Say I have some function that I'm applying every element in a list to... if that element matches some criteria:

If[#==<condition>,{#}, <Do nothing> ] & /@ LongList

Is there a way to do something like this? I want the result to just not have that element.

  • $\begingroup$ Welcome to Mathematica.SE, Andrew! I'm curious, how did you find this site? $\endgroup$
    – Szabolcs
    Commented Mar 24, 2012 at 19:30
  • $\begingroup$ It's not clear to me why DeleteCases and similar functions are not the right tools for the job. Or you might perhaps prefer something like this: Flatten[If[# > 1, #, {Sequence[]}] & /@ {1, 2, 3, 0}] $\endgroup$ Commented Mar 24, 2012 at 19:30
  • $\begingroup$ I think this should be closed as a duplicate of this question but I don't want to cast a moderator super-vote. $\endgroup$
    – Mr.Wizard
    Commented Mar 24, 2012 at 20:57
  • $\begingroup$ @Mr.Wizard I agree. I'm not voting because I've also answered it. $\endgroup$
    – rm -rf
    Commented Mar 24, 2012 at 21:14
  • 1
    $\begingroup$ It is in substance but several of the answers to that question were specific to the implementation, testing for Image heads or Missing. Noone used DeleteCases. I think this question would be a better canonical version. $\endgroup$
    – Verbeia
    Commented Mar 24, 2012 at 23:20

4 Answers 4


This specific behaviour can be achieved using

If[condition, something, Unevaluated@Sequence[]]& /@ list

The key is Sequence[]. Unevaluated prevents it from disappearing from inside the If.

Alternatively you can use Cases (or many other solutions shown in other answers and comments---some of these solutions may be better suited for the problem but Sequence[] has its place too).

Cases[list, element_ /; condition :> something]
  • $\begingroup$ This is exactly what I was looking for. The Select answer I was aware of, but this is much more what I was looking for, I wasn't aware of the "Sequence[]" function. $\endgroup$ Commented Mar 24, 2012 at 23:39
  • $\begingroup$ @AndrewSpott Though I think this use of Sequence is OK here, this function has some peculiarities that you should be aware off, especially in combination with Part. Try, e.g., the following code s = {1, 2, 3} <ret> s[[1]] = Sequence[] <ret> s[[1]] <ret> s[[2]] <ret> s[[2]] = 1000 <ret> s[[2]] <ret> s <ret> and examine closely the output of the last two lines (every instruction should be on a separate line but that doesn't work here in comments). It will amaze you. $\endgroup$ Commented Mar 25, 2012 at 20:50
  • $\begingroup$ Weird, is that a bug? Also, it seems to add some sort of history to the list, which would be ridiculously hard to debug, is there a way to remove that history? $\endgroup$ Commented Mar 27, 2012 at 18:10

You can combine Map (aka /@) and Select:

In[242]:= Sqrt /@ Select[{1, -2, 3, -4, 5}, # > 0 &]
Out[242]= {1, Sqrt[3], Sqrt[5]}

enter image description here


In a similar vein to Szabolcs' first solution, you can use SlotSequence to achieve the same effect.

If[condition, <do something>, ## &[]] & /@ list
  • 2
    $\begingroup$ ##&[] evaluates to Sequence[] as well. The difference is that ##&[] is not expanded in heads with HoldAll, while Sequence[] is (it needs HoldSequence to prevent that). $\endgroup$
    – Szabolcs
    Commented Mar 24, 2012 at 23:24

In versions 11+, you can use Nothing for <do nothing>:

If[Mod[#, 3] == 0, #, Nothing] & /@ Range[20]

{3, 6, 9, 12, 15, 18}


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