Map

Map and (f[#]&) can be used to map any function (f) to each argument under  any Head.  In the next Cell (Exp[_]) mapped to each term in a sum.  Many  users understand what Map does with the default level specification.  In the  first example the default level specification is used, and Exp[_] is mapped  to each expression at the first level.  The #& notation is explained in the  discussion of Function.

Clear[x] ;  demo = x + Cos[(x^(1/2) π)/3] + Sin[(x^(1/2) π)/6] ; <br /> Map[ Exp[#] &, demo]

^x + ^Cos[(π x^(1/2))/3] + ^Sin[(π x^(1/2))/6]


Note:  The shorthand notation for Map is   /@ .  This shorthand notation is
used to do the same as the previous line.  At first this notation seems very
cryptic.

Exp[#] &/@demo

^x + ^Cos[(π x^(1/2))/3] + ^Sin[(π x^(1/2))/6]

If you need to specify a level specification the short hand notation is  not at all convenient.  Different variations of level specification are demonstrated in the lines below.  Many other commands allow a user to  specify the level specification, and the conventions are always the same.   Level specification {3} means to only map the function to Level 3.  In the  next Cell we Map
(# + 1)& to all subexpressions in (demo) at Level {3}.

Map[(# + z) &, demo, {3}]

x + Cos[(1/3 + z) (π + z) (x^(1/2) + z)] + Sin[(1/6 + z) (π + z) (x^(1/2) + z)]

Level specification {-1} refers to the smallest subexpressions (the  atoms).  In the next cell we add (1) to each atom in (demo).  It should be  pointed out that the list of atoms includes 1/6, 1/3,  and 1/2, instead of 1,  2, 3, 6.  This is because since Mathematica rational numbers as atoms.  Complex numbers are also considered atoms.  
    

Map[ (# + 1) &, demo, {-1}]

1 + x + Cos[4/3 (1 + π) (1 + x)^(3/2)] + Sin[7/6 (1 + π) (1 + x)^(3/2)]


Level specification (2) refers to all levels from (Level 1) to (Level 2).  
Now we Map ( #/z&) to all subexpressions of (demo) from (Level 1) to (Level
2).


Map[#/z&, demo, 2]

x/z + Cos[(π x^(1/2))/(3 z)]/z + Sin[(π x^(1/2))/(6 z)]/z


The next line Maps (#+z&) to all subexpressions of (demo) at (Level 2) and
(Level 3).


Map[# + z&, demo, {2, 3}]

x + Cos[z + (1/3 + z) (π + z) (x^(1/2) + z)] + Sin[z + (1/6 + z) (π + z) (x^(1/2) + z)]

Heads Option


Map has a Heads option like several other functions, but it's hard to think
of a practical use for this feature.  By default Map uses the setting (Heads
→False).  If your writing a program and you need to make it full proof
you should use the form  Map[f, expr, Heads→False]  instead of the
shorter (and normally equivalent) form  f/@expr.  The reason is that the user
may have changed the default setting via SetOptions[Map, Heads→True]
which would change the behavior of  f/@expr.  Now consider the next cell to
see how Map uses the Heads option.

ClearAll[f, h] ;  expr = {h[1], {2, 3}} ;  Map[f, expr, {2}, HeadsFalse]

{h[f[1]], {f[2], f[3]}}


In the cell above (f) was mapped to expressions at positions {1,1}, {2,1},
and {2,2} since they are all the expressions at level 2.  In the next cell
the same example is evaluated with the setting (Heads→True) and (f) is
also mapped to every head at level 2. In this case the heads at level 2 are
(h) at position {1,0} and List at position {2,0}.

Map[f, expr, {2}, HeadsTrue]

{f[h][f[1]], f[List][f[2], f[3]]}

Created by Mathematica  (May 16, 2004)

Back to Ted’s Tricks index page