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I am confused by the commands @ and @@. From the documentation, I learnt that @@ is the function Apply. However, if I input Cos@@5 the output is 5, while if I input Cos@5, I get Cos[5]. By this it seems that @ rather than @@ has the function of "apply".

When I tried the function Plus, Plus@@{1,2} gave me 3 as desired, while Plus@{1,2} just gave me {1,2}. Could anyone help me with the difference (or relation) between @ and @@?

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    $\begingroup$ These are discussed in the answers to this question: mathematica.stackexchange.com/questions/5432/… The @@ is described at length in the documentation for Apply. @ is harder to find in the documentation: see Prefix, $\endgroup$ – Michael E2 Nov 14 '13 at 20:03
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    $\begingroup$ @Kuba, actually the documentation isn't great on these shortcuts. Nintety five percent of the time, I access the documentation on a function via ?FunctionName. But when you do type in ?@ you get nothing useful, likewise for @@. So my next step is usually a web search, but "at symbol mathematica" doesn't really return any useful results in Google either. $\endgroup$ – Jason B. Nov 14 '13 at 20:07
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    $\begingroup$ Welcome to Mathematica. Your question is very reasonable, as Jason points out. @ is the only function in Mathematica I can think of that apparently has no name in English. $\endgroup$ – DavidC Nov 14 '13 at 20:13
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    $\begingroup$ @Kuba, when I type ?@@ inside a notebook, that is not what I get as the output. To find that, I have to specifically open up the documentation center from the Help menu and type it into the search box. $\endgroup$ – Jason B. Nov 14 '13 at 20:22
  • $\begingroup$ @Hector I like the discussion but it is offtopic so I vote for deleting those comments to not confuse others. I admit that documentation is not perfect :). $\endgroup$ – Kuba Nov 14 '13 at 20:31
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I can see the source of your confusion: If you use Head[f[x]] and Head[5] you get f and Integer respectively. Then, you read the documentation

Apply[f,expr] or f@@expr replaces the head of expr by f.

and you expect Cos@@5 to replace the Integer head by Cos. The way I explain it to myself is by saying Mathematica has two (types of) heads ;-) One type is for expressions such as f[x] and the other is for expressions such as 5. I even have names for them: explicit and implicit heads. Then I conclude: Apply replaces explicit heads only.

Now, in {1,2}, the head is an explicit one. Thus, Plus@@{1,2}=Plus@@List[1,2]=Plus[1,2]=3.

The @ symbol is easier to understand. f@x is just f[x]. So, Plus@{1,2}=Plus[List[1,2]] and the result is List[1,2] because you are not adding anything to List[1,2]. If you want to add something to List[1,2], it must be included as another parameter to Plus. Try Plus[List[1,2],a].


As mentioned in the comments, the documentation does not warn about these two types of heads. If you look closely, you will find a warning in Possible Issues in Apply and in the 3rd statement within details in AtomQ. I would have expected some clarification in Everything Is an Expression. A naive reading of it suggests that 5 and Integer[5] are the same thing.

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    $\begingroup$ The chief difference between the two is the whether AtomQ returns True or not. If it does, then Apply doesn't work on it. $\endgroup$ – rcollyer Nov 14 '13 at 20:28
  • $\begingroup$ Even understanding that distinction, is there some sense to f@@5 simply discarding the f ? Some example where that is actually used? $\endgroup$ – george2079 Nov 14 '13 at 21:25
  • $\begingroup$ @Hector Thanks for your comment. It makes sense to me. Could you say more about the meaning of 'Head'. The help document gives some examples, but doesn't give a description or a definition. $\endgroup$ – user10568 Nov 14 '13 at 21:35
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    $\begingroup$ @george2079: In f@@5, the problem is not about "discarding" the f but rather that there is no (explicit) head to be replaced. Again, f@@Integer[5] = f[5] because Integer[5] has an explicit head Integer. On the other hand, 5 also has head Integer but it is implicit. Implicit heads are useful for pattern matching, but they are not really there ... $\endgroup$ – Hector Nov 14 '13 at 21:35
  • $\begingroup$ @hongchaniyi: After you read this link: Everything is an Expression, play with TreeForm. $\endgroup$ – Hector Nov 14 '13 at 21:41
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I like to think about @@ as a Frankstein decapitation operator. It take out the Head of the old expression and replace by the new one. And @@@ as a mass Frankstein decapitation operator. It get inside each list element and apply @@ to each element inside the list.

To understand what Head means, use FullForm. For example, in the list l={1,2,3} if you apply FullForm@lyou get List[1,2,3], where List is the Head of the expression. If you want to sum the list, you can use Plus@@l, so you change List[1,2,3] by Plus[1,2,3]. List was decapted and replaced by Plus.

If you have l={{1,2},{2,3}} (that is equivalent to List[List[1,2],List[2,3]]), if you use Plus@@@l, you get {3,5}. Plus get into the list, and execute @@ in each element.

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You may think of it as follows: @ applies a single-parameter function to a single argument, and @@ applies a multi-parameter function to a list of arguments (effectively, replacing its head List with the function). And, as a bonus, @@ works not only on lists with the List head, it can replace any head of a structured expression (but not heads of atomic expressions like numbers).

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    $\begingroup$ This description hides the fundamental feature of Apply that it replaces one head with another. Technically, a multi-parameter function is applied to a Sequence of arguments, not a List (or any other head there was around them, which gets "eaten up" by Apply). $\endgroup$ – Leonid Shifrin Nov 14 '13 at 20:51
  • $\begingroup$ @Vladimir Thanks for your reply. However, if I input Permutations@{x,y}, I have {{x,y},{y,x}}. It seems @ also apply to a set of argument. $\endgroup$ – user10568 Nov 14 '13 at 21:06
  • $\begingroup$ @hongchaniyi It's just that some functions accept a signle argument that is itself a list. $\endgroup$ – ěŕëĺíüęŕ͘  ěţěëŕ Nov 14 '13 at 21:07
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    $\begingroup$ @hongchaniyi: It might help if you visualize the {} as having an explicit head: Permutations@{x,y} = Permutations@List[x,y] = Permutations[List[x,y]] $\endgroup$ – Hector Nov 14 '13 at 21:19

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