chen li wrote:
>
> --- Tom Phoenix <[EMAIL PROTECTED]> wrote:
>
>>The expression in a map is evaluated in a list
>>context. The list that
>>it returns is included in the result list.
>
>>The expression in a grep is a Boolean value, so it's
>>evaluated in a
>>scalar context. If it's true, the corresponding
>>value from the list
>>(that is, the value of $_ inside the expression) is
>>included in the
>>result list.
>>
>>These properties make grep useful for simply
>>selecting elements from a
>>list, while map is able to transform a list in a
>>more general way.
>
> So map function returns the transformed or changed
> elements but not the original ones and grep still
> returns the original ones? For example after certain
> operation A changes to B, in case of map the return
> is B but in case grep the return is still A. Is that
> right?
Perl provides four basic list transformation functions:
NEW_LIST = grep EXPRESSION, OLD_LIST;
NEW_LIST = map EXPRESSION, OLD_LIST;
NEW_LIST = reverse OLD_LIST;
NEW_LIST = sort SUB OLD_LIST;
They all take an old list and potentialy modify it and return a different
list. The ONLY thing that they have in common is that (just like the
for/foreach loops/statement modifiers) they alias $_ so if $_ is modified the
ORIGINAL DATA IS MODIFIED.
$ perl -le'
my @a = qw/ 6 3 8 4 1 0 4 3 /;
print "[EMAIL PROTECTED] = @a";
my @b = sort @a;
print "[EMAIL PROTECTED] = @a [EMAIL PROTECTED] = @b";
my @c = map { $_ *= 2 } sort @a;
print "[EMAIL PROTECTED] = @a [EMAIL PROTECTED] = @c";
my @d = grep { $_ -= 1 } reverse @a;
print "[EMAIL PROTECTED] = @a [EMAIL PROTECTED] = @d";
'
@a = 6 3 8 4 1 0 4 3
@a = 6 3 8 4 1 0 4 3 @b = 0 1 3 3 4 4 6 8
@a = 12 6 16 8 2 0 8 6 @c = 0 2 6 6 8 8 12 16
@a = 11 5 15 7 1 -1 7 5 @d = 5 7 -1 1 7 15 5 11
Because of the way that grep evaluates its expression it cannot return
anything that is not in the original list. map however can return anything:
my @x = map { ( 'X' ) x 3 } 1 .. 3;
@x now contains ( 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X', 'X' ).
map can imitate grep:
NEW_LIST = grep EXPRESSION, OLD_LIST;
NEW_LIST = map EXPRESSION ? $_ : (), OLD_LIST;
But grep CANNOT imitate map.
John
--
use Perl;
program
fulfillment
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