Re: Assigning an array to a dereferenced hash slice - syntax!
On Sat, Mar 14, 2009 at 01:06, Chap Harrison c...@pobox.com wrote: On Mar 13, 2009, at 7:52 AM, Jenda Krynicky wrote: That's prettymuch it, except that it's not an array of aliases, but LIST of aliases. The difference is subtle, but important. See http://perldoc.perl.org/perlfaq4.html#What-is-the-difference- between-a-list-and-an-array%3f and http://www.perlmonks.org/?node_id=130861 Thanks - I've read the FAQ several times in the past without really getting it, but the perlmonks thread looks promising. It IS subtle, all right. :-/ snip The eureka moment for me was when I realized that lists are a data type (like number or string) and arrays and hashes are containers that hold that data type. Once you realize that it is easy to see why they have different behaviours. -- Chas. Owens wonkden.net The most important skill a programmer can have is the ability to read. -- To unsubscribe, e-mail: beginners-unsubscr...@perl.org For additional commands, e-mail: beginners-h...@perl.org http://learn.perl.org/
Re: Assigning an array to a dereferenced hash slice - syntax!
From: Chap Harrison c...@pobox.com
Let me break this expression down according to my current understanding.
@{$hash{adams...@keys} = @values;
parses as follows:
adams : short for 'adams', a string literal, being used as a hash key
$hash{adams} : the value in the hash associated with 'adams', a scalar
(always), and a ref to a hash in this case
{$hash{adams}} : extra braces, used to force correct binding (?)
with surrounding context, I expect.
{$hash{adams...@keys} : the hashref is dereferenced, and a set of keys
(provided by an array) is looked up in the hash, resulting in a set of
values; or, more precisely, *aliases* of values.
@{$hash{adams...@keys} : an array of aliases of the hash values
associated with the keys supplied in @keys.
@{$hash{adams...@keys} = @values : behaves much like...
( $hash{adams}{a}, $hash{adams}{ar}, $hash{adams}{af}, $hash{adams}
{aw} ) = ( 1, 19, 13, 11 )
That's prettymuch it, except that it's not an array of aliases, but
LIST of aliases. The difference is subtle, but important.
See http://perldoc.perl.org/perlfaq4.html#What-is-the-difference-
between-a-list-and-an-array%3f and
http://www.perlmonks.org/?node_id=130861
And it doesn't really make sense to add only the {...@keys} without the
@. You have to consider them both at the same time, because the sigil
you use affects how is the @keys evaluated!
See
%hash = (a = 999, b = 888, c = 777, d = 666, 3 = 123456);
@keys = ('a','b','c');
print $ha...@keys},\n;
print @ha...@keys},\n;
$href = \%hash;
print ${$hre...@keys},\n;
print @{$hre...@keys},\n;
HTH, Jenda
= je...@krynicky.cz === http://Jenda.Krynicky.cz =
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to get drunk and croon as much as they like.
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Re: Assigning an array to a dereferenced hash slice - syntax!
On Mar 13, 2009, at 7:52 AM, Jenda Krynicky wrote:
That's prettymuch it, except that it's not an array of aliases, but
LIST of aliases. The difference is subtle, but important.
See http://perldoc.perl.org/perlfaq4.html#What-is-the-difference-
between-a-list-and-an-array%3f and
http://www.perlmonks.org/?node_id=130861
Thanks - I've read the FAQ several times in the past without really
getting it, but the perlmonks thread looks promising. It IS subtle,
all right. :-/
And it doesn't really make sense to add only the {...@keys} without the
@. You have to consider them both at the same time, because the sigil
you use affects how is the @keys evaluated!
Yes, I see your point.
That exercise helped to clarify it a bit more. Thanks to everyone for
helping me!
Chap
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Re: Assigning an array to a dereferenced hash slice - syntax!
On Mar 12, 2009, at 12:52 AM, Chas. Owens wrote:
On Thu, Mar 12, 2009 at 01:27, Chap Harrison c...@pobox.com wrote:
It's still not intuitive to me why we FIRST convert the hash to
an array,
and THEN ask for keys - keys being hash-ish, rather than array-ish
sorts of
things. (I've said that badly.) What exactly are the elements of
the array
@{$hash{adams...@keys} ?
snip
It isn't really becoming an array. A better way to think of it is
that $ means we expect one value back from the data structure:
Let me break this expression down according to my current understanding.
@{$hash{adams...@keys} = @values;
parses as follows:
adams : short for 'adams', a string literal, being used as a hash key
$hash{adams} : the value in the hash associated with 'adams', a scalar
(always), and a ref to a hash in this case
{$hash{adams}} : extra braces, used to force correct binding (?)
with surrounding context, I expect.
{$hash{adams...@keys} : the hashref is dereferenced, and a set of keys
(provided by an array) is looked up in the hash, resulting in a set of
values; or, more precisely, *aliases* of values.
@{$hash{adams...@keys} : an array of aliases of the hash values
associated with the keys supplied in @keys.
@{$hash{adams...@keys} = @values : behaves much like...
( $hash{adams}{a}, $hash{adams}{ar}, $hash{adams}{af}, $hash{adams}
{aw} ) = ( 1, 19, 13, 11 )
So that's my own badly-described understanding of how that expression
does what it does. I'd greatly appreciate any corrections.
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Assigning an array to a dereferenced hash slice - syntax!
Hello,
I have a hash of hashes, and I'm trying to bulk-assign some key/value
pairs to the referenced hash, using a technique I read on the
perlmonks list (hash slices, see link below). I've never been good at
working out syntax where complex data structures are concerned, and
this eludes me completely. Could someone please help me with the
syntax of the assignment, and retrieval of key/value pairs?
Thanks - I appreciate it!
Chap
*
#!/usr/bin/perl -d
use strict;
use warnings;
my %schoolprops;
# +
-
# | %schoolprops - a hash of references to hashes
# |
# | key valuehash
# | ---
---
# | Adams-- {a = 1, ar = 19, af = 13, aw = 11}
# | Baker-- {b = 2, bq = 20}
# | Charlie -- {c = 3, cc = 33, cu = 38}
# +
-
$schoolprops{Adams} = \{}; # map 'Adams' to a ref to an empty hash
my $hashref = $schoolprops{Adams}; # create a ref to a hash.
my @keys = ( 'a', 'ar', 'af', 'aw' );
my @values = ( 1, 19, 13, 11);
#
# The following is trying to assign an array to a hash slice.
# (courtesy http://www.perlmonks.org/?node_id=4402, who did a simpler
# version that did not involve references.)
#
@{{$hashref}-{...@keys}} = @values;# don't know if this is right
while (my ($k, $v) = each %{$hashref}) { # this is just one of many
things
print $k = $v\n; # I've tried; this one fails at
runtime.
}
#EOF
*
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Re: Assigning an array to a dereferenced hash slice - syntax!
On Thu, Mar 12, 2009 at 00:19, Chap Harrison c...@pobox.com wrote:
Hello,
I have a hash of hashes, and I'm trying to bulk-assign some key/value pairs
to the referenced hash, using a technique I read on the perlmonks list (hash
slices, see link below). I've never been good at working out syntax where
complex data structures are concerned, and this eludes me completely. Could
someone please help me with the syntax of the assignment, and retrieval of
key/value pairs?
Thanks - I appreciate it!
