processing-scripts
processing-scripts > RE: Fisher Transformation
Jun 8, 2011 01:06 PM | Xi-Nian Zuo
RE: Fisher Transformation
Hi Ali,
Mathematically, yes, your are right. But, in practice, it gets same outputs. It's a trick about using 3dcalc, which gives you the real number if the true is a complex number, the real part of the complex number. Better to change it to the correct form to avoid the confusion. Hope this explaination clear.
Best, XN.
Originally posted by Ali Bayram:
Mathematically, yes, your are right. But, in practice, it gets same outputs. It's a trick about using 3dcalc, which gives you the real number if the true is a complex number, the real part of the complex number. Better to change it to the correct form to avoid the confusion. Hope this explaination clear.
Best, XN.
Originally posted by Ali Bayram:
Dear Experts,
In the 6_singlesubjectRSFC.sh file,
****
## 3. Z-transform correlations
echo Z-transforming correlations for seed ${seed_name}
3dcalc -a ${RSFC_dir}/${seed_name}_corr.nii.gz -expr 'log((a+1)/(a-1))/2' -prefix ${RSFC_dir}/${seed_name}_Z.nii.gz
****
This step applies Fisher Transformation, and if I'm not wrong the definition of this transformation is log((1+a)/(1-a))/2, why (a-1) used instead?
Thank you very much.
Regards,
Ali Bayram
In the 6_singlesubjectRSFC.sh file,
****
## 3. Z-transform correlations
echo Z-transforming correlations for seed ${seed_name}
3dcalc -a ${RSFC_dir}/${seed_name}_corr.nii.gz -expr 'log((a+1)/(a-1))/2' -prefix ${RSFC_dir}/${seed_name}_Z.nii.gz
****
This step applies Fisher Transformation, and if I'm not wrong the definition of this transformation is log((1+a)/(1-a))/2, why (a-1) used instead?
Thank you very much.
Regards,
Ali Bayram
Threaded View
| Title | Author | Date |
|---|---|---|
| Ali Bayram | Jun 8, 2011 | |
| Xi-Nian Zuo | Jun 8, 2011 | |
| Ali Bayram | Jun 9, 2011 | |