2
$\begingroup$

when we wish to use one list to select items from another using something like Pick, writing the following code, for example, causes us to lose the 1-to-1 correspondence between items in the two lists:

dataArray = {"A","B","C","D","E","F","G"};
testArray = {0.223,0.3,1.2,0.44,4,0.24449,1.01};
dataArray = Pick[dataArray, #>= 1 &/@ testArray];

output = {"C", "E", "G"}

Without having to make a copy of anything, how do we safely prune items from, here testArray, to restore the previous 1-to-1 correspondence between elements in testArray and elements in dataArray? For example, if B in dataArray corresponds to 0.3 in testArray (based on its index), it should again do so after the Pick pruning step.

$\endgroup$
  • 1
    $\begingroup$ I'm not sure what is you goal at the end but maybe you can just work on pairs? Select[Transpose[{dataArray, testArray}], #[[2]] > 1 &] $\endgroup$ – Kuba Apr 4 '14 at 20:06
3
$\begingroup$

There are surely many ways to approach this problem. Which is best likely (again) depends on your data. I will illustrate three variants.

Paired data (a la decorate-and-sort)

We can do as Kuba did and merge the two lists into one to keep the elements together:

Select[{dataArray, testArray}\[Transpose], #[[2]] >= 1 &]
{{"C", 1.2}, {"E", 4}, {"G", 1.01}}

You can finish with a second Transpose to separate the data into two lists.

Reused mask

A typically faster method is to simply construct the mask once and then reuse it in Pick as needed:

mask = UnitStep[testArray - 1];

Pick[#, mask, 1] & /@ {dataArray, testArray}
{{"C", "E", "G"}, {1.2, 4, 1.01}}

Note that I converted your test to a vectorized numeric form for better performance.

Index-based filtering

Perhaps the top performing method for filter reapplication (especially in version 7 before Pick was better optimized) is to create a list of positions you wish to keep, then extract them using Part or Extract. Faster than Position, when applicable, is SparseArray, using the undocumented Properties method:

fastpos = SparseArray[#]["AdjacencyLists"] &;

idx = fastpos @ UnitStep[testArray - 1]
{3, 5, 7}
#[[idx]] & /@ {dataArray, testArray}
{{"C", "E", "G"}, {1.2, 4, 1.01}}

You can also process multiple lists at once with the help of All, like this:

{dataArray, testArray}[[All, idx]]
{{"C", "E", "G"}, {1.2, 4, 1.01}}
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.