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Re: @@@ for Version 3??
*To*: mathgroup at smc.vnet.net
*Subject*: [mg23123] Re: [mg22967] @@@ for Version 3??
*From*: Hartmut Wolf <hwolf at debis.com>
*Date*: Wed, 19 Apr 2000 02:30:52 -0400 (EDT)
*Organization*: debis Systemhaus
*References*: <200004050241.WAA01036@smc.vnet.net> <8ch9iv$95n@smc.vnet.net> <Pine.GSO.4.21.0004061746200.23405-100000@flip.eecs.umich.edu> <200004070654.CAA16541@smc.vnet.net>
*Sender*: owner-wri-mathgroup at wolfram.com
See below for more on @@@ and Version 3:
Hartmut Wolf schrieb:
>
> Daniel Reeves schrieb:
> >
> > > (If you don't have version 4, then replace @@@ by the alternative
> > > input form:
> > >
> > > In[5]:= Hold[F[x] @@@ k] // InputForm
> > > Out[5]//InputForm=
> > > Hold[Apply[F[x], k, {1}]]
> > > )
> >
> > Or you can actually replace the
> > @@@
> > with
> > @@#&/@
> >
> > Ugly, but I do it that way so it's easy to switch over to the ver4 way
> > sometime in the near future when I can feel sure that my code won't ever
> > need to run on ver3 anymore.
> >
>
> Hallo Daniel,
>
> nice ideom! Yet you have to be cautious when applying it:
>
> In[1]:= k = {{a1, b1}, {a2, b2}, {a3, b3}}
>
> In[2]:= f @@@ k
> Out[2]= {f[a1, b1], f[a2, b2], f[a3, b3]}
>
> In[3]:= g /@ f @@@ k
> Out[3]= {g[f[a1, b1]], g[f[a2, b2]], g[f[a3, b3]]}
>
> In[4]:= f @@#&/@ k
> Out[4]= {f[a1, b1], f[a2, b2], f[a3, b3]}
>
> In[5]:= g /@ f @@#&/@ k
> Out[5]= {f[g[a1], g[b1]], f[g[a2], g[b2]], f[g[a3], g[b3]]}
>
> So @@#&/@ is not a "read macro" to be textually substituted by @@@ when
> transiting from version 3 to 4 (or vice versa). Of course the correct
> expression for In[5] would be:
>
> In[6]:= g /@ (f @@ # &) /@ k
> Out[6]= {g[f[a1, b1]], g[f[a2, b2]], g[f[a3, b3]]}
>
> The full form Apply[f, k, {1}] however remains correct in any case for
> both versions.
>
> If you want to have a true syntactical equivalent substition for @@@ you
> may have one; define
>
> In[16]:= aaa = ( Function[{e}, #1 @@ e ] /@ #2 &)
>
> In[17]:= f ~aaa~ k
> Out[17]= {f[a1, b1], f[a2, b2], f[a3, b3]}
>
> In[18]:= g /@ f ~aaa~ k
> Out[18]= {g[f[a1, b1]], g[f[a2, b2]], g[f[a3, b3]]}
>
Dear Daniel,
this was one of my replies I soon felt uncomfortable with after having
pressed the send button too early. First of all, I missed to include a
neccessary wink ;-) Second, although I raised a problem, this is not
the solution, not at all a "true syntactical equivalent". What is still
flawed are the associativity rules for @@@, @@, /@. In Version 4 these
operators all are of equal precedence and group to the right:
With aaa = Apply[#1, #2, {1}] & however ~aaa~ groups to the left, and
the precedence is one above @@ and /@, see:
In[21]:= g @@@ f @@@ k === g ~aaa~ f ~aaa~ k
Out[21]= False
In[22]:= g @@@ f @@@ k === g ~aaa~ (f ~aaa~ k)
Out[22]= True
In[23]:= g @@@ f @@ k === g ~aaa~ f @@ k
Out[23]= False
In[24]:= g @@@ f @@ k === g ~aaa~ (f @@ k)
Out[24]= True
In[25]:= g @@@ f /@ k === g ~aaa~ f /@ k
Out[25]= False
In[26]:= g @@@ f /@ k === g ~aaa~ (f /@ k)
Out[26]= True
However, when you always explicitly parenthesize your expressions you
can safely use your suggested (elegant I think it is) ideom:
In[27]:= g @@@ f @@@ k === g ~aaa~ (f @@ # & /@ k)
Out[27]= True
For some days I pondered the question: can we do better?