Chap
*
#!/usr/bin/perl -d
use strict;
use warnings;
my %schoolprops;
# +
-
# | %schoolprops - a hash of references to hashes
# |
# | key value hash
# | ---
---
# | Adams -- {a = 1, ar = 19, af = 13, aw = 11}
# | Baker -- {b = 2, bq = 20}
# | Charlie -- {c = 3, cc = 33, cu = 38}
# +
-
$schoolprops{Adams} = \{}; # map 'Adams' to a ref to an empty hash
my $hashref = $schoolprops{Adams}; # create a ref to a hash.
my @keys = ( 'a', 'ar', 'af', 'aw' );
my @values = ( 1, 19, 13, 11);
#
# The following is trying to assign an array to a hash slice.
# (courtesy http://www.perlmonks.org/?node_id=4402, who did a simpler
# version that did not involve references.)
#
@{{$hashref}-{...@keys}} = @values; # don't know if this is right
while (my ($k, $v) = each %{$hashref}) { # this is just one of many things
print $k = $v\n; # I've tried; this one fails at
runtime.
}
#EOF
*
Dereference the hashref as an arrayref then ask for the keys:
#!/usr/bin/perl
use strict;
use warnings;
my %hash = ( adams = {} );
my @keys = qw/a ar af aw/;
my @values = (1, 19, 13, 11);
@{$hash{adams...@keys} = @values;
use Data::Dumper;
print Dumper \%hash;
--
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Re: Assigning an array to a dereferenced hash slice - syntax!
On Mar 11, 2009, at 11:51 PM, Chas. Owens wrote:
Dereference the hashref as an arrayref then ask for the keys:
#!/usr/bin/perl
use strict;
use warnings;
my %hash = ( adams = {} );
my @keys = qw/a ar af aw/;
my @values = (1, 19, 13, 11);
@{$hash{adams...@keys} = @values;
use Data::Dumper;
print Dumper \%hash;
Thank you for both a solution and several other useful tips as well!
It's still not intuitive to me why we FIRST convert the hash to an
array, and THEN ask for keys - keys being hash-ish, rather than array-
ish sorts of things. (I've said that badly.) What exactly are the
elements of the array @{$hash{adams...@keys} ?
Thanks,
Chap.
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Re: Assigning an array to a dereferenced hash slice - syntax!
Chap Harrison wrote:
Hello,
Hello,
I have a hash of hashes, and I'm trying to bulk-assign some key/value
pairs to the referenced hash, using a technique I read on the perlmonks
list (hash slices, see link below). I've never been good at working out
syntax where complex data structures are concerned, and this eludes me
completely. Could someone please help me with the syntax of the
assignment, and retrieval of key/value pairs?
Thanks - I appreciate it!
Chap
*
#!/usr/bin/perl -d
use strict;
use warnings;
my %schoolprops;
# +
-
# | %schoolprops - a hash of references to hashes
# |
# | key valuehash
# | ---
---
# | Adams-- {a = 1, ar = 19, af = 13, aw = 11}
# | Baker-- {b = 2, bq = 20}
# | Charlie -- {c = 3, cc = 33, cu = 38}
# +
-
$schoolprops{Adams} = \{};# map 'Adams' to a ref to an empty hash
{} *is* a reference to an empty hash, \{} is a reference to a reference
to an empty hash.
my $hashref = $schoolprops{Adams}; # create a ref to a hash.
It doesn't create it, it just copies it.
my @keys = ( 'a', 'ar', 'af', 'aw' );
my @values = ( 1, 19, 13, 11);
#
# The following is trying to assign an array to a hash slice.
# (courtesy http://www.perlmonks.org/?node_id=4402, who did a simpler
# version that did not involve references.)
#
@{{$hashref}-{...@keys}} = @values;# don't know if this is right
If you change = \{} to = {} then:
@{ $hashref }{ @keys } = @values;
Otherwise it won't work.
while (my ($k, $v) = each %{$hashref}) { # this is just one of many things
print $k = $v\n; # I've tried; this one fails at
runtime.
}
John
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Re: Assigning an array to a dereferenced hash slice - syntax!
Chap Harrison wrote:
On Mar 11, 2009, at 11:51 PM, Chas. Owens wrote:
Dereference the hashref as an arrayref then ask for the keys:
#!/usr/bin/perl
use strict;
use warnings;
my %hash = ( adams = {} );
my @keys = qw/a ar af aw/;
my @values = (1, 19, 13, 11);
@{$hash{adams...@keys} = @values;
use Data::Dumper;
print Dumper \%hash;
Thank you for both a solution and several other useful tips as well!
It's still not intuitive to me why we FIRST convert the hash to an
array, and THEN ask for keys - keys being hash-ish, rather than
array-ish sorts of things. (I've said that badly.) What exactly are
the elements of the array @{$hash{adams...@keys} ?
@{$hash{adams...@keys} is *not* an array, it is a hash slice.
John
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Re: Assigning an array to a dereferenced hash slice - syntax!
On Thu, Mar 12, 2009 at 01:27, Chap Harrison c...@pobox.com wrote:
On Mar 11, 2009, at 11:51 PM, Chas. Owens wrote:
Dereference the hashref as an arrayref then ask for the keys:
#!/usr/bin/perl
use strict;
use warnings;
my %hash = ( adams = {} );
my @keys = qw/a ar af aw/;
my @values = (1, 19, 13, 11);
@{$hash{adams...@keys} = @values;
use Data::Dumper;
print Dumper \%hash;
Thank you for both a solution and several other useful tips as well!
It's still not intuitive to me why we FIRST convert the hash to an array,
and THEN ask for keys - keys being hash-ish, rather than array-ish sorts of
things. (I've said that badly.) What exactly are the elements of the array
@{$hash{adams...@keys} ?
snip
It isn't really becoming an array. A better way to think of it is
that $ means we expect one value back from the data structure:
my $scalar = $array[0];
my $scalar = $hash{foo};
and @ means we expect to get many values back from the data structure
my @result = @array[0 .. 5];
my @result = @hash{qw/foo bar baz/};
of course, you have to look at the context to determine what happens:
my $length = @array;
--
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The most important skill a programmer can have is the ability to read.
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hash slice??
Is there such a thing?
I'm trying to take a HoH and make a reference to a sub-part of the hash.
This doesn't work:
my %sub_hash = $main_hash{'sub_hash'};
I get the following error:
Reference found where even-sized list expected at ./my_buggy_program line 30.
Any quick tips on how to reference and assign this sub-hash?
Thanks,
- Travis
Re: hash slice??
On Mon, Nov 10, 2008 at 09:52, Travis Thornhill
[EMAIL PROTECTED] wrote:
Is there such a thing?
I'm trying to take a HoH and make a reference to a sub-part of the hash.
This doesn't work:
my %sub_hash = $main_hash{'sub_hash'};
I get the following error:
Reference found where even-sized list expected at ./my_buggy_program line 30.
Any quick tips on how to reference and assign this sub-hash?
snip
#!/usr/bin/perl
use strict;
use warnings;
use Data::Dumper;
my %name_to_num = (
one = 1,
two = 2,
three = 3,
four = 4,
five = 5,
six = 6
);
#get keys whose values are odd
my @keys = grep { $name_to_num{$_} % 2 } keys %name_to_num;
#make a hash of the odd keys and values with a hash slice
my %odd_name_to_num;
@[EMAIL PROTECTED] = @[EMAIL PROTECTED];
print Dumper \%name_to_num, \%odd_name_to_num;
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Re: hash slice??
On Mon, Nov 10, 2008 at 3:52 PM, Travis Thornhill [EMAIL PROTECTED]
wrote:
Is there such a thing?
I'm trying to take a HoH and make a reference to a sub-part of the hash.
This doesn't work:
my %sub_hash = $main_hash{'sub_hash'};
I get the following error:
Reference found where even-sized list expected at ./my_buggy_program line
30.
Any quick tips on how to reference and assign this sub-hash?
