibrated or non calibrated image?!

G.

Ron Zhu

Re: Image transform

In reply to this post by Manuel Urrutia Avisrror

I am new to this list and ImageJ.

Is there pluggin that can be used for 2D/2D imaging registration for
digital reconstructed radiograph (DRR) and a digital x-ray image (both in
DICOM)?

Thanks.

Ron Zhu

Liu, Dongfang (NIH/NIAID) [F]

Re: Image transform

In reply to this post by Manuel Urrutia Avisrror

Dear friends,

Thank you for your help, Gabriel.
I am using a non-calibrated image. I don't know how to calibrate a image and what is function of this calibration? You mean the spatial calibration? In our system, we use the 10.35 microns pixel size of CCD and 100*objective. So I think of the pixel size of image is 100nm/pixel. If I know the Center of Mass, such as 100, 100., then I can know the coordiante of XY is 10000, 10000, we also can know the displacemnet of two dots. Is right?

Thank you very much.
I am looking forward to your reply,

df

> ----------
> From: Gabriel Landini
> Reply To: List IMAGEJ
> Sent: Friday, March 3, 2006 9:48 AM
> To: List IMAGEJ
> Subject: Re: Image transform
>
> On Friday 03 March 2006 14:40, Liu, Dongfang (NIH/NIAID) [E] wrote:
> > Could you tell me what is the unit of "center of mass" in the ImageJ? The
> > unit of "Center of Mass" is pixels?
>
> The coordinates in the image.
>
> How can we know whether you are using a calibrated or non calibrated image?!
>
> G.
>
>

Gabriel Landini

Re: Image transform

On Friday 03 March 2006 15:40, Liu, Dongfang (NIH/NIAID) [E] wrote:
> I am using a non-calibrated image. I don't know how to calibrate a image
> and what is function of this calibration? You mean the spatial calibration?

Yes. The result is the coordinates in your image.
If you have a calibrated image (you do this by going to Image>Properties and
set the width and height of you pixels.)

> In our system, we use the 10.35 microns pixel size of CCD and
> 100*objective. So I think of the pixel size of image is 100nm/pixel.

And not other optics in between? Is it a microscope? if so, it is very likely
that there are other correcting lenses in between.

I can't tell if your calculation is correct. You need to put a known sized
object under your imaging system and see how many pixels it spans. Then you
estimate the size of your pixels dividing the known size of your object by
the number of pixels it occupies.

The XM and YM coordinates are relative to the image coordinates. Since the
resolution of the image is given in pixels, there is no advantage in
calculating the coordinates in calibrated units. The result is not any more
accurate than in pixel units.

Cheers,

Gabriel

Wayne Rasband

Re: Image transform

In reply to this post by Liu, Dongfang (NIH/NIAID) [F]

> Could you tell me what is the unit of "center of mass" in the ImageJ?
> The unit of "Center of Mass" is pixels? When I try to track a dot, we
> can get the "center of mass", such as XM and YM.
> I don't know the unit of "center of mass" .
> Thank you very much.

The unit of length of is displayed at the top of each image. If it says
something like "512x512 pixels" then the image is not spatially
calibrated and the unit is pixels. If is says "240.00x240.00 mm
(512x512)" then the unit is millimeters. Use the Analyze>Set Scale or
Image>Properties command to specify the scale and unit. Check "Global"
and ImageJ will use the same scale and unit for all images you open
during the current session.

-wayne

Liu, Dongfang (NIH/NIAID) [F]

Re: Image transform

In reply to this post by Manuel Urrutia Avisrror

Dear Gabriel,

Thank you for your help.
Yes, the system is a miroscope. Could you tell what other optics I should consider to calibrate a image? Could you give me more detail information about this?

