the entirety of datasets through identifiers.org

[yarikoptic]

• An example record for a dataset pointing to the individual files: http://identifiers.org/rest/metadata/reactome:R-HSA-446203
• Their REST API implementation: https://github.com/identifiers-org/rest
• List of their services endpoints: http://identifiers.org/restws

Cons

• ATM I do not see anything neuroimaging (according to search) but situation might change, especially if we register our datalad: prefix ("Register" DataLad with identifiers.org? datalad-registry#67)

Analysis/possible difficulties

• I do not see yet how to discover individual IDs/datasets for a particular prefix (sent out a question via their web interface; the answer was: not at the moment, but it sounded to them as an interesting feature so might come at some point)
• Not all prefixes relate to "datasets", but some are known as "(data) collections": https://www.ebi.ac.uk/miriam/main/collections/
• I do not think there is any versioning, but most probably it is assumed that an identifier points to immutable dataset
• There will be a lot of datasets. So we would need some sensible structure/hierarchy. First level would be the identifier. Then we could partition even further splitting IDs on / and -.
• There seems to be no "filename" information provided. So we would have choices:
  • like a default git-annex behavior - just use the entire url to compose a unique filename
  • one from the URL (often from Content-Disposition header field) - but that might lead to conflicts since we would allow only for a flat structure:
    • we could preanalyze the entire list of those first and see if conflicts arise. If there are conflicts, try to deduce somehow disambiguating structure. but that is unreliable in case a dataset record changes with more files etc
    • just add an arbitrary, or based on some metadata?, numeric index in addition

[chrisgorgo]

BTW NeuroVault uses identifiers.org: http://identifiers.org/neurovault.collection and http://identifiers.org/neurovault.image. Happy to answer any questions I can.