Re: For Routine to Map Routine

*To*: mathgroup at smc.vnet.net*Subject*: [mg70417] Re: For Routine to Map Routine*From*: "Ray Koopman" <koopman at sfu.ca>*Date*: Mon, 16 Oct 2006 02:34:11 -0400 (EDT)*References*: <egse9i$css$1@smc.vnet.net>

Benedetto Bongiorno wrote: > To All, > > Dimensions of ans01 is 2185850 by 5. > I am looking to convert my For routine to the more efficient Map routine > > n=Length[ans01] > fo={}; > > For[i=1, i<n, i++, > If[ans01[[i,1]] == ans01[[i+1,1]], > AppendTo[fo, ans01[[1]]]]; > > I tried > Map[If[#[[1]] == #[[1]],#]&,ans01]; > > But it does not work. > > Thank you > > Ben I interpret your code as saying that you want the rows of ans01 whose first element is the same as the first element in the next row. In a previous thread, Carl Woll wrote: | The fastest approach for selecting a conditional subset of a large | data set usually works as follows: | 1. Construct a list of 0's and 1's corresponding to which elements | you want to keep. | 2. Extract the position of the 0's or 1's. | 3. Use Part to obtain the subset. Also, operating on entire vectors is usually faster than operating on elements of vectors inside a loop. Applying those recommendations: c = ans01[[All,1]] gives the first column of ans01. d = Most@c-Rest@c gives the vector of differences between successive terms in c. This is coded below as (Most@#-Rest@#&)[c]. u = UnitStep[-Abs[d]] gives 1 where d=0, and gives 0 everywhere else. p = SparseArray[u]/.SparseArray[_,_,_,p_]:>Flatten[p[[2,2]]] gives the positions of the 1's in u. ans01[[p]] gives the rows of ans01 that you want. Putting it all together: ans01[[SparseArray[UnitStep[-Abs[(Most@#-Rest@#&)[ans01[[All,1]]]]]]/. SparseArray[_,_,_,p_]:>Flatten[p[[2,2]]]]]