open-discussion
open-discussion > RE: Reproducibility analysis mICA
Apr 10, 2018 02:04 AM | Will Khan - King's College London
RE: Reproducibility analysis mICA
Hi Tawfik,
Thanks very much for your response.
I did think about this yesterday - I initially did not try it because it would be very computationally expensive, particularly when using the HCP dataset.
For example, if using the 50 repetitions generated by split half that would mean 2*50 sub-folders = 100 ICA's. I was planning to run anywhere between 2-20 dimensionalities which would mean running 1000 group ICA's. That is quite a lot!
I saw in your recent HBM paper you ran 3-50 components on 50 HCP subjects using 50 repetitions from split-half. I suspect you ran this on a colossal supercomputer - Do you remember how long it took you?
Also, would you suggest running this on a single repetition, i.e. 2*1 sub-folder = 2 * 10 ICA analyses = 20 ICA's. Do you think the reproducibility script should work?
Thanks very much,
Cheers!
Will
Originally posted by Tawfik Moher Alsady:
Thanks very much for your response.
I did think about this yesterday - I initially did not try it because it would be very computationally expensive, particularly when using the HCP dataset.
For example, if using the 50 repetitions generated by split half that would mean 2*50 sub-folders = 100 ICA's. I was planning to run anywhere between 2-20 dimensionalities which would mean running 1000 group ICA's. That is quite a lot!
I saw in your recent HBM paper you ran 3-50 components on 50 HCP subjects using 50 repetitions from split-half. I suspect you ran this on a colossal supercomputer - Do you remember how long it took you?
Also, would you suggest running this on a single repetition, i.e. 2*1 sub-folder = 2 * 10 ICA analyses = 20 ICA's. Do you think the reproducibility script should work?
Thanks very much,
Cheers!
Will
Originally posted by Tawfik Moher Alsady:
Hi Will,
Question 1:
The split-half script generates 2*N sub-folders where N is the number of repitions. For each subfolder you have to call the mica command in order to calculate mICA for that random subgroup.
In your example, those subfolders should lie in $working_dir/results/out_50
Step 2 should be like that:
export MICADIR=/Users/wasimkhan/Documents/mICA_Toolbox
Then start the following command for each sample 1-50 and each group 1-2:
./mica $working_dir/input_list $working_dir/results/out_50/sample_XX/groupXX_input.txt $working_dir/results/out_50/sample_XX/groupXX 2,4,6,8,10 -mask $working_dir/masks/PCC_and_Precuneus_mask_thr20_Harvard.nii.gz -extra -a concat -- sep_vn --nobet --bgthreshold=10 --tr=1 --report --mmthresh=0.5 --Oall
I guess performing the previous steps as suggested here will allow py/ic_corr.py be able to calculate the reproducibilty.
Let me know if it works.
Best,
Tawfik
Question 1:
The split-half script generates 2*N sub-folders where N is the number of repitions. For each subfolder you have to call the mica command in order to calculate mICA for that random subgroup.
In your example, those subfolders should lie in $working_dir/results/out_50
Step 2 should be like that:
export MICADIR=/Users/wasimkhan/Documents/mICA_Toolbox
Then start the following command for each sample 1-50 and each group 1-2:
./mica $working_dir/input_list $working_dir/results/out_50/sample_XX/groupXX_input.txt $working_dir/results/out_50/sample_XX/groupXX 2,4,6,8,10 -mask $working_dir/masks/PCC_and_Precuneus_mask_thr20_Harvard.nii.gz -extra -a concat -- sep_vn --nobet --bgthreshold=10 --tr=1 --report --mmthresh=0.5 --Oall
I guess performing the previous steps as suggested here will allow py/ic_corr.py be able to calculate the reproducibilty.
Let me know if it works.
Best,
Tawfik
Threaded View
Title | Author | Date |
---|---|---|
Will Khan | Mar 22, 2018 | |
Florian Beissner | Mar 22, 2018 | |
Will Khan | Mar 27, 2018 | |
Tawfik Moher Alsady | Mar 27, 2018 | |
Will Khan | Mar 29, 2018 | |
Will Khan | Mar 29, 2018 | |
Tawfik Moher Alsady | Mar 29, 2018 | |
Will Khan | Apr 4, 2018 | |
Tawfik Moher Alsady | Apr 9, 2018 | |
Will Khan | Apr 10, 2018 | |