Thanks,
- Travis
Hi Travis,
The problem you are facing is that $main_hash{'sub_hash'} does not return a
hash at all but a reference. Which is exactly what perl is telling you. You
can simply tell perl that it should dereference the reference to a hash, you
could do that like this:
#!/usr/bin/perl
use warnings;
use strict;
use Data::Dumper;
my %hash = ( 1 = 'one',
2 = 'two',
3 = { 3.1 = 'three point one',
3.2 = 'three point two' },
4 = 'four' );
my %hash_slice = %{$hash{3}};
print Dumper %hash_slice;
Reagrds,
Rob
Re: hash slice??
Travis Thornhill wrote:
Is there such a thing?
Yes there is.
I'm trying to take a HoH and make a reference to a sub-part of the hash.
This doesn't work:
my %sub_hash = $main_hash{'sub_hash'};
I get the following error:
Reference found where even-sized list expected at ./my_buggy_program line 30.
Any quick tips on how to reference and assign this sub-hash?
my %sub_hash;
@sub_hash{ 'sub_hash' } = @main_hash{ 'sub_hash' };
John
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can special-order certain sorts of tools at low cost and
in short order.-- Larry Wall
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Re: hash slice??
Travis Thornhill wrote:
Is there such a thing?
I'm trying to take a HoH and make a reference to a sub-part of the hash.
This doesn't work:
my %sub_hash = $main_hash{'sub_hash'};
I get the following error:
Reference found where even-sized list expected at ./my_buggy_program line 30.
Any quick tips on how to reference and assign this sub-hash?
There is certainly such a thing as a hash slice, but I think it is probably not
what you want here. A slice will let you extract multiple values from the hash
simultaneously, something like
my @subhashes = @main_hash{'key1', 'key2', 'key3'};
and it doesn't look to me like that's what you're trying to achieve.
If you have a hash of hashes then $main_hash{'sub_hash'} is already a reference,
as the error message has told you, so you should write
my $subhash_ref = $main_hash{'sub_hash'};
after which you can access the subsidiary hash with constructs like
foreach my $key (keys %$subhash_ref) {
printf %s = %s\n, $key, $subhash_ref-{$key};
}
HTH,
Rob
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Re: populating a hash slice from a filehandle
On Fri, Jun 20, 2008 at 4:52 PM, Bryan R Harris
[EMAIL PROTECTED] wrote:
Bryan R Harris wrote:
John W. Krahn wrote:
Bryan R Harris wrote:
John W. Krahn wrote:
The left hand side of the assignment determines context so the @l2r{...}
part.
That strikes me as odd... When perl goes to populate @l2r{a,b}, it
seems to me that it would go through this process:
- I have a slice here, so I'll loop over the slice elements
- The first is a, so I'll pull a scalar off the list and assign it to
$l2r{a}
- The second is b, so I'll pull another scalar off the list and assign
it
to $l2r{b}
- Remaining scalars in the list are discarded
Correct, except for the loop part.
Why would $l2r{a} here be considered list context?
It isn't, unless it's written as ( $l2r{a} ), then it's a list with
one element.
So I still don't understand what about @l2r{a,b} makes it evaluate the
first (FILE... in list context instead of scalar context.
The '@' sigil at the front of the variable name says that it is either
an array or a slice and so it forces list context on the right hand side
of the assignment.
I think it finally clicked!
It makes more sense to me that (FILE,FILE) is kind of the same thing as
saying (@a,@b). In list context @a returns the array as a list, but in
scalar context @a returns the number of elements. Obviously (@a,@b) returns
the union of the two lists, not two scalars. FILE is treated the same
way.
Bryan,
That isn't really as obvious as you think. There are probably
situations where (@a, @b) would be a list of scalars. I'll say it
again: the function of the parentheses in a particular situation
(grouping, precedence, capturing, list assignment) is assign *after*
context is determined. Context has to do with the result you are
asking for, not the data you are inputting. Just because you are
working with a list doesn't mean your expression will be evaluated in
list context. The important thing to keep sight of is that it is what
happens on the lefthand side of the equals sign that forces the
context, here, and there is nothing on the righthand side to
explicitly force a change.
You can think of this as an apples to apples issue. Lists can be
assigned to lists, and processed as lists. When you perform an
operation on a list, you aren't always iterating over the elements in
sequence. A list is a collection of scalars, true, but it's also a
fundamental data type in its own right. A slice is a type of list, so
Perl looks at the slice assignment and says I need a list. It then
evaluates the expression on the right with an eye toward making a
list. *Then* it worries about how the elements of the list on the
right will be affected by whatever has to happen to the list on the
left.
HTH,
-- j
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Re: populating a hash slice from a filehandle
From: Bryan R Harris [EMAIL PROTECTED]
Bryan R Harris wrote:
John W. Krahn wrote:
Bryan R Harris wrote:
John W. Krahn wrote:
The left hand side of the assignment determines context so the @l2r{...}
part.
That strikes me as odd... When perl goes to populate @l2r{a,b}, it
seems to me that it would go through this process:
- I have a slice here, so I'll loop over the slice elements
- The first is a, so I'll pull a scalar off the list and assign it to
$l2r{a}
- The second is b, so I'll pull another scalar off the list and assign
it
to $l2r{b}
- Remaining scalars in the list are discarded
Correct, except for the loop part.
Why would $l2r{a} here be considered list context?
It isn't, unless it's written as ( $l2r{a} ), then it's a list with
one element.
So I still don't understand what about @l2r{a,b} makes it evaluate the
first (FILE... in list context instead of scalar context.
The '@' sigil at the front of the variable name says that it is either
an array or a slice and so it forces list context on the right hand side
of the assignment.
I think it finally clicked!
It makes more sense to me that (FILE,FILE) is kind of the same thing as
saying (@a,@b). In list context @a returns the array as a list, but in
scalar context @a returns the number of elements. Obviously (@a,@b) returns
the union of the two lists, not two scalars. FILE is treated the same
way.
Almost. It still depends on the left hand side. Try this:
@a = (10,20,30);
@b = (40,50,60);
$s = (@a,@b);
print $s\n;
#versus
($s) = (@a,@b);
print $s\n;
In the first case the @a and @b are evaluated in scalar context. Even
though they are enclosed in braces.
Dang!
But your example helped -- this one makes sense to me, at least:
perl -e '@a=(1,2,3);@b=(4,5,6,7);@l{1,2}=(@a,@b);print $l{1}\t$l{2}\n;'
Should I think of this as: ($l{1}, $l{2}) = (@a,@b)
?
If so, then that makes a lot more sense to me.
Thanks for helping me along on this one...
- Bryan
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Re: populating a hash slice from a filehandle
From:Bryan R Harris [EMAIL PROTECTED]
Jenda wrote:
From: Bryan R Harris [EMAIL PROTECTED]
It makes more sense to me that (FILE,FILE) is kind of the same thing as
saying (@a,@b). In list context @a returns the array as a list, but in
scalar context @a returns the number of elements. Obviously (@a,@b)
returns
the union of the two lists, not two scalars. FILE is treated the same
way.
Almost. It still depends on the left hand side. Try this:
@a = (10,20,30);
@b = (40,50,60);
$s = (@a,@b);
print $s\n;
#versus
($s) = (@a,@b);
print $s\n;
In the first case the @a and @b are evaluated in scalar context. Even
though they are enclosed in braces.
Dang!
But your example helped -- this one makes sense to me, at least:
perl -e '@a=(1,2,3);@b=(4,5,6,7);@l{1,2}=(@a,@b);print $l{1}\t$l{2}\n;'
Should I think of this as: ($l{1}, $l{2}) = (@a,@b)
?
If so, then that makes a lot more sense to me.
Right. That's it.