Best regards
df

> ----------
> From: Gabriel Landini
> Reply To: List IMAGEJ
> Sent: Friday, March 3, 2006 10:53 AM
> To: List IMAGEJ
> Subject: Re: Image transform
>
> On Friday 03 March 2006 15:40, Liu, Dongfang (NIH/NIAID) [E] wrote:
> > I am using a non-calibrated image. I don't know how to calibrate a image
> > and what is function of this calibration? You mean the spatial calibration?
>
> Yes. The result is the coordinates in your image.
> If you have a calibrated image (you do this by going to Image>Properties and
> set the width and height of you pixels.)
>
> > In our system, we use the 10.35 microns pixel size of CCD and
> > 100*objective. So I think of the pixel size of image is 100nm/pixel.
>
> And not other optics in between? Is it a microscope? if so, it is very likely
> that there are other correcting lenses in between.
>
> I can't tell if your calculation is correct. You need to put a known sized
> object under your imaging system and see how many pixels it spans. Then you
> estimate the size of your pixels dividing the known size of your object by
> the number of pixels it occupies.
>
> The XM and YM coordinates are relative to the image coordinates. Since the
> resolution of the image is given in pixels, there is no advantage in
> calculating the coordinates in calibrated units. The result is not any more
> accurate than in pixel units.
>
> Cheers,
>
> Gabriel
>
>

Gabriel Landini

Re: Image transform

On Friday 03 March 2006 16:25, Liu, Dongfang (NIH/NIAID) [E] wrote:
> Thank you for your help.
> Yes, the system is a miroscope. Could you tell what other optics I should
> consider to calibrate a image?

No, I couldn't because I do not know what kind of optics your microscope has.

You have to do this:

> > You need to put a known
> > sized object under your imaging system and see how many pixels it spans.
> > Then you estimate the size of your pixels dividing the known size of your
> > object by the number of pixels it occupies.

Cheers,

Gabriel

Christophe CHAMOT

Re: Image transform

In reply to this post by Liu, Dongfang (NIH/NIAID) [F]

Liu, Dongfang (NIH/NIAID) [E] a écrit :

> Dear Gabriel,
>
> Thank you for your help.
> Yes, the system is a miroscope. Could you tell what other optics I should consider to calibrate a image? Could you give me more detail information about this?
>
> Best regards
> df
>
>
>>----------

Hello,

The formula is :

(pixel size) = (pixel camera * binning) / (total magnification)

As an example, when I use my CoolSnap (6.45µm pixel) with a 2x2 binning,
I have a "virtual" pixel of (6.45*2) 12.9 µm. This "pixel" gather a
distance through my 100x objective of 12.9/100 = 0.129 µm.
This is the pixel size I give in Image>Properties as Gabriel said
previously.

I said "total magnification" because some people use optional lenses
(zoom or optovars). I got a 1.5x on my Leica but it decreases
dramatically the intensity of fluorescence. In the previous example,
with the optovar 1.5x, pixel size will be (6.45*2)/(100*1.5)=0.086 µm

Hope it helps.

--
CHAMOT Christophe
---------------------------------------------------------------------
Plate-Forme de Recherche IFR117
"Imageries des Processus Dynamiques
en Biologie Cellulaire et Biologie du Développement "
Institut Jacques Monod, CNRS, Universités Paris 6 et 7
2, place Jussieu - Tour 43
75251 Paris cedex 05
Tel: 01 44 27 57 84
http://www.ijm.jussieu.fr/
---------------------------------------------------------------------

Liu, Dongfang (NIH/NIAID) [F]

Questions about analysis "radial intensity distribution"

In reply to this post by Manuel Urrutia Avisrror

Dear friends,
I am new user of Image J.
I am sorry to bother you.
I am looking for your help about the analysis of "radial intensity distribution" . The radial intensity distribution I(r) of the flourescence dot was fit for each frame with a nonlinear Levenberg-Marquardt routine to the Gaussian:I(r)=Io exp(-R^2/w^2) +BG, where R is the distance of each pixel to the center of mass. The fitting parameters are Io, w, and BG, where Io is the peak intensity, BG is the background intensity, and w is the measure of the width (the Gauss width). It is very important for us to get the peak intensity and width. some people suggest me to use Plot Profile(Analyze -> Plot.
But I think the "plot profile" is not good. Because the concept of "plot profile"---Displays a two-dimensional graph of the intensities of pixels along a line within the image. The x-axis represents distance along the line and the y-axis is the pixel intensity. For rectangular selections, displays a "column average plot", where the x-axis represents the horizontal distance through the selection and the y-axis the vertically averaged pixel intensity
I was confused that how to calculate the "radial intensity distribution" in Image Pro Plus or Image J or other software. Could you help me? Advice would be greatly appreciated.
I am looking forward to your reply.

