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?
Look at this
In[119]:= FoldList[Plus, a, {b, c, d, f}]/Range[5] Out[119]= {a, (a + b)/2, 1/3 (a + b + c), 1/4 (a + b + c + d), 1/5 (a + b + c + d + f)}
with other words
FoldList[Plus, First[data2], Rest[data2]]/Range[Length[data2]]
should do it.
Somewhat faster, by keeping the arrays packed:
myList = Accumulate[data2]/N[Range[Length[data2]]]; myList == MyList2
takes about 0.01 seconds on my machine.
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.
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.