The fact that there is the same number of things in the braces on
both sides of the equals might be a bit confusing at first, but it's
irrelevant. Since in this case the equals provide a list context for
the (@a,@b), the two arrays are flattened into one list. So
($l{1}, $l{2}) = (@a,@b,@c);
or
($l{1}, $l{2}, $l{4}) = (@a,@b);
would be just as OK. The commas are not matched, the elements of the
two lists are. The list of specified elements of %l and the list of
all elements of @a and @b.
Jenda
= [EMAIL PROTECTED] === http://Jenda.Krynicky.cz =
When it comes to wine, women and song, wizards are allowed
to get drunk and croon as much as they like.
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Re: populating a hash slice from a filehandle
From: Bryan R Harris [EMAIL PROTECTED]
Bryan R Harris wrote:
John W. Krahn wrote:
Bryan R Harris wrote:
John W. Krahn wrote:
The left hand side of the assignment determines context so the @l2r{...}
part.
That strikes me as odd... When perl goes to populate @l2r{a,b}, it
seems to me that it would go through this process:
- I have a slice here, so I'll loop over the slice elements
- The first is a, so I'll pull a scalar off the list and assign it to
$l2r{a}
- The second is b, so I'll pull another scalar off the list and assign
it
to $l2r{b}
- Remaining scalars in the list are discarded
Correct, except for the loop part.
Why would $l2r{a} here be considered list context?
It isn't, unless it's written as ( $l2r{a} ), then it's a list with
one element.
So I still don't understand what about @l2r{a,b} makes it evaluate the
first (FILE... in list context instead of scalar context.
The '@' sigil at the front of the variable name says that it is either
an array or a slice and so it forces list context on the right hand side
of the assignment.
I think it finally clicked!
It makes more sense to me that (FILE,FILE) is kind of the same thing as
saying (@a,@b). In list context @a returns the array as a list, but in
scalar context @a returns the number of elements. Obviously (@a,@b) returns
the union of the two lists, not two scalars. FILE is treated the same
way.
Almost. It still depends on the left hand side. Try this:
@a = (10,20,30);
@b = (40,50,60);
$s = (@a,@b);
print $s\n;
#versus
($s) = (@a,@b);
print $s\n;
In the first case the @a and @b are evaluated in scalar context. Even
though they are enclosed in braces.
Jenda
= [EMAIL PROTECTED] === http://Jenda.Krynicky.cz =
When it comes to wine, women and song, wizards are allowed
to get drunk and croon as much as they like.
-- Terry Pratchett in Sourcery
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Re: populating a hash slice from a filehandle
Bryan R Harris wrote:
John W. Krahn wrote:
Bryan R Harris wrote:
Jenda Krynicky wrote:
Context. The FILEHANDLE returns a single line in scalar context and
a list of all lines in a list context. And there is no such thing as
a two-item-list context.
So in the first case the assignment to @l2r{a,b} provides a list
context so the very first FILE reads all the lines left in FILE,
the second and third return an empty list. The first two lists are
concatenated together and the first two items of the resulting list
are assigned to $l2r{a} and $l2r{b}. And the rest of the list is
forgotten. You'd have to do something like
@l2r{a,b} = (scalar(FILE), scalar(FILE));
or
@l2r{a,b} = (FILE.'', FILE.'');
to ensure that the FILE is evaluated in scalar context.
Which part is forcing the list context? The fact that the FILE is inside
parenthesis () or the @l2r{...} part?
The left hand side of the assignment determines context so the @l2r{...}
part.
That strikes me as odd... When perl goes to populate @l2r{a,b}, it
seems to me that it would go through this process:
- I have a slice here, so I'll loop over the slice elements
- The first is a, so I'll pull a scalar off the list and assign it to
$l2r{a}
- The second is b, so I'll pull another scalar off the list and assign it
to $l2r{b}
- Remaining scalars in the list are discarded
Correct, except for the loop part.
Why would $l2r{a} here be considered list context?
It isn't, unless it's written as ( $l2r{a} ), then it's a list with
one element.
So I still don't understand what about @l2r{a,b} makes it evaluate the
first (FILE... in list context instead of scalar context.
What am I missing?
Hard to say? :-)
Indeed.
Maybe this one's just too complex for my little brain.
- B
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Re: populating a hash slice from a filehandle
Bryan R Harris wrote:
John W. Krahn wrote:
Bryan R Harris wrote:
John W. Krahn wrote:
The left hand side of the assignment determines context so the @l2r{...}
part.
That strikes me as odd... When perl goes to populate @l2r{a,b}, it
seems to me that it would go through this process:
- I have a slice here, so I'll loop over the slice elements
- The first is a, so I'll pull a scalar off the list and assign it to
$l2r{a}
- The second is b, so I'll pull another scalar off the list and assign it
to $l2r{b}
- Remaining scalars in the list are discarded
Correct, except for the loop part.
Why would $l2r{a} here be considered list context?
It isn't, unless it's written as ( $l2r{a} ), then it's a list with
one element.
So I still don't understand what about @l2r{a,b} makes it evaluate the
first (FILE... in list context instead of scalar context.
The '@' sigil at the front of the variable name says that it is either
an array or a slice and so it forces list context on the right hand side
of the assignment.
John
--
Perl isn't a toolbox, but a small machine shop where you
can special-order certain sorts of tools at low cost and
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Re: populating a hash slice from a filehandle
Bryan R Harris wrote:
John W. Krahn wrote:
Bryan R Harris wrote:
John W. Krahn wrote:
The left hand side of the assignment determines context so the @l2r{...}
part.
That strikes me as odd... When perl goes to populate @l2r{a,b}, it
seems to me that it would go through this process:
- I have a slice here, so I'll loop over the slice elements
- The first is a, so I'll pull a scalar off the list and assign it to
$l2r{a}
- The second is b, so I'll pull another scalar off the list and assign it
to $l2r{b}
- Remaining scalars in the list are discarded
Correct, except for the loop part.
Why would $l2r{a} here be considered list context?
It isn't, unless it's written as ( $l2r{a} ), then it's a list with
one element.
So I still don't understand what about @l2r{a,b} makes it evaluate the
first (FILE... in list context instead of scalar context.
The '@' sigil at the front of the variable name says that it is either
an array or a slice and so it forces list context on the right hand side
of the assignment.
I think it finally clicked!
It makes more sense to me that (FILE,FILE) is kind of the same thing as
saying (@a,@b). In list context @a returns the array as a list, but in
scalar context @a returns the number of elements. Obviously (@a,@b) returns
the union of the two lists, not two scalars. FILE is treated the same
way.
Thanks!
- Bryan
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Re: populating a hash slice from a filehandle
From: Bryan R Harris [EMAIL PROTECTED]
Given an open filehandle, why don't these two things do the same thing?
**
@l2r{a,b} = (FILE, FILE);
$c = FILE;
**
$l2r{a} = FILE;
$l2r{b} = FILE;
$c = FILE;
**
The first seems to be slurping the whole file into $l2r{b} and leaving $c
undefined... The second does what I want. Doesn't seem to make sense.
Context. The FILEHANDLE returns a single line in scalar context and
a list of all lines in a list context. And there is no such thing as
a two-item-list context.
So in the first case the assignment to @l2r{a,b} provides a list
context so the very first FILE reads all the lines left in FILE,
the second and third return an empty list. The first two lists are
concatenated together and the first two items of the resulting list
are assigned to $l2r{a} and $l2r{b}. And the rest of the list is
forgotten. You'd have to do something like
@l2r{a,b} = (scalar(FILE), scalar(FILE));
or
@l2r{a,b} = (FILE.'', FILE.'');
to ensure that the FILE is evaluated in scalar context.
Which part is forcing the list context? The fact that the FILE is inside
parenthesis () or the @l2r{...} part?
Is every element of a list () contextualized as a list itself?
Thanks for the responses, Jeff and Jenda!