Thank you very much.
Best regards
df

> ----------
> From: Gabriel Landini
> Reply To: List IMAGEJ
> Sent: Friday, March 3, 2006 12:10 PM
> To: List IMAGEJ
> Subject: Re: Image transform
>
> On Friday 03 March 2006 16:25, Liu, Dongfang (NIH/NIAID) [E] wrote:
> > Thank you for your help.
> > Yes, the system is a miroscope. Could you tell what other optics I should
> > consider to calibrate a image?
>
> No, I couldn't because I do not know what kind of optics your microscope has.
>
> You have to do this:
>
> > > You need to put a known
> > > sized object under your imaging system and see how many pixels it spans.
> > > Then you estimate the size of your pixels dividing the known size of your
> > > object by the number of pixels it occupies.
>
> Cheers,
>
> Gabriel
>
>

Felipe J. Álvarez M.

Re: Questions about analysis "radial intensity distribution"

Dear Liu:
I don't know if this could be of help, but there is a plug-in called "radial profile" which maybe could help you.

fjam

"Liu, Dongfang (NIH/NIAID) [E]" <[hidden email]> escribió:
Dear friends,
I am new user of Image J.
I am sorry to bother you.
I am looking for your help about the analysis of "radial intensity distribution" . The radial intensity distribution I(r) of the flourescence dot was fit for each frame with a nonlinear Levenberg-Marquardt routine to the Gaussian:I(r)=Io exp(-R^2/w^2) +BG, where R is the distance of each pixel to the center of mass. The fitting parameters are Io, w, and BG, where Io is the peak intensity, BG is the background intensity, and w is the measure of the width (the Gauss width). It is very important for us to get the peak intensity and width. some people suggest me to use Plot Profile(Analyze -> Plot.
But I think the "plot profile" is not good. Because the concept of "plot profile"---Displays a two-dimensional graph of the intensities of pixels along a line within the image. The x-axis represents distance along the line and the y-axis is the pixel intensity. For rectangular selections, displays a "column average plot", where the x-axis represents the horizontal distance through the selection and the y-axis the vertically averaged pixel intensity
I was confused that how to calculate the "radial intensity distribution" in Image Pro Plus or Image J or other software. Could you help me? Advice would be greatly appreciated.
I am looking forward to your reply.

Thank you very much.
Best regards
df

---------------------------------

LLama Gratis a cualquier PC del Mundo.
Llamadas a fijos y móviles desde 1 céntimo por minuto.
http://es.voice.yahoo.com

Martin du Saire

Re: Questions about analysis "radial intensity distribution"

If I have understood correctly, you need to get
the profiles along each radius "by hand". For example:

for (j=0; j<nAngles; j++) {
makeLine(fX,fY,fX+(profileRadius*cosA[j]),fY+(profileRadius*sinA[j]));
// (fX, fY) are the coordinates of the center
of the object or region you want to profile
// cosA and sinA are lookup tables, or you could calculate them on the fly
profile = getProfile();
// profile now contains the intensities along
the selected direction for you to use as needed
}

M

At 11:03 AM 3/6/2006, you wrote:

>Dear Liu:
> I don't know if this could be of help, but
> there is a plug-in called "radial profile" which maybe could help you.
>
> fjam
>
>"Liu, Dongfang (NIH/NIAID) [E]" <[hidden email]> escribió:
> Dear friends,
>I am new user of Image J.
>I am sorry to bother you.
>I am looking for your help about the analysis of
>"radial intensity distribution" . The radial
>intensity distribution I(r) of the flourescence
>dot was fit for each frame with a nonlinear
>Levenberg-Marquardt routine to the
>Gaussian:I(r)=Io exp(-R^2/w^2) +BG, where R is
>the distance of each pixel to the center of
>mass. The fitting parameters are Io, w, and BG,
>where Io is the peak intensity, BG is the
>background intensity, and w is the measure of
>the width (the Gauss width). It is very
>important for us to get the peak intensity and
>width. some people suggest me to use Plot Profile(Analyze -> Plot.
>But I think the "plot profile" is not good.
>Because the concept of "plot profile"---Displays
>a two-dimensional graph of the intensities of
>pixels along a line within the image. The x-axis
>represents distance along the line and the
>y-axis is the pixel intensity. For rectangular
>selections, displays a "column average plot",
>where the x-axis represents the horizontal
>distance through the selection and the y-axis
>the vertically averaged pixel intensity
>I was confused that how to calculate the "radial
>intensity distribution" in Image Pro Plus or
>Image J or other software. Could you help me?
>Advice would be greatly appreciated.
>I am looking forward to your reply.
>
>Thank you very much.
>Best regards
>df
>
>
> > ----------
> > From: Gabriel Landini
> > Reply To: List IMAGEJ
> > Sent: Friday, March 3, 2006 12:10 PM
> > To: List IMAGEJ
> > Subject: Re: Image transform
> >
> > On Friday 03 March 2006 16:25, Liu, Dongfang (NIH/NIAID) [E] wrote:
> > > Thank you for your help.
> > > Yes, the system is a miroscope. Could you tell what other optics I should
> > > consider to calibrate a image?
> >
> > No, I couldn't because I do not know what
> kind of optics your microscope has.
> >
> > You have to do this:
> >
> > > > You need to put a known
> > > > sized object under your imaging system
> and see how many pixels it spans.
> > > > Then you estimate the size of your pixels
> dividing the known size of your
> > > > object by the number of pixels it occupies.
> >
> > Cheers,
> >
> > Gabriel
> >
> >
>
>
>
>---------------------------------
>
>LLama Gratis a cualquier PC del Mundo.
>Llamadas a fijos y móviles desde 1 céntimo por minuto.
>http://es.voice.yahoo.com

