The problem is now fixed. You should be able to download the documentation at the following link
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I can see the problem with handling all those files! I was lucky enough to have only 54 realizations (thus 432=54*8 csv files) for 4 POE, which I later post-processed. The way I handled the files, though, was a bit lame, meaning that I performed the analyses one POE at a time, such that I had the results in separate folders. I am unfamiliar with the HDF5 format, but for the csv format I agree that maybe some results could be joined together, but not necessarily per column. For instance, since the POE are usually limited (maybe corresponding to return periods of 2500, 1460, 75, 50 years and so on) you could separate those results per sheet. If only one POE is used, then the csv would have one sheet, and so on. This way, considering the example you illustrated you would have 70000/5 = 14000 files, which is still a huge number, but more manageable. If the number of sites and disaggregation types needed by the user are also low, then this sheet methodology could be also used, but it would fail otherwise, as you would have a csv file with N*P*t sheets (where t is the number of disaggregation types you need). On the other hand, if you have M>P, so more IMTs than POEs, you would have 70000/10 = 7000 files each one with 10 sheets, for M=10.
I do not know if it is possible and how intuitive might be, but maybe the user could have the option to decide how to join the results. In my case, it would have been great to have the results of different POEs in different sheets. Another user maybe would find it more interesting to join them for site (e.g. they only have 1 POE).
I think that a couple of csv files containing the information of ALL the results might be confusing, unless the columns are really well organized. In addition, there is the parsing problem you mentioned.
However, if the mean disaggregation is done internally, there would be no need at all to export all the realizations, unless for verification problems. As it is now, I think is a good compromise, as you divide the total number by R.