- Bryan
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Re: populating a hash slice from a filehandle
Bryan R Harris wrote:
From: Bryan R Harris [EMAIL PROTECTED]
Given an open filehandle, why don't these two things do the same thing?
**
@l2r{a,b} = (FILE, FILE);
$c = FILE;
**
$l2r{a} = FILE;
$l2r{b} = FILE;
$c = FILE;
**
The first seems to be slurping the whole file into $l2r{b} and leaving $c
undefined... The second does what I want. Doesn't seem to make sense.
Context. The FILEHANDLE returns a single line in scalar context and
a list of all lines in a list context. And there is no such thing as
a two-item-list context.
So in the first case the assignment to @l2r{a,b} provides a list
context so the very first FILE reads all the lines left in FILE,
the second and third return an empty list. The first two lists are
concatenated together and the first two items of the resulting list
are assigned to $l2r{a} and $l2r{b}. And the rest of the list is
forgotten. You'd have to do something like
@l2r{a,b} = (scalar(FILE), scalar(FILE));
or
@l2r{a,b} = (FILE.'', FILE.'');
to ensure that the FILE is evaluated in scalar context.
Which part is forcing the list context? The fact that the FILE is inside
parenthesis () or the @l2r{...} part?
The left hand side of the assignment determines context so the @l2r{...}
part.
Is every element of a list () contextualized as a list itself?
A list is composed of zero or more scalars, if that's what you mean?
John
--
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can special-order certain sorts of tools at low cost and
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Re: populating a hash slice from a filehandle
Bryan R Harris wrote:
From: Bryan R Harris [EMAIL PROTECTED]
Given an open filehandle, why don't these two things do the same thing?
**
@l2r{a,b} = (FILE, FILE);
$c = FILE;
**
$l2r{a} = FILE;
$l2r{b} = FILE;
$c = FILE;
**
The first seems to be slurping the whole file into $l2r{b} and leaving $c
undefined... The second does what I want. Doesn't seem to make sense.
Context. The FILEHANDLE returns a single line in scalar context and
a list of all lines in a list context. And there is no such thing as
a two-item-list context.
So in the first case the assignment to @l2r{a,b} provides a list
context so the very first FILE reads all the lines left in FILE,
the second and third return an empty list. The first two lists are
concatenated together and the first two items of the resulting list
are assigned to $l2r{a} and $l2r{b}. And the rest of the list is
forgotten. You'd have to do something like
@l2r{a,b} = (scalar(FILE), scalar(FILE));
or
@l2r{a,b} = (FILE.'', FILE.'');
to ensure that the FILE is evaluated in scalar context.
Which part is forcing the list context? The fact that the FILE is inside
parenthesis () or the @l2r{...} part?
The left hand side of the assignment determines context so the @l2r{...}
part.
That strikes me as odd... When perl goes to populate @l2r{a,b}, it
seems to me that it would go through this process:
- I have a slice here, so I'll loop over the slice elements
- The first is a, so I'll pull a scalar off the list and assign it to
$l2r{a}
- The second is b, so I'll pull another scalar off the list and assign it
to $l2r{b}
- Remaining scalars in the list are discarded
Why would $l2r{a} here be considered list context?
What am I missing?
- B
ps. I'm often surprised at how little I seem to know even after 8 years of
perl scripting and monitoring this list...
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Re: populating a hash slice from a filehandle
Bryan R Harris wrote:
John W. Krahn wrote:
Bryan R Harris wrote:
Jenda Krynicky wrote:
Context. The FILEHANDLE returns a single line in scalar context and
a list of all lines in a list context. And there is no such thing as
a two-item-list context.
So in the first case the assignment to @l2r{a,b} provides a list
context so the very first FILE reads all the lines left in FILE,
the second and third return an empty list. The first two lists are
concatenated together and the first two items of the resulting list
are assigned to $l2r{a} and $l2r{b}. And the rest of the list is
forgotten. You'd have to do something like
@l2r{a,b} = (scalar(FILE), scalar(FILE));
or
@l2r{a,b} = (FILE.'', FILE.'');
to ensure that the FILE is evaluated in scalar context.
Which part is forcing the list context? The fact that the FILE is inside
parenthesis () or the @l2r{...} part?
The left hand side of the assignment determines context so the @l2r{...}
part.
That strikes me as odd... When perl goes to populate @l2r{a,b}, it
seems to me that it would go through this process:
- I have a slice here, so I'll loop over the slice elements
- The first is a, so I'll pull a scalar off the list and assign it to
$l2r{a}
- The second is b, so I'll pull another scalar off the list and assign it
to $l2r{b}
- Remaining scalars in the list are discarded
Correct, except for the loop part.
Why would $l2r{a} here be considered list context?
It isn't, unless it's written as ( $l2r{a} ), then it's a list with
one element.
What am I missing?
Hard to say? :-)
John
--
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can special-order certain sorts of tools at low cost and
in short order.-- Larry Wall
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populating a hash slice from a filehandle
Given an open filehandle, why don't these two things do the same thing?
**
@l2r{a,b} = (FILE, FILE);
$c = FILE;
**
$l2r{a} = FILE;
$l2r{b} = FILE;
$c = FILE;
**
The first seems to be slurping the whole file into $l2r{b} and leaving $c
undefined... The second does what I want. Doesn't seem to make sense.
TIA.
- Bryan
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Re: populating a hash slice from a filehandle
On Thu, Jun 19, 2008 at 5:50 AM, Bryan R Harris
[EMAIL PROTECTED] wrote:
Given an open filehandle, why don't these two things do the same thing?
**
@l2r{a,b} = (FILE, FILE);
$c = FILE;
because @l2r{...} is a list, right?
so the statement above is in a list context.
so the second FILE will slurp all file content.
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Re: populating a hash slice from a filehandle
From: Bryan R Harris [EMAIL PROTECTED]
Given an open filehandle, why don't these two things do the same thing?
**
@l2r{a,b} = (FILE, FILE);
$c = FILE;
**
$l2r{a} = FILE;
$l2r{b} = FILE;
$c = FILE;
**
The first seems to be slurping the whole file into $l2r{b} and leaving $c
undefined... The second does what I want. Doesn't seem to make sense.
Context. The FILEHANDLE returns a single line in scalar context and
a list of all lines in a list context. And there is no such thing as
a two-item-list context.
So in the first case the assignment to @l2r{a,b} provides a list
context so the very first FILE reads all the lines left in FILE,
the second and third return an empty list. The first two lists are
concatenated together and the first two items of the resulting list
are assigned to $l2r{a} and $l2r{b}. And the rest of the list is
forgotten. You'd have to do something like
@l2r{a,b} = (scalar(FILE), scalar(FILE));
or
@l2r{a,b} = (FILE.'', FILE.'');
to ensure that the FILE is evaluated in scalar context.
HTH, Jenda
= [EMAIL PROTECTED] === http://Jenda.Krynicky.cz =
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to get drunk and croon as much as they like.
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Re: Hash slice
On 9 Nov 2007 at 20:04, Jenda Krynicky wrote:
From: Beginner [EMAIL PROTECTED]
On 9 Nov 2007 at 16:35, Jenda Krynicky wrote:
From: Beginner [EMAIL PROTECTED]
#!/bin/perl
use strict;
use warnings;
use Data::Dumper;
my @keys = qw(fe fi fo thumb);
my @valone = 1..4;
my @valtwo = 10..14;
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED],@valtwo];
[...] creates an array reference. You want
@[EMAIL PROTECTED] = ( [EMAIL PROTECTED],[EMAIL PROTECTED]);
Probably. Though it will only set the values for 'fe' and 'fi',
because I only specify two values.