Liu, Dongfang (NIH/NIAID) [F]

Re: Questions about analysis "radial intensity distribution"

In reply to this post by Liu, Dongfang (NIH/NIAID) [F]

Dear friends;

Where I can get the plugin called "radial profile"?

Best regards
df

> ----------
> From: Felipe J. Álvarez M.
> Reply To: List IMAGEJ
> Sent: Monday, March 6, 2006 12:03 PM
> To: List IMAGEJ
> Subject: Re: Questions about analysis "radial intensity distribution"
>
> Dear Liu:
> I don't know if this could be of help, but there is a plug-in called "radial profile" which maybe could help you.
>
> fjam
>
> "Liu, Dongfang (NIH/NIAID) [E]" <[hidden email]> escribió:
> Dear friends,
> I am new user of Image J.
> I am sorry to bother you.
> I am looking for your help about the analysis of "radial intensity distribution" . The radial intensity distribution I(r) of the flourescence dot was fit for each frame with a nonlinear Levenberg-Marquardt routine to the Gaussian:I(r)=Io exp(-R^2/w^2) +BG, where R is the distance of each pixel to the center of mass. The fitting parameters are Io, w, and BG, where Io is the peak intensity, BG is the background intensity, and w is the measure of the width (the Gauss width). It is very important for us to get the peak intensity and width. some people suggest me to use Plot Profile(Analyze -> Plot.
> But I think the "plot profile" is not good. Because the concept of "plot profile"---Displays a two-dimensional graph of the intensities of pixels along a line within the image. The x-axis represents distance along the line and the y-axis is the pixel intensity. For rectangular selections, displays a "column average plot", where the x-axis represents the horizontal distance through the selection and the y-axis the vertically averaged pixel intensity
> I was confused that how to calculate the "radial intensity distribution" in Image Pro Plus or Image J or other software. Could you help me? Advice would be greatly appreciated.
> I am looking forward to your reply.
>
> Thank you very much.
> Best regards
> df
>
>
> > ----------
> > From: Gabriel Landini
> > Reply To: List IMAGEJ
> > Sent: Friday, March 3, 2006 12:10 PM
> > To: List IMAGEJ
> > Subject: Re: Image transform
> >
> > On Friday 03 March 2006 16:25, Liu, Dongfang (NIH/NIAID) [E] wrote:
> > > Thank you for your help.
> > > Yes, the system is a miroscope. Could you tell what other optics I should
> > > consider to calibrate a image?
> >
> > No, I couldn't because I do not know what kind of optics your microscope has.
> >
> > You have to do this:
> >
> > > > You need to put a known
> > > > sized object under your imaging system and see how many pixels it spans.
> > > > Then you estimate the size of your pixels dividing the known size of your
> > > > object by the number of pixels it occupies.
> >
> > Cheers,
> >
> > Gabriel
> >
> >
>
>
>
> ---------------------------------
>
> LLama Gratis a cualquier PC del Mundo.
> Llamadas a fijos y móviles desde 1 céntimo por minuto.
> http://es.voice.yahoo.com
>
>

Felipe J. Álvarez M.

