I don't know why Y does not get its type inferred correctly. But you can do: eltype(df[2])[df[i,2] for i in 1:size(df,1)]
On Thu, 2015-06-18 at 20:21, SG <[email protected]> wrote: > #= This line adds functions to take > an AR(2) model for US inflation > =# > > using DataFrames > > function lag0(x,p) > R::Int32=size(x,1) > C::Int32=size(x,2) > > # Take the first R-p rows of matrix x > x1=x[1:(R-p),:] > return out=[zeros(p,C); x1] > > end > > > #load inflation datatable > #df=readtable("inflation.csv") > > df=DataFrame() > df[:A]=1:8 > df[:B]=10:17 > > ################################## > # Way of converting DataArray to Array # > ################################## > Y=[df[i,2] for i in [1:size(df,1)]] > > ## >>>> type of (Y) is now "Any" > > > T=size(Y,1) > X=[ones(T,1) lag0(Y,1) lag0(Y,2)] > inv(X'*X) > > ========================================================== > > Hi All, > > I am posting again the issue. I found when I create an Array as shown > above, the type of Array > changed to "Any" which is not acceptable for the base Inverse function(an > error occurred). > I am wondering why Julia does not assume the type of original data, here in > the example, Int64. > > Many thanks, > > > > > > > On Thursday, June 18, 2015 at 7:23:16 PM UTC+2, SG wrote: >> >> Thank you, I will cut it down and post it again soon. >> >> On Thursday, June 18, 2015 at 7:01:16 PM UTC+2, Stefan Karpinski wrote: >>> >>> I think these sample programs may be too big for people to review for >>> you. If you can pare the problem down to an example that can be posted in >>> an email, you've more likely to get help. >>> >>> On Thu, Jun 18, 2015 at 11:11 AM, SG <[email protected]> wrote: >>> >>>> >>>> Hi All, >>>> >>>> I am a novice to Julia. >>>> >>>> While I am running the attached julia program, I found "inverse" >>>> function throws an error? >>>> However it worked well in Matlab. Can you help me? Many thanks. >>>> >>>> >>>> >>>> >>>