What I was attempting was to have each key to be assigned the
coresponding items from the array. So it might look like something
like:
$VAR1 = {
'fe' = [
1,
10
],
'fi' = [
2,
11,
],
'fo' = [
3,
13,
],
'thumb' = [
4,
14,
]
};
@[EMAIL PROTECTED] = map [$valone[$_], $valtwo[$_]] (0..$#valone);
The map produces a list of arrayrefs, each referenced array contains
one item from @valone and one from @valtwo. The 0th element of the
result, the 0th elements of @valone and @valtwo, etc.
Now there you go again with a beautifully simple use of map that does
exactly what I want (do you need a comma before (0..$#valone)?).
I notice now that I wrote my example wrong. Each array was meant to
have 4 items in (sorry Rob). So it should have been
my @valone = 1..4;
my @valtwo = 11..14;
There seems to be a bit of confusion over what I was trying to
achieve. I typed out the output from Dumper I was expecting because I
was/am not entirely sure what terms to use. The above operation looks
like a hash slice to me albeit with another operator (map) involved.
Thanx for the help.
Dp.
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Re: Hash slice
From: Beginner [EMAIL PROTECTED]
On 9 Nov 2007 at 20:04, Jenda Krynicky wrote:
@[EMAIL PROTECTED] = map [$valone[$_], $valtwo[$_]] (0..$#valone);
The map produces a list of arrayrefs, each referenced array contains
one item from @valone and one from @valtwo. The 0th element of the
result, the 0th elements of @valone and @valtwo, etc.
Now there you go again with a beautifully simple use of map that does
exactly what I want (do you need a comma before (0..$#valone)?).
Yes, I was missing a comma. I usually use the
map {} (list)
format that doesn't require the comma ... this is what happens if I
change the habbits and don't test my code :-)
I notice now that I wrote my example wrong. Each array was meant to
have 4 items in (sorry Rob). So it should have been
my @valone = 1..4;
my @valtwo = 11..14;
There seems to be a bit of confusion over what I was trying to
achieve. I typed out the output from Dumper I was expecting because I
was/am not entirely sure what terms to use. The above operation looks
like a hash slice to me albeit with another operator (map) involved.
Yep, the Dumper output was exactly the right thing to use :-)
Jenda
= [EMAIL PROTECTED] === http://Jenda.Krynicky.cz =
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to get drunk and croon as much as they like.
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Hash slice
Hi all,
Is it possible to make a hash slice like so
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED];
My efforts suggest not:
#!/bin/perl
use strict;
use warnings;
use Data::Dumper;
my @keys = qw(fe fi fo thumb);
my @vals = 1..4;
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED];
print Dumper(\%hash);
$VAR1 = {
'ARRAY(0x226d54)' = [
1,
2,
3,
4
]
};
Am I missing something or isn't this possible?
TIA.
Dp.
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Re: Hash slice
Beginner wrote:
Hi all,
Is it possible to make a hash slice like so
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED];
My efforts suggest not:
#!/bin/perl
use strict;
use warnings;
use Data::Dumper;
my @keys = qw(fe fi fo thumb);
my @vals = 1..4;
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED];
print Dumper(\%hash);
$VAR1 = {
'ARRAY(0x226d54)' = [
1,
2,
3,
4
]
};
Am I missing something or isn't this possible?
Hey Dermot
It's certainly possible, but I'm not sure why you've taken a reference
to your key and value arrays. [EMAIL PROTECTED] is a single scalar value, as is
[EMAIL PROTECTED], so you're creating a single hash element. Perl has had to
stringify the reference to @keys as Perl hash keys must be strings.
The hash value is a reference you your @vals array which contains the
values 1 through 4 as Dumper shows.
That you need is simply
@[EMAIL PROTECTED] = @vals;
which you will find has the effect you expect.
HTH,
Rob
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Re: Hash slice
From: Beginner [EMAIL PROTECTED] #!/bin/perl use strict; use warnings; use Data::Dumper; my @keys = qw(fe fi fo thumb); my @valone = 1..4; my @valtwo = 10..14; my %hash; @[EMAIL PROTECTED] = [EMAIL PROTECTED],@valtwo]; [...] creates an array reference. You want @[EMAIL PROTECTED] = ( [EMAIL PROTECTED],[EMAIL PROTECTED]); Probably. Though it will only set the values for 'fe' and 'fi', because I only specify two values. Jenda = [EMAIL PROTECTED] === http://Jenda.Krynicky.cz = When it comes to wine, women and song, wizards are allowed to get drunk and croon as much as they like. -- Terry Pratchett in Sourcery -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Hash slice
On 9 Nov 2007 at 14:59, Rob Dixon wrote:
Beginner wrote:
Hi all,
Is it possible to make a hash slice like so
Hey Dermot
Hi Rob,
It's certainly possible, but I'm not sure why you've taken a reference
to your key and value arrays. [EMAIL PROTECTED] is a single scalar value, as
is
[EMAIL PROTECTED], so you're creating a single hash element. Perl has had to
stringify the reference to @keys as Perl hash keys must be strings.
The hash value is a reference you your @vals array which contains the
values 1 through 4 as Dumper shows.
I see, scalar used where list expected.
How about this:
#!/bin/perl
use strict;
use warnings;
use Data::Dumper;
my @keys = qw(fe fi fo thumb);
my @valone = 1..4;
my @valtwo = 10..14;
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED],@valtwo];
print Dumper(\%hash);
$VAR1 = {
'fo' = undef,
'fi' = undef,
'fe' = [
1,
2,
3,
4,
10,
11,
12,
13,
14
],
'thumb' = undef
};
I can't see why you can't create a slice hows values are arrays or
why all the values are assigned to the first key.
Not a biggy. I can work around it but I'm interested to know.
Thanx,
Dp.
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Re: Hash slice
On 9 Nov 2007 at 16:35, Jenda Krynicky wrote:
From: Beginner [EMAIL PROTECTED]
#!/bin/perl
use strict;
use warnings;
use Data::Dumper;
my @keys = qw(fe fi fo thumb);
my @valone = 1..4;
my @valtwo = 10..14;
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED],@valtwo];
[...] creates an array reference. You want
@[EMAIL PROTECTED] = ( [EMAIL PROTECTED],[EMAIL PROTECTED]);
Probably. Though it will only set the values for 'fe' and 'fi',
because I only specify two values.
What I was attempting was to have each key to be assigned the
coresponding items from the array. So it might look like something
like:
$VAR1 = {
'fe' = [
1,
10
],
'fi' = [
2,
11,
],
'fo' = [
3,
13,
],
'thumb' = [
4,
14,
]
};
(yes I typed that by hand) and not
$VAR1 = {
'fo' = undef,
'fi' = [
10,
11,
12,
13,
14
],
'fe' = [
1,
2,
3,
4
],
'thumb' = undef
};
But I see why it's hasn't worked.
Dp,
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Re: Hash slice
On Nov 9, 4:09 pm, [EMAIL PROTECTED] (Beginner) wrote: On 9 Nov 2007 at 16:35, Jenda Krynicky wrote: What I was attempting was to have each key to be assigned the coresponding items from the array. So it might look like something like: Right, so your question has nothing to do with hash slices. Learning to break down your problem is and important skill in programming. What you seek is a way swap (transpose) the axes of a list of array references. ie. you want the list ([1,2],[3,4],[5,6]) to become ([1,3,5],[2,4,6]) There's no built-in way to do that but I'm prepared to bet there are several modules on CPAN that provide this (I just don't seem to be able to find them right now). I have a strange feeling there is a way to do this in Perl6. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Hash slice
On Nov 9, 3:35 pm, [EMAIL PROTECTED] (Jenda Krynicky) wrote: @[EMAIL PROTECTED] = ( [EMAIL PROTECTED],[EMAIL PROTECTED]); Note that _can_ also be written @[EMAIL PROTECTED] = \( @valone, @valtwo); But IMNSHO it this syntax should _only_ be used in code that is intended as part of a response to the question what's the most confusing feature of Perl5 syntax? -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/
Re: Hash slice
Beginner wrote:
On 9 Nov 2007 at 14:59, Rob Dixon wrote:
Beginner wrote:
Is it possible to make a hash slice like so
It's certainly possible, but I'm not sure why you've taken a reference
to your key and value arrays. [EMAIL PROTECTED] is a single scalar value, as is
[EMAIL PROTECTED], so you're creating a single hash element. Perl has had to
stringify the reference to @keys as Perl hash keys must be strings.
