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I'm trying to convert one list to another by applying a condition on the list's elements. But I can't figure out how to do this in mathematica's language. Python example below

l1 = [0, 1, 2, 3, 4]
l2 = []
for i in l1:
    if i % 2 == 0:
        l2.append(i)
    else:
        l2.append(0)
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  • $\begingroup$ {0, 1, 2, 3, 4} // Select[Mod[#, 2] == 0 &] (* {0, 2, 4} *) $\endgroup$ – alancalvitti Nov 25 '14 at 17:06
  • $\begingroup$ @alancalvitti this way you miss appending zeros $\endgroup$ – funnyp0ny Nov 25 '14 at 17:40
  • $\begingroup$ @funnypony, thanks. {0, 1, 2, 3, 4} // Map[If[Mod[#, 2] == 0, #, 0] &] $\endgroup$ – alancalvitti Nov 25 '14 at 18:52
  • $\begingroup$ @alancalvitti that's nice! you can also freely remove Map, which seems unnecessary $\endgroup$ – funnyp0ny Nov 25 '14 at 20:01
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For the specific condition in OP's example, you can also use

f = (1 - Mod[#, 2]) #  &;
f@l1
(* {0, 0, 2, 0, 4} *)
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Table[If[Mod[l, 2] == 0, l, 0], {l, 0, 4}]

or 

Table[If[Mod[l1[[i]], 2] == 0, l1[[i]], 0], {i, Length@l1}]

for arbitrary list l1

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  • 3
    $\begingroup$ What about Table[If[Mod[l, 2] == 0, l, 0], {l, l1}] for arbitrary l1? Remember that you can simply iterate through l1 with this, when you don't need the index of the current element. $\endgroup$ – halirutan♦ Dec 25 '14 at 22:50
  • $\begingroup$ @halirutan very nice! thanks! Didn't know that.. $\endgroup$ – funnyp0ny Dec 26 '14 at 13:09
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Piecewise[{{#,Mod[#,2]==0},{0,True}},#]&/@ l1
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  • $\begingroup$ Why is your Piecewise so complicated? You only need one condition which is Mod[#,2]==0. If this is not satisfied, that you can use the default value. So either Piecewise[{{#, Mod[#, 2] == 0}}, 0] & or even shorter Piecewise[{{#, Mod[#, 2] == 0}}] & should be enough. $\endgroup$ – halirutan♦ Dec 25 '14 at 22:47
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(Assuming you work only with integers in the list)

 l1={0,1,2,3,4}

and then

 l2=l1 /. {x_?OddQ -> 0}
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  • $\begingroup$ This is wrong. If you have sublists this will do all sorts of unpredictable stuff. Consider using Replace[list, rules, {1}] $\endgroup$ – Carlo Jan 24 '15 at 23:43
  • $\begingroup$ @Carlo I don't claim that to be a universal solution. I just claim that it works for the list that Erik asks about... Thanks for your clarification, though. $\endgroup$ – mickep Jan 25 '15 at 6:53

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