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Possible Duplicate:
How to avoid returning a Null if there is no “else” condition in an If contruct

I know there are other ways to solve this problem. Just because I'm curious I tried:

Map[If[# == 1, , #] &, {2, 1, 3, 1, 4}]

This should return a list without the value 1, because when the statement is true it should do nothing. The solution should be:

{2,3,4}

and not

{2,Null,3,Null,4}

Why is it not working as I expect?

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  • $\begingroup$ DeleteCases[{2, 1, 3, 1, 4},1] $\endgroup$ – ssch Dec 21 '12 at 10:03
  • $\begingroup$ thank you ssch. I know that this works. Just wanted to know if there is a possibility to get the solution out of the if directly. $\endgroup$ – RMMA Dec 21 '12 at 10:06
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    $\begingroup$ Map[If[# == 1, Unevaluated[Sequence[]], #] &, {2, 1, 3, 1, 4}] in that case :) $\endgroup$ – ssch Dec 21 '12 at 10:08
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    $\begingroup$ Frink, please try to make an effort to find existing questions which address your problem before asking a new one. In this case a search for "If Null" would have brought up the proper question as the first result. $\endgroup$ – Mr.Wizard Dec 21 '12 at 15:03
  • $\begingroup$ Another duplicate: mathematica.stackexchange.com/q/3447/5 $\endgroup$ – rm -rf Dec 21 '12 at 15:50
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Your If statement is actually

If[# == 1, Null, #]

so it is working perfectly. Try evaluating

Hold@Map[If[# == 1, , #] &, {2, 1, 3, 1, 4}]

to see for yourself. To get what you want, try

Map[If[# == 1, Unevaluated@Sequence[], #] &, {2, 1, 3, 1, 4}]
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  • $\begingroup$ thx, exactly what I was looking for :-) $\endgroup$ – RMMA Dec 21 '12 at 10:22
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    $\begingroup$ Maybe mentioning If[x > 1, , 2] // FullForm, which evaluates to If[Greater[x, 1], Null, 2], is also useful $\endgroup$ – acl Dec 21 '12 at 10:31
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    $\begingroup$ Incidentally, Unevaluated@Sequence[] can be replaced with ##&[] which I call a "vanishing function." $\endgroup$ – Mr.Wizard Dec 21 '12 at 15:01
  • $\begingroup$ @Mr.Wizard. ##&[] is a neat trick! $\endgroup$ – m_goldberg Dec 21 '12 at 17:21

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