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How to improve efficiency of Table expression

Gerald Feigin
Posted 10 years ago

The following code is correct but is highly inefficient:

data2 = RandomVariate[NormalDistribution[0, 1], 10^6];
MyList2 = Table[Mean[Take[data2, i]], {i, Length[data2]}];

Is there a concise way of doing this more efficiently?
POSTED BY: Gerald Feigin
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Gerald Feigin
Gerald Feigin
Posted 10 years ago
POSTED BY: Gerald Feigin
POSTED BY: Udo Krause

Somewhat faster, by keeping the arrays packed: 

myList = Accumulate[data2]/N[Range[Length[data2]]];

myList == MyList2

takes about 0.01 seconds on my machine.
POSTED BY: Ilian Gachevski
Jim Baldwin
Jim Baldwin, Retired
Posted 10 years ago

If you're wanting to calculate the sequential means of observations 1, 1+2, 1+2+3, etc., then

MyEfficientList = Accumulate[data2]/Table[i, {i, Length[data2]}]

is much faster.

If you're then going to do something with those sequential means, there might be more efficient ways to get to that final product.
POSTED BY: Jim Baldwin

Seems to me the mean of k+1 terms can be calculated from the mean of k terms and the k+1 term, if you're willing to put up with round off errors. You could do that sequentially starting with the first term and that should be faster.
POSTED BY: Frank Kampas
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