Re: set versus set-delay

• To: mathgroup at smc.vnet.net
• Subject: [mg75243] Re: set versus set-delay
• From: Bill Rowe <readnewsciv at sbcglobal.net>
• Date: Sat, 21 Apr 2007 23:15:55 -0400 (EDT)

```On 4/20/07 at 12:43 AM, siewsk at bp.com wrote:

>SET versus SETDELAY

>Newbie are always told to use SETDELAY instead of SET when they
>create a mathematica function.

This is generally good advice for someone starting out with
Mathematica since functions defined using SetDelayed (not
SETDELAY) will operate closer to what most people expect.

>SET and SETDELAY using ordinary language (what linguistics call
>vernacular language)

>The main differences between those two are:

>SET       means    CALCULATE FIRST THEN CHANGE
>SETDELAY  means    CHANGE FIRST THEN CALCULATE

>To explain, consider the following two functions f ang g

>f[x_] = Expand[(x+7)^2]           (* Using SET       *) g[x_]:=
>Expand[(x+7)^2]          (* Using SETDELAY  *)

>The two functions are identical except f uses SET while g uses
>SETDELAY. In normal usage there is no difference in the two
>functions. However they produce their results using different steps.

The explanation that followed (and I have snipped) is close to
the mark but not quite there.

The key difference is when the rhs of the expression gets
evaluated. This is readily shown using Trace.

=46irst with your f as above

In[1]:=
(f[x_]=Expand[(x+7)^2])//Trace

Out[1]=
{{{{HoldForm[x + 7], HoldForm[x + 7]},
HoldForm[(x + 7)^2]}, HoldForm[Expand[(x + 7)^2]],
HoldForm[x^2 + 14*x + 49]},
HoldForm[f[x_] = x^2 + 14*x + 49],
HoldForm[x^2 + 14*x + 49]}

That is the result Expand[(x+7)^2] is evaluated and the result
assigned to f. Using your terminology above, there is a
calculation and a change. f has been changed.

Then some time later

In[2]:=
f[4]//Trace

Out[2]=
{HoldForm[f[4]], HoldForm[49 + 14*4 + 4^2],
{HoldForm[14*4], HoldForm[56]}, {HoldForm[4^2],
HoldForm[16]}, HoldForm[49 + 56 + 16], HoldForm[121]}

All parts of the expression that had x are immediately replaced
with 4, then the resulting expression is evaluated

Contrast this with

In[3]:=
(g[x_]:=Expand[(x+7)^2])//Trace

Out[3]=
{HoldForm[g[x_] := Expand[(x + 7)^2]], HoldForm[Null]}

No evaluation of the expression has occurred

Then later

In[4]:=
g[4]//Trace

Out[4]=
{HoldForm[g[4]], HoldForm[Expand[(4 + 7)^2]],
{{HoldForm[4 + 7], HoldForm[11]}, HoldForm[11^2],
HoldForm[121]}, HoldForm[Expand[121]], HoldForm[121]}

The problem beginners usually have with using Set instead of
SetDelayed is illustrated by

In[17]:=
Clear[f,x];
x=14;
f[x_]=Expand[(x+7)^2];
f[4]

Out[20]=
441

Since x had a defined value when f was defined Expand[(x+7)^2]
evaluated to 441 and that was assigned to f. So, f now returns
the constant 441 for all arguments, something a beginner
definitely doesn't expect.

Contrast this with

In[21]:=
Clear[g,x];
x=14;
g[x_]:=Expand[(x+7)^2];
g[4]

Out[24]=
121

which returns the desired value even though x was previously defined.

The time to use Set is when you are the function is defined in
terms of something Mathematica requires significant CPU time to
compute and the result is used in a loop. Clearly, if the rhs
includes say a complex integral, you don't want to solve the
integral repeatedly as would occur with SetDelayed.

But it is important to understand what is being evaluated and
when. The particular function you chose illustrates using Set
instead of SetDelayed will not always speed up evaluation. In
this particular example, expansion of (x+7)^2 results in an
expression that requires doing one multiply, one exponentiation
and one addition to get the result. Leaving (x+7)^2 unevaluated
until x is defined eliminates the multiply, i.e., SetDelayed
results in fewer operations to get the same result.
--
To reply via email subtract one hundred and four

```

• Prev by Date: Re: ContourPlot3D Problem
• Next by Date: Re: how to get the table
• Previous by thread: Re: set versus set-delay
• Next by thread: Re: Importing and retaining graphics