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

I need to use If but with only one option that is if "a" then do "b", else do nothing. So I wrote If[a,b] but the problem is that if it is not a it returns Null in my output. How to avoid this?

Here is the specific example I was working on.

I am looking for the number of comparisons performed by the quick sort algorithm. Here is the code from Rosetta code with my additions

QuickSort[x_List] :=
Module[{pivot, aa = 0, bb = 0}, If[Length@x <= 1, Return[x]];
pivot = First[x];
aa = If [Length[Cases[x, j_ /; j < pivot]] > 1,
Length[Cases[x, j_ /; j < pivot]] - 1 , Sequence[]];
bb = If [Length[Cases[x, j_ /; j < pivot]] > 1,
Length[Cases[x, j_ /; j > pivot]] - 1 , Sequence[]];
count = count + aa + bb;
Flatten@{QuickSort[Cases[x, j_ /; j < pivot]],
Cases[x, j_ /; j == pivot],
QuickSort[Cases[x, j_ /; j > pivot]]} ; Return[count] ]


now if you run QuickSort[{4, 3, 2, 1, 5}] you will get 2+2 Null instead of 4

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What is the problem with that, returning Null is usually no problem, the FrontEnd will usually not even create an output cell for that return value... –  Albert Retey Mar 30 '12 at 8:43
ok i my case i could just add 0, in the "else" case. but i wonder why proposal of "kguler" and "ruebenko" that i also tested didn't work in my case –  Raph Mar 30 '12 at 9:33

## 7 Answers

It depends what you consider nothing, but you could try something like this

If[a, b, Unevaluated[Sequence[]]]


for example

3 + If[False, 1, Unevaluated[Sequence[]]]


returns 3. Wrapping an argument of a function in Unevaluated is effectively the same as temporarily setting the attribute Hold for that argument meaning that the argument isn't evaluated until after it's inserted in the definition of that function.

By the way, in your definition of QuickSort you're calling Cases[x, j_ /; j < pivot] six times. It's probably more efficient to assign Cases[x, j_ /; j < pivot] to a dummy variable and use that instead.

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Cases[x, j_ /; j < pivot] 4 times, Cases[x, j_ /; j == pivot] and Cases[x, j_ /; j > pivot] each once. But, a variable is definitely needed. –  rcollyer Mar 30 '12 at 11:24
@rcollyer you're right, I was a bit hasty in scanning the code. –  Heike Mar 30 '12 at 11:36
Unevaluated[] == Unevaluated[Sequence[]]? –  chyaong Jul 19 '13 at 6:43
@chyanog I believe this is true: one may use Unevaluated[] in any place were Unevaluated[Sequence[]] is used. I did not recommend it because ##&[] is shorter, but it should be understood that ##&[] and Unevaluated[] behave differently, though both are appropriate for this problem. I added a section to my answer that hopefully explains this. –  Mr.Wizard Aug 13 '13 at 1:13

If does not have the attribute SequenceHold; use the "vanishing function" ##&[] instead of Sequence[]:

If[# > 5, #, ## &[]] & /@ Range[10]

{6, 7, 8, 9, 10}


See this and this for other uses, and SlotSequence if you are confused by ##.

To further explain the method above, Sequence is flattened when it appears as an argument (level one) in an expression with a head that does not have either the SequenceHold or HoldAllComplete attributes.

For our purpose in If one can use any expression which:

• is held unevaluated by Hold (More specifically the HoldRest attribute.)
• does not have the head Sequence
• evaluates to Sequence[]

All of these heads meet this requirement:

• ## &
• Unevaluated
• Symbol (e.g. seq = Sequence[])
• Apply (e.g. Sequence @@ {})

Observe:

seq = Sequence[8, 9];
h = Hold[1, Sequence[2, 3], ## &[4, 5], Unevaluated[6, 7], seq, Sequence @@ {10, 11}]

Hold[1, 2, 3, (##1 &)[4, 5], Unevaluated[6, 7], seq, Sequence @@ {10, 11}]


Note that Sequence[2, 3] is flattened while all others remain. An application of List will invoke additional evaluation:

List @@ h

{1, 2, 3, 4, 5, Unevaluated[6, 7], 8, 9, 10, 11}


One sees that Unevaluated[6, 7] remains, which is an important distinction, but this is acknowledged behavior* for inert functions like List and it will evaluate in other applications:

Plus @@ h == Plus @@ Range[11]


True

Note: You may notice red highlighting in the Front End in the expression Unevaluated[6, 7]; the multi-argument (and null-argument) form of Unevaluated is not documented but it is used internally by Mathematica. You can adjust the highlighting to match by setting:

Unprotect[Unevaluated];
SyntaxInformation[Unevaluated] = {"ArgumentsPattern" -> {___}};
Protect[Unevaluated];

