RE: Pure Functions in rules
- To: mathgroup at smc.vnet.net
- Subject: [mg15976] RE: [mg15932] Pure Functions in rules
- From: "Ersek, Ted R" <ErsekTR at navair.navy.mil>
- Date: Fri, 19 Feb 1999 03:26:59 -0500
- Sender: owner-wri-mathgroup at wolfram.com
When you use {1,2,3}/. (m_List->(2*#& /@ m))
(2*#& /@ m) evaluates before the replacement is made.
To see what it evaluates to see Out[1].
In[1]:=
Clear[m];
2*#&/@m
Out[1]=
m
So what you end up doing is simply
{1,2,3}/.m_List->m
________________________________
Instead you should use (:>) instead
of (->) as in In[2]. When you do it
this way (2*#& /@ m) doesn't evaluate
until m={1,2,3}.
In[2]:=
{1,2,3}/.(m_List:>(2*#& /@ m))
Out[2]=
{2,4,6}
_____________________
Now I give you another reason to
prefer (:>) over (->). Suppose (m) has a global value as below.
In[4]:=
m=Log[5];
In[5]:=
....
....
Then you do a lot of work, and forget about your previous use of (m).
....
....
Now you try to use (->) in a case where it worked before, and your stunned
by the result at Out[87]. I have yet to find a book about Mathematica that
warns users about this, and I am puzzled way none of the authors discuss
this point.
In[87]:=
{1,2,3}/.m_List->2*m
Out[87]=
2 Log[5]
_______________________
If you use (:>) you get the right result even
though (m) has a value assigned to it.
In[88]:=
{1,2,3}/.m_List:>2*m
Out[88]=
{2,4,6}
In general you shouldn't use (lhs->rhs) if (lhs) contains a named pattern
and the name of the pattern is used in (rhs). Notice the next line comes
out right because (m) isn't used on the right hand side of (->). You only
have trouble when the name of the pattern is used on the right hand side of
(->).
In[89]:=
{1,2,3,4,5,6}/.m_?EvenQ->0
Out[89]=
{1,0,3,0,5,0}
Actually there is a way to always get the right result using (lhs->rhs).
You have to use Block (as below), but this is more complicated than
(lhs:>rhs).
In[90]:=
Block[{m},{1,2,3}/.m_List->2*m]
Out[90]=
{2,4,6}
In[91]:=
m
Out[91]=
Log[5]
At Out[90] we got the right result even though (m) still has the value
Log[5].
Regards,
Ted Ersek
_______________________
Will Self wrote:
It appears that I cannot depend on using a pure function
in a pattern-matching rule.
Here I am trying to convince reluctant students that they're
better off learning to use Mathematica than doing things
by hand, and we run across something like this, and in a
much more complicated situation where the trouble was
hard to isolate.
I am quite frankly incensed by the behavior shown in
In/Out 80, below. Look at these examples:
In[73]:= {1,2,3}/.(m_List->7)
Out[73]= 7
In[74]:= {1,2,3}/.(m_List->(2*m))
Out[74]= {2,4,6}
In[75]:= 2*#& /@ {1,2,3}
Out[75]= {2,4,6}
In[77]:= f[m_List]:=2*#& /@ m
In[78]:= f[{1,2,3}]
Out[78]= {2,4,6}
In[79]:= {1,2,3}/.m_List->f[m]
Out[79]= {2,4,6}
Now try this:
In[80]:= {1,2,3}/.(m_List->(2*#& /@ m))
Out[80]= {1,2,3}
Does anyone (say, at WRI for example) care to comment on
this?
Will Self