First of all, I think we cannot define @@@ as new operator in Version 3.
The front end scanner will separate this into the two tokens @@ and @ on
the fly on sequential input (this is from version 3):
f @@ @ g
(Only input cell, key [Enter] not pressed yet.) If we activate Show
Expression from the Format menu (or press [Ctrl]+[Shift]+[e]) we see
(and still nothing evaluated):
Cell[BoxData[
RowBox[{"f", " ", "@@",
RowBox[{"@", " ", "g"}]}]], "Input"]
(You can see from that expression, that no Space was input between the
second and third '@'.)
An idea would be to use Low-Level Input Rules (as explained in §2.8.17);
however we must first find an input string which we can use for
reworking the parsed input. After some futile trials I came up with the
sequence
f @$@ g
Cell[BoxData[
RowBox[{"f", " ", "@",
RowBox[{"$", "@", " ", "g"}]}]], "Input"]
which interpreted in FullForm is of course just
f[$[g]]
I think, you will never come across the symbol $ to have a value,
contrary to 'q' (which else I would have preferred for input ergonomy)
<< now quit the kernel and restart >>
In[1]:= MakeExpression[ RowBox[{f_, "@",
RowBox[{"$", "@", g_}]}], StandardForm] :=
MakeExpression[RowBox[{"mapply", "[", f, ",", g, "]"}],
StandardForm]
We try it:
In[2]:= k = {{a1, b1}, {a2, b2}, {a3, b3}}
In[3]:= f @$@ k
Out[3]= mapply[f, {{a1, b1}, {a2, b2}, {a3, b3}}]
In[4]:= g @$@ f @$@ k
Out[4]= mapply[g, mapply[f, {{a1, b1}, {a2, b2}, {a3, b3}}]]
So all what we have to do is define mapply:
In[5]:= mapply[f_, g_] := Apply[f, g, {1}]
In[6]:= f @$@ k
Out[6]= {f[a1, b1], f[a2, b2], f[a3, b3]}
So it works! Grouping turned out to be right (automatically, as a
consequence of the grouping of @).
Now how to get the right precedence with @@ and /@ ? Again we locally
manipulate the parse tree:
In[8]:=
MakeExpression[RowBox[{
RowBox[{f_, "@",
RowBox[{"$", "@", g_}]}], "@@", h_}], StandardForm] :=
MakeExpression[RowBox[{"mapply", "[",
RowBox[{f, ",",
RowBox[{g, "@@", h}]}], "]"}], StandardForm]
In[9]:=
MakeExpression[RowBox[{
RowBox[{f_, "@",
RowBox[{"$", "@", g_}]}], "/@", h_}], StandardForm] :=
MakeExpression[RowBox[{"mapply", "[",
RowBox[{f, ",",
RowBox[{g, "/@", h}]}], "]"}], StandardForm]
With that things seem to work:
In[10]:= g @@@ f @@@ k === g @$@ f @$@ k
Out[10]= True
In[11]:= g @@@ f @@ k === g @$@ f @@ k
Out[11]= True
In[12]:= g @@@ f /@ k === g @$@ f /@ k
Out[12]= True
Of course also
In[13]:= g /@ f @$@ k
Out[13]= {g[f[a1, b1]], g[f[a2, b2]], g[f[a3, b3]]}
In[14]:= g @@ f @$@ k
Out[14]= g[f[a1, b1], f[a2, b2], f[a3, b3]]
I tried more cases, they all worked. I would be interested if someone
finds a hole in it!
Kind regards, Hartmut
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