Re: Questions about analysis "radial intensity distribution"

Dear friend:

You can find the plug-in in the following web address:
http://rsb.info.nih.gov/ij/plugins/radial-profile.html

Best regards
fjam

"Liu, Dongfang (NIH/NIAID) [E]" <[hidden email]> escribió:
Dear friends;

Where I can get the plugin called "radial profile"?

Best regards
df

escribió:

> Dear friends,
> I am new user of Image J.
> I am sorry to bother you.
> I am looking for your help about the analysis of "radial intensity distribution" . The radial intensity distribution I(r) of the flourescence dot was fit for each frame with a nonlinear Levenberg-Marquardt routine to the Gaussian:I(r)=Io exp(-R^2/w^2) +BG, where R is the distance of each pixel to the center of mass. The fitting parameters are Io, w, and BG, where Io is the peak intensity, BG is the background intensity, and w is the measure of the width (the Gauss width). It is very important for us to get the peak intensity and width. some people suggest me to use Plot Profile(Analyze -> Plot.
> But I think the "plot profile" is not good. Because the concept of "plot profile"---Displays a two-dimensional graph of the intensities of pixels along a line within the image. The x-axis represents distance along the line and the y-axis is the pixel intensity. For rectangular selections, displays a "column average plot", where the x-axis represents the horizontal distance through the selection and the y-axis the vertically averaged pixel intensity
> I was confused that how to calculate the "radial intensity distribution" in Image Pro Plus or Image J or other software. Could you help me? Advice would be greatly appreciated.
> I am looking forward to your reply.
>
> Thank you very much.
> Best regards
> df
>
>
> > ----------
> > From: Gabriel Landini
> > Reply To: List IMAGEJ
> > Sent: Friday, March 3, 2006 12:10 PM
> > To: List IMAGEJ
> > Subject: Re: Image transform
> >
> > On Friday 03 March 2006 16:25, Liu, Dongfang (NIH/NIAID) [E] wrote:
> > > Thank you for your help.
> > > Yes, the system is a miroscope. Could you tell what other optics I should
> > > consider to calibrate a image?
> >
> > No, I couldn't because I do not know what kind of optics your microscope has.
> >
> > You have to do this:
> >
> > > > You need to put a known
> > > > sized object under your imaging system and see how many pixels it spans.
> > > > Then you estimate the size of your pixels dividing the known size of your
> > > > object by the number of pixels it occupies.
> >
> > Cheers,
> >
> > Gabriel
> >
> >
>
>
>
> ---------------------------------
>
> LLama Gratis a cualquier PC del Mundo.
> Llamadas a fijos y móviles desde 1 céntimo por minuto.
> http://es.voice.yahoo.com
>
>

---------------------------------

LLama Gratis a cualquier PC del Mundo.
Llamadas a fijos y móviles desde 1 céntimo por minuto.
http://es.voice.yahoo.com

Liu, Dongfang (NIH/NIAID) [F]

Re: Questions about analysis "radial intensity distribution"

In reply to this post by Martin du Saire

Dear friend;

People tell me to calculate as following:
Another method of measuring the radial sweep I(r) for single fluorescent spots, more complicated but more precise, is to use many circles, centered on the target spot, with different diameters (i.e., the distance from the spot center, r) spaced in an increment of 1pixel. The average intensity in each circle can be independently measured, among which I(0) just reflects the intensity of the center of the target spot. All measurements could be used to construct the radial intensity distribution against different r values. This method was used in Zenisek's paper (2002, neuron), and you could find some details in his paper which has been attached to this mail.

Could you give me some advice? Thank you very much.

Best regards
df

________________________________

From: Martin du Saire [mailto:[hidden email]]
Sent: Mon 3/6/2006 1:04 PM
To: List IMAGEJ
Subject: Re: Questions about analysis "radial intensity distribution"

If I have understood correctly, you need to get
the profiles along each radius "by hand". For example:

for (j=0; j<nAngles; j++) {
makeLine(fX,fY,fX+(profileRadius*cosA[j]),fY+(profileRadius*sinA[j]));
// (fX, fY) are the coordinates of the center
of the object or region you want to profile
// cosA and sinA are lookup tables, or you could calculate them on the fly
profile = getProfile();
// profile now contains the intensities along
the selected direction for you to use as needed
}

M

At 11:03 AM 3/6/2006, you wrote: