I am trying to find a more efficient way to shift the columns of the partitions of an array according to a defined slope and fill in the shifted regions with zeros I have done it below but, I know it could probably done in a way that manages memory better and is probably faster too. Memory comes in to play for me when the array is really large. See the example code below.

columns = 40;
rows = 9;
data = Table[i*100 + j, {i, rows}, {j, columns}];
dt = 2;
steplength = 5*dt;
steps = Floor[columns/steplength];
shifts = Floor[steplength/Abs[dt]];
partdata = Partition[data, {rows, steplength}];
Do[step[i] = partdata[[1]][[i]], {i, steps}];
Do[dtstep[i] = Partition[step[i], {rows, Abs[dt]}], {i, steps}];
Do[shiftdtstep[i][j] = ArrayPad[dtstep[i][[1]][[j]], {If[dt >= 0, {shifts - j, j - 1}, {j - 1, shifts - j}],{0,0}}], {i, steps}, {j, shifts}];
Do[shiftedstep[i] = ArrayReshape[Transpose[Table[shiftdtstep[i][j], {j, shifts}]], {rows + shifts - 1,Abs[dt]*shifts}], {i, steps}];
shifteddata = ArrayReshape[Transpose[Table[shiftedstep[i], {i, steps}]], {rows + shifts - 1, steps*steplength}];
data // MatrixForm
shifteddata // MatrixForm

The important thing for me is that the variable dt can be either positive or negative and any value I choose within reason of the original data.


This could be a rough short way you can build on:

datapart = Partition[Transpose[data], 10];
shifteddata2 = 
        datapart[[i]][[#]], {4 - Floor[#/2.1], Floor[#/2.1]}] & /@ 
      Range[10], {i, 1, 4}], 1]];

shifteddata2 === shifteddata
| improve this answer | |
  • $\begingroup$ This works with the same data. However, if I change the number of columns to 20, I don't get the same results. Also, if a negative dt is given, it doesn't work at all. $\endgroup$ – Kane Dec 20 '14 at 1:31

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