Interesting but, if I'm understanding this well, to get a normal LI, the --paired-diff-norm should be divided by 2 not multiplied, isn't it?
standardLI = (lh-rh)/(lh+rh)
FS_LI = (lh-rh)/((lh+rh)/2) <-- as you are dividing the denominator, your LI will be twice an standard one.
Further, this division should be done before the smoothing, right?
If I do use the LI maps to perform my stats, which would be the difference in interpretation between the results of using the --paired-diff maps, and the standarized LI maps?
Regards,
Gabriel.
El 29/03/13, Douglas N Greve <gr...@nmr.mgh.harvard.edu> escribió:
The --xhemi flag causes both the left and right hemispheres of each
subject to be stacked into the output file. So the output file will have
number of subjects x2 frames. The order is subject1.lh, subject1.rh,
subject2.lh, subject2.rh, ...
If you add the --paired-diff, then you will get number of frames =
number of subjects, and each frame will be subject1.lh-rh,
subject2.lh-rh, etc
If you use the --paired-diff-norm instead, then you will get what you
want subject1.(lh-rh)/((lh+rh)/2), subject2.(lh-rh)/((lh+rh)/2), etc.
Note that most laterality indices (LI) are (lh-rh)/(lh+rh), so you would
need to multiply the paired-diff-norm by 2.
Finally, I want to point out that it may be better to smooth before
computing the LI because the LI computation is non-linear and it has the
potential to divide by a noisy number. To do this, run without the
--paired-diff flags, then smooth, then run
mri_concat yourfile.smoothed.mgh --paired-diff-norm --o
yourfile.smoothed.LI.mgh
In my study, I ran it both ways and it did not make a difference, but I
think smoothing before LI is the safer bet.
doug
On 03/29/2013 12:33 PM, Ejoe Yizhou Ma wrote:
> Hi freesurfer experts,
>
> I'm investigating lh-to-rh asymmetry and am following instructions on
> this page :http://surfer.nmr.mgh.harvard.edu/fswiki/Xhemi.
> My question is, what does the "--xhemi" flag do in the "mris_preproc"
> command? It seems to me that it calculates (lh-rh) value at each
> vertex for every subject. (I tried to use the same subject for each
> pair of input for "--paired-diff", and the result is not a all-zero
> output.)
>
> It would also be nice if someone can tell me how to realize my final
> goal, which is to get the (lh-rh)/[(lh+rh)/2] value at each vertex for
> each subject in a .mgh file.
>
> Thanks,
> Cherry
>
>
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Douglas N. Greve, Ph.D.
MGH-NMR Center
gr...@nmr.mgh.harvard.edu
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--------------------------
PhD. student Gabriel González-Escamilla
Laboratory of Functional Neuroscience
Department of Physiology, Anatomy, and Cell Biology
University Pablo de Olavide
Ctra. de Utrera, Km.1
41013 - Seville
- Spain -
Email: ggon...@upo.es
http://www.upo.es/neuroaging/es/
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