Skip to main content

performance tuning - Speeding up Import and Export in CSV format


I am handling large numerical data in Mathematica. In smaller problems everything worked fine using Export and Import with the parameter csv and nothing more.


Now I am facing a much larger data volume and plain Export and Import is way too slow for CSV format.


What I want to do: First, exporting a numerical list of approximately $1400 \cdot 260$. Then I perform some calculations outside of mathematica and finally I import a csv file back using Import.



In this question I read how to improve the speed of Export with CSV.


I tried


Export["data.csv", 
  ExportString[Transpose[temp], "CSV", "FieldSeparators" -> ", "], 
  "Table"];

for a toy-example of dimension $9 \cdot 6$. This could be


temp = RandomReal[{0, 1000}, {9, 6}];

The problem is that I got additional blank lines in my CSV file. How can I avoid those blank lines? I did not have them using plain Export.



Second part of the question: can I use a similar approach to speed up Import for CSV files?


I work in English locale in Windows 7 and Mathematica 8. My data should be comma-separated.


The result of the above looks like this:


155.9418457227914,427.72566448956945,462.4370455183139,434.230107096781,377.73423736605037,457.7044624877774,229.5721937681028,453.6973831247924,827.5146478718962

656.9573702857699,975.1048399942904,716.715190156526,67.07781324817643,168.78248854317894,863.1953962590844,997.7580107302701,427.94798294100747,565.2955778916687

192.3648037459477,435.0418975785194,126.17228368369842,772.0737083559297,453.73573640921836,957.9178360741387,920.4158275934401,234.75353158374764,162.82606110943834

841.7132070637356,799.1268178998612,931.2448706410551,950.7753472229233,114.01596316796622,145.0771999411104,287.47149951303663,786.9008107323455,99.09420650484662


116.9916885502289,715.7594598282562,970.6252946068753,654.1742185278038,262.3778046629968,200.13980161337577,347.24862854841354,314.5612015073982,241.11046402342163

203.65015448763597,952.1236458849723,578.2673369638862,527.2990305555661,655.1228742370724,318.81372163827496,311.2738362265584,315.97629887850667,514.7676854642548

EDIT: A small example of the data that I want to import (2nd step) can be found here. The true problem size is here.



Answer



Here's a much faster, purely Mathematica way than using Import to import your data:


UPDATE


As Leonid mentioned the previous code doesn't exactly replicate Import. The truth is I was only trying to retrieve the numerical part. Here's an updated version that tries to replicate the output from Import.



readYourCSV2[file_String?FileExistsQ, n_Integer] := Module[{str = OpenRead[file], data}, 
  data = ReadList[str, Table[Record, {n}], RecordSeparators -> {",", "\n"}]; 
  Close[str]; 
  ReleaseHold[ToExpression[data, InputForm, Hold] /. {Plus[Times[x_, E | e], y_] :> x * 10 ^ y}] 
 ]

Here, n is the number of columns.


UPDATE 2


Now for the Export, here's a fast, again, purely Mathematica way to export in CSV format.


writeYourCSV[file_String, list_List?MatrixQ] := 

 With[{str = OpenWrite[file, PageWidth -> Infinity], len = Length[ list[[1]] ]},
      Scan[Write[str, Sequence @@ (Flatten[Table[{FortranForm[ #[[i]] ], OutputForm[","]}, 
              {i, len - 1}]]) ~ Join ~ { FortranForm[ #[[len]] ] }] &, list]; Close[str];
]

This takes less than 10 seconds to write your large data back to CSV format:


writeYourCSV["testcsv.csv", databig] // AbsoluteTiming


{9.921969, Null}




Comments

Post a Comment

Popular posts from this blog

mathematical optimization - Minimizing using indices, error: Part::pkspec1: The expression cannot be used as a part specification

I want to use Minimize where the variables to minimize are indices pointing into an array. Here a MWE that hopefully shows what my problem is. vars = u@# & /@ Range[3]; cons = Flatten@ { Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; Minimize[{Total@((vec1[[#]] - vec2[[u[#]]])^2 & /@ Range[1, 3]), cons}, vars, Integers] The error I get: Part::pkspec1: The expression u[1] cannot be used as a part specification. >> Answer Ok, it seems that one can get around Mathematica trying to evaluate vec2[[u[1]]] too early by using the function Indexed[vec2,u[1]] . The working MWE would then look like the following: vars = u@# & /@ Range[3]; cons = Flatten@{ Table[(u[j] != #) & /@ vars[[j + 1 ;; -1]], {j, 1, 3 - 1}], 1 vec1 = {1, 2, 3}; vec2 = {1, 2, 3}; NMinimize[ {Total@((vec1[[#]] - Indexed[vec2, u[#]])^2 & /@ R...
Post a Comment
Read more

functions - Get leading series expansion term?

Given a function f[x] , I would like to have a function leadingSeries that returns just the leading term in the series around x=0 . For example: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x)] x and leadingSeries[(1/x + 2 + (1 - 1/x^3)/4)/(4 + x)] -(1/(16 x^3)) Is there such a function in Mathematica? Or maybe one can implement it efficiently? EDIT I finally went with the following implementation, based on Carl Woll 's answer: lds[ex_,x_]:=( (ex/.x->(x+O[x]^2))/.SeriesData[U_,Z_,L_List,Mi_,Ma_,De_]:>SeriesData[U,Z,{L[[1]]},Mi,Mi+1,De]//Quiet//Normal) The advantage is, that this one also properly works with functions whose leading term is a constant: lds[Exp[x],x] 1 Answer Update 1 Updated to eliminate SeriesData and to not return additional terms Perhaps you could use: leadingSeries[expr_, x_] := Normal[expr /. x->(x+O[x]^2) /. a_List :> Take[a, 1]] Then for your examples: leadingSeries[(1/x + 2)/(4 + 1/x^2 + x), x] leadingSeries[Exp[x], x] leadingSeries[(1/x + 2 + (1 - 1/x...
Post a Comment
Read more

What is and isn't a valid variable specification for Manipulate?

I have an expression whose terms have arguments (representing subscripts), like this: myExpr = A[0] + V[1,T] I would like to put it inside a Manipulate to see its value as I move around the parameters. (The goal is eventually to plot it wrt one of the variables inside.) However, Mathematica complains when I set V[1,T] as a manipulated variable: Manipulate[Evaluate[myExpr], {A[0], 0, 1}, {V[1, T], 0, 1}] (*Manipulate::vsform: Manipulate argument {V[1,T],0,1} does not have the correct form for a variable specification. >> *) As a workaround, if I get rid of the symbol T inside the argument, it works fine: Manipulate[ Evaluate[myExpr /. T -> 15], {A[0], 0, 1}, {V[1, 15], 0, 1}] Why this behavior? Can anyone point me to the documentation that says what counts as a valid variable? And is there a way to get Manpiulate to accept an expression with a symbolic argument as a variable? Investigations I've done so far: I tried using variableQ from this answer , but it says V[1...
Post a Comment
Read more