Is anyone working with large datasets ( 200M) in LabVIEW?

Question

I am working with external Bioinformatics databasesa and find the datasets to be quite large (2 files easily come out at 50M or more). Is anyone working with large datasets like these? What is your experience with performance?

Answer 1

Colby, it all depends on how much memory you have in your system. You could be okay doing all that with 1GB of memory, but you still have to take care to not make copies of your data in your program. That said, I would not be surprised if your code could be written so that it would work on a machine with much less ram by using efficient algorithms. I am not a statistician, but I know that the averages & standard deviations can be calculated using a few bytes (even on arbitrary length data sets). Can't the ANOVA be performed using the standard deviations and means (and other information like the degrees of freedom, etc.)? Potentially, you could calculate all the various bits that are necessary and do the F-test with that information, and not need to ever have the entire data set in memory at one time. The tricky part for your application may be getting the desired data at the necessary times from all those different sources. I am usually working with files on disk where I grab x samples at a time, perform the statistics, dump the samples and get the next set, repeat as necessary. I can calculate the average of an arbitrary length data set easily by only loading one sample at a time from disk (it's still more efficient to work in small batches because the disk I/O overhead builds up).
Let me use the calculation of the mean as an example (hopefully the notation makes sense): see the jpg. What this means in plain english is that the mean can be calculated solely as a function of the current data point, the previous mean, and the sample number. For instance, given the data set [1 2 3 4 5], sum it, and divide by 5, you get 3. Or take it a point at a time: the average of [1]=1, [2+1*1]/2=1.5, [3+1.5*2]/3=2, [4+2*3]/4=2.5, [5+2.5*4]/5=3. This second method required far more multiplications and divisions, but it only ever required remembering the previous mean and the sample number, in addition to the new data point. Using this technique, I can find the average of gigs of data without ever needing more than three doubles and an int32 in memory. A similar derivation can be done for the variance, but it's easier to look it up (I can provide it if you have trouble finding it). Also, I think this funtionality is built into the LabVIEW pt by pt statistics functions.
I think you can probably get the data you need from those db's through some carefully crafted queries, but it's hard to say more without knowing a lot more about your application.
Hope this helps!
Chris
Attachments:
Mean Derivation.JPG ‏20 KB