The hash value is a reference you your @vals array which contains the
values 1 through 4 as Dumper shows.
I see, scalar used where list expected.
Nope, not this time! But you used a hash slice of one element:
@[EMAIL PROTECTED] = [EMAIL PROTECTED];
is the same as
my ($k, $v) = ([EMAIL PROTECTED], [EMAIL PROTECTED]);
@hash{$k} = $v;
which is ok, but does the same thing as
$hash{$k} = $v;
How about this:
#!/bin/perl
use strict;
use warnings;
use Data::Dumper;
my @keys = qw(fe fi fo thumb);
my @valone = 1..4;
my @valtwo = 10..14;
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED],@valtwo];
print Dumper(\%hash);
$VAR1 = {
'fo' = undef,
'fi' = undef,
'fe' = [
1,
2,
3,
4,
10,
11,
12,
13,
14
],
'thumb' = undef
};
Alright, that's equivalent to
my @vals = (@valone, @valtwo);
my $v = [EMAIL PROTECTED];
@[EMAIL PROTECTED] = $v;
and because there are four keys in the slice and only one value, the
trailing three get set to undef.
I can't see why you can't create a slice hows values are arrays or
why all the values are assigned to the first key.
Not a biggy. I can work around it but I'm interested to know.
Each hash element must have exactly one key and one value, but you've
got an array of four keys and a list of nine values (1, 2, 3, 4, 10,
11, 12, 13, 14). Four keys can't be paired with nine values! What do
you actually want here?
@[EMAIL PROTECTED] = (@valone, @valtwo);
pairs the first four values to the keys in the array and throws away
the remaining five.
Don't forget that a reference is a single scalar value, so both
[EMAIL PROTECTED]
and
[EMAIL PROTECTED], @valtwo]
are one scalar.
HTH,
Rob
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Re: Hash slice
On Nov 9, 11:09 am, [EMAIL PROTECTED] (Beginner) wrote:
What I was attempting was to have each key to be assigned the
coresponding items from the array.
Why didn't you just say that in the first place, rather than letting
everyone guess as to what you wanted?
So it might look like something
like:
$VAR1 = {
'fe' = [
1,
10
],
'fi' = [
2,
11,
],
'fo' = [
3,
13,
],
'thumb' = [
4,
14,
]
};
#!/usr/bin/perl
use strict;
use warnings;
use Data::Dumper;
use List::MoreUtils 'each_array';
my @keys = qw/fe fi fo fum/;
my @one = (1..4);
my @two = (11..14);
my $it = each_array @one, @two;
my %h;
@[EMAIL PROTECTED] = map { [ $it-() ] } [EMAIL PROTECTED];
print Dumper(\%h);
__END__
Paul Lalli
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Re: Hash slice
From: Beginner [EMAIL PROTECTED]
On 9 Nov 2007 at 16:35, Jenda Krynicky wrote:
From: Beginner [EMAIL PROTECTED]
#!/bin/perl
use strict;
use warnings;
use Data::Dumper;
my @keys = qw(fe fi fo thumb);
my @valone = 1..4;
my @valtwo = 10..14;
my %hash;
@[EMAIL PROTECTED] = [EMAIL PROTECTED],@valtwo];
[...] creates an array reference. You want
@[EMAIL PROTECTED] = ( [EMAIL PROTECTED],[EMAIL PROTECTED]);
Probably. Though it will only set the values for 'fe' and 'fi',
because I only specify two values.
What I was attempting was to have each key to be assigned the
coresponding items from the array. So it might look like something
like:
$VAR1 = {
'fe' = [
1,
10
],
'fi' = [
2,
11,
],
'fo' = [
3,
13,
],
'thumb' = [
4,
14,
]
};
@[EMAIL PROTECTED] = map [$valone[$_], $valtwo[$_]] (0..$#valone);
The map produces a list of arrayrefs, each referenced array contains
one item from @valone and one from @valtwo. The 0th element of the
result, the 0th elements of @valone and @valtwo, etc.
Jenda
= [EMAIL PROTECTED] === http://Jenda.Krynicky.cz =
When it comes to wine, women and song, wizards are allowed
to get drunk and croon as much as they like.
-- Terry Pratchett in Sourcery
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Re: hash slice/exists vs. grep benchmark weirdness...
JupiterHost.Net ha scritto:
In benchmarking some code I've come across something I did not expect:
slice:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
my %n;@[EMAIL PROTECTED] = @k;
print hi if exists $n{1};
print hi if exists $n{3};
print hi if exists $n{5};
print hi if exists $n{7};
print hi if exists $n{9};
print hi if exists $n{11};
grep:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
print hi if grep /^1$/, @k;
print hi if grep /^3$/, @k;
print hi if grep /^5$/, @k;
print hi if grep /^7$/, @k;
print hi if grep /^9$/, @k;
print hi if grep /^11$/, @k;
Benchmark: timing 500 iterations of grep, slice...
grep: 3.65945 wallclock secs ( 2.33 usr + 0.04 sys = 2.37 CPU) @
2109704.64/s (n=500)
slice: 2.37966 wallclock secs ( 2.52 usr + -0.01 sys = 2.51 CPU) @
1992031.87/s (n=500)
Rate slice grep
slice 1992032/s-- -6%
grep 2109705/s6%--
I would've thought the slice and then use exists would have been
faster then greping the entire array each time and using regexes when
you check it. but its consistently faster by an average 6-10%
Any ideas why that might be?
Just for curiosity, I ran a slightly different version of your benchmarks:
code source=slice-bench.pl
use strict;
use warnings;
use diagnostics;
my @k=qw(1 2 3 4 5 6);
my %n;
@[EMAIL PROTECTED] = @k;
my $times = 500;
my $dummy;
while($times--) {
$dummy=1 if exists $n{1};
$dummy=1 if exists $n{3};
$dummy=1 if exists $n{5};
$dummy=1 if exists $n{7};
$dummy=1 if exists $n{9};
$dummy=1 if exists $n{11};
}
/code
code source=grep-bench.pl
use strict;
use warnings;
use diagnostics;
my @k=qw(1 2 3 4 5 6);
my $times = 500;
my $dummy;
while($times--) {
$dummy=1 if grep /^1$/, @k;
$dummy=1 if grep /^3$/, @k;
$dummy=1 if grep /^5$/, @k;
$dummy=1 if grep /^7$/, @k;
$dummy=1 if grep /^9$/, @k;
$dummy=1 if grep /^11$/, @k;
}
/code
The results were:
log shelltype=bash
[EMAIL PROTECTED] time perl slice-bench.pl
9.15user 0.01system 0:09.89elapsed 92%CPU (0avgtext+0avgdata 0maxresident)k
0inputs+0outputs (0major+1015minor)pagefaults 0swaps
[EMAIL PROTECTED] time perl grep-bench.pl
67.52user 0.01system 1:12.97elapsed 92%CPU (0avgtext+0avgdata 0maxresident)k
0inputs+0outputs (0major+1015minor)pagefaults 0swaps
/log
The slice version took about 10 seconds, the grep one took more than 1
minute.