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If you look at it with Trace, you will see that you are just hiding it: If[# > 5, #, ## &[]] & /@ Range[10] // Trace. In have a year you will not remember what your code does. –  user21 Mar 30 '12 at 10:24
@ruebenko I really don't understand what you are trying to say. Using & effectively prevents (holds) evaluation in If. In a year I most certainly will remember what this code does, barring traumatic brain injury. What's your point? –  Mr.Wizard Mar 30 '12 at 10:28
I have the feeling this is in essence replacing the Null with a Sequence; and subjectively makes it harder to read. –  user21 Mar 30 '12 at 12:24
@ruebenko re: "in essence replacing the Null with a Sequence" -- isn't that the idea? The OP has If[. . ., Sequence[]] but it undesirably evaluates to If[. . ., Null]. Using ##&[] fixes it. What am I still missing? –  Mr.Wizard Mar 30 '12 at 12:50
you are not missing anything - I just find that neither Null nor Sequence should be inserted into the list. I find If[cond, a=result]; a cleaner (a little off the track: it can work with packed arrays). That's all I am trying to say. –  user21 Mar 30 '12 at 13:00

I note that all answers so far try to solve the problem of assigning a potential Null value by manipulating the return value. I feel it would be more appropriate to make the whole assignment conditional. Like this:

If[condition, aa = value]


There's also a small bug in your program (count isn't initialized), and, of course, it doesn't sort at the moment. I assume that you aware of that and that the Return value is used for testing.

The code would then be:

QuickSort[x_List] :=
Module[{pivot, aa = 0, bb = 0, count = 0},
If[Length@x <= 1, Return[x]];
pivot = First[x];
If[Length[Cases[x, j_ /; j < pivot]] > 1,
aa = Length[Cases[x, j_ /; j < pivot]] - 1
];
If[Length[Cases[x, j_ /; j < pivot]] > 1,
bb = Length[Cases[x, j_ /; j > pivot]] - 1
];
count = count + aa + bb;
Flatten@{QuickSort[Cases[x, j_ /; j < pivot]],
Cases[x, j_ /; j == pivot], QuickSort[Cases[x, j_ /; j > pivot]]};
Return[count]]


Remove ;Return[count] to let it sort again.

I'm not sure about the j < pivot test in the second If. It depends on your intention with bb, but I guess the test should be j > pivot.

-

In Mathematica, “not returning anything” is not possible. An expression that does not return anything has a value of Null, even though you only actually see this Null in certain circumstances:

Null
is a symbol used to indicate the absence of an expression or a result. It is not displayed in ordinary output.

When Null appears as a complete output expression, no output is printed.

So, Null is always returned, but whether it's displayed depend on what you do with this returned value. For example:

In[1]:= If[x > 0, 1]; /. x -> -1
(* nothing is output *)
In[2]:= {If[x > 0, 1]; /. x -> -1, 42}
Out[2]= {Null, 42}

-

The semantics for if is If[cond,t,f] if f (False) is not given Null is returned.

I am guessing that you need this in a pattern, then you can use Condition:

f[x_] := ppp[x] /; x > 0
f[5]
f[-6]


OK, here is a different version of quick sort (I found on my disk - I am not sure how the author is...)

ClearAll[quickSort1]
quickSort1[lst0_] :=
Block[{left, right, pivot, lst, $RecursionLimit = 10^6}, pivot = First[lst0]; lst = Rest[lst0]; left = Select[lst, # <= pivot &]; If[left =!= {}, left = quickSort1[left]]; right = Select[lst, # > pivot &]; If[right =!= {}, right = quickSort1[right]]; Join[left, {pivot}, right] ]  If someone feels like adding the counter... my brain feels like gum this morning... please go ahead. The point in the above code is to not use the return value of If but to set it to a variable. - I usually use: a = {1, 2, Null}; a = DeleteCases[a, Null];  May not be efficient, but it works. - Indeed it does work. Also, since Null can be implicit you can use a = DeleteCases[a,] which is rather nice to those of us who like terse code. – Mr.Wizard Jul 18 '13 at 23:46 Wow, Looks like I'm more than a year late to this party, but I thought this would be nice to post. The Sequence @@ {} trick. Here it is using Mr. Wizard's example: If[# > 5, #, Sequence @@ {}] & /@ Range[10]  {6, 7, 8, 9, 10} - This does work, and was recommended by Leonid at one time but isn't it better to use ##&[] now? Do you prefer Sequence @@ {} or are you just posting it for completeness? – Mr.Wizard Sep 21 '13 at 23:30 @Mr.Wizard, I didn't know about it until recently I needed to do a time consuming computation and found it in Michael Trott's Programming guidebook. I came across this post but realized it wasn't posted so decided to post it for completeness. I actually liked your ##&[] trick better. – RunnyKine Sep 22 '13 at 1:05 Okay. Completeness is good, so +1. (I also included this in my answer which I am attempting to make complete itself.) – Mr.Wizard Sep 22 '13 at 1:09 @Mr.Wizard. Do you know why none of those methods work with Compile? Is there a way to get around it? – RunnyKine Sep 22 '13 at 1:28 Without trying it I expect that it won't work with Compile because the possible return types must match. A work-around would be to use something like -1 (for a list of otherwise positive integers) or perhaps Developer$MaxMachineInteger for the "sequence" return and then DeleteCases` at the end. This can create edge cases but I don't know of a better way off-hand. –  Mr.Wizard Sep 22 '13 at 4:18