Marcello
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Re: hash slice/exists vs. grep benchmark weirdness...
The slice version took about 10 seconds, the grep one took more than 1 minute. Marcello Thanks Marcello! Your example, John's. and renards have been very helpful to help see the point someone else was trying to make about watching what you include in your benchmark. I appreciate your time :) -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: hash slice/exists vs. grep benchmark weirdness...
renard wrote: BE AWARE THAT THE BENCHMARK PROVIDES INCORRECT RESULTS. Thnaks, I was aware of that, but still wanted a general idea :) The tested code is within an anonymous subroutine while this does executes, the results differ dramitcally from the results when the tested code is enclosed within single quotation marks.!!! I do not know why the great difference ... perhaps the experts might be able to explain the difference. This is not the first time that I observe this divergence. The test results below show when the tested code is enclosed within single quotation marks, the 'exists' code has a 40% advantage. over the 'grep_e' code but has a 1100% advantage over the grep_e code when enclosed within an anonymous subroutine. The code enclosed inside the anonympous subroutine takes substantially longer to execute. I tend to believe that the results for the single quotation mark is more accurate. I'll use that info in the future, thanks! So buyer beware ... [snip] Benchmark: timing 100 iterations of exists, grep_e, grep_r, hash... exists: 2.99903 wallclock secs ( 2.97 usr + 0.01 sys = 2.98 CPU) @ 335232.99/s (n=100) grep_e: 4.14286 wallclock secs ( 4.16 usr + 0.00 sys = 4.16 CPU) @ 240615.98/s (n=100) grep_r: 3.93865 wallclock secs ( 3.92 usr + 0.00 sys = 3.92 CPU) @ 254971.95/s (n=100) hash: 3.01261 wallclock secs ( 3.00 usr + 0.00 sys = 3.00 CPU) @ 33.33/s (n=100) Rate grep_e grep_r hash exists grep_e 240616/s ---6% -28% -28% grep_r 254972/s 6% -- -24% -24% hash 33/s39%31% ---1% exists 335233/s39%31% 1% -- Benchmark: timing 100 iterations of exists, grep_e, grep_r, hash... exists: 7.63617 wallclock secs ( 7.56 usr + 0.00 sys = 7.56 CPU) @ 13.66/s (n=100) grep_e: 14.9662 wallclock secs (14.72 usr + 0.00 sys = 14.72 CPU) @ 67939.40/s (n=100) grep_r: 95.0288 wallclock secs (93.33 usr + 0.03 sys = 93.36 CPU) @ 10711.34/s (n=100) hash: 15.6237 wallclock secs (15.50 usr + 0.02 sys = 15.52 CPU) @ 64449.60/s (n=100) Rate grep_r hash grep_e exists grep_r 10711/s -- -83% -84% -92% hash64450/s 502% ---5% -51% grep_e 67939/s 534% 5% -- -49% exists 132223/s 1134% 105%95% -- Excellent! I beleive from all the different tests I will be going with exists *and* more craefull how/what I benchmark :) Thanks for your time! -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: hash slice/exists vs. grep benchmark weirdness...
You have to stop spending so much time playing with all this bogus
benchmarking :)
It not bogus :) Its an example to find the best method for a project.
If you prefer I'll ask the question I was trying to answer with the
benchmark:
Assuming you have an array of 0-15 elements is it quicker/more efficient to:
1) create a hash slice and use exists when checking for specific ones
or
2) grep a regex out of the array when checking for specific ones
slice:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
my %n;@[EMAIL PROTECTED] = @k;
print hi if exists $n{1};
print hi if exists $n{3};
print hi if exists $n{5};
print hi if exists $n{7};
print hi if exists $n{9};
print hi if exists $n{11};
grep:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
print hi if grep /^1$/, @k;
print hi if grep /^3$/, @k;
print hi if grep /^5$/, @k;
print hi if grep /^7$/, @k;
print hi if grep /^9$/, @k;
print hi if grep /^11$/, @k;
Benchmark: timing 500 iterations of grep, slice...
grep: 3.65945 wallclock secs ( 2.33 usr + 0.04 sys = 2.37 CPU)
@ 2109704.64/s (n=500)
slice: 2.37966 wallclock secs ( 2.52 usr + -0.01 sys = 2.51 CPU)
@ 1992031.87/s (n=500)
Rate slice grep
slice 1992032/s-- -6%
grep 2109705/s6%--
I would've thought the slice and then use exists would have been
faster then greping the entire array each time and using regexes when
you check it. but its consistently faster by an average 6-10%
Any ideas why that might be?
Well, first -- you're creating the hash inside the benchmark loop,
which is not particularly light-weight.
Good point,
although in the real world app I'll have an array already so I want to
see if its quicker to:
- make the hash via a slice and use exists
or
- just grep the whole array each time with a regex
If I use
my %hash = qw(1..2);
and exists
vs.
my @array = qw(1..1);
IE - no slice
then exists is a bit faster 2% or so, the problem is I start with an
array I have no control over.
So, I'd have to do a slice to get the hash (or loop through it or
otherwise do something to get the hash)
Second: You're using pathologically small lists. Try running it with
10,000 elements instead of ten.
that makes them about even, with slice being 1% faster at times, however
the size of the array in the real world app are similar to the original
benchmark
Third: Premature optimization is a terrible thing.
Premature? Could you elaborate?
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Re: hash slice/exists vs. grep benchmark weirdness...
- Original Message -
From: JupiterHost.Net [EMAIL PROTECTED]
Date: Tuesday, January 25, 2005 11:01 am
Subject: Re: hash slice/exists vs. grep benchmark weirdness...
You have to stop spending so much time playing with all this bogus
benchmarking :)
It not bogus :) Its an example to find the best method for a project.
If you prefer I'll ask the question I was trying to answer with
the
benchmark:
Assuming you have an array of 0-15 elements is it quicker/more
efficient to:
1) create a hash slice and use exists when checking for specific
onesor
2) grep a regex out of the array when checking for specific ones
slice:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
my %n;@[EMAIL PROTECTED] = @k;
print hi if exists $n{1};
print hi if exists $n{3};
print hi if exists $n{5};
print hi if exists $n{7};
print hi if exists $n{9};
print hi if exists $n{11};
grep:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
print hi if grep /^1$/, @k;
print hi if grep /^3$/, @k;
print hi if grep /^5$/, @k;
print hi if grep /^7$/, @k;
print hi if grep /^9$/, @k;
print hi if grep /^11$/, @k;
Benchmark: timing 500 iterations of grep, slice...
grep: 3.65945 wallclock secs ( 2.33 usr + 0.04 sys =
2.37 CPU)
@ 2109704.64/s (n=500)
slice: 2.37966 wallclock secs ( 2.52 usr + -0.01 sys =
2.51 CPU)
@ 1992031.87/s (n=500)
Rate slice grep
slice 1992032/s-- -6%
grep 2109705/s6%--
I would've thought the slice and then use exists would have
been
faster then greping the entire array each time and using
regexes when
you check it. but its consistently faster by an average 6-10%
Any ideas why that might be?
Well, first -- you're creating the hash inside the benchmark loop,
which is not particularly light-weight.
Good point,
although in the real world app I'll have an array already so I
want to
see if its quicker to:
- make the hash via a slice and use exists
or
- just grep the whole array each time with a regex
If I use
my %hash = qw(1..2);
and exists
vs.
my @array = qw(1..1);
IE - no slice
then exists is a bit faster 2% or so, the problem is I start with
an
array I have no control over.
So, I'd have to do a slice to get the hash (or loop through it or
otherwise do something to get the hash)
Second: You're using pathologically small lists. Try running
it with
10,000 elements instead of ten.
that makes them about even, with slice being 1% faster at times,
however
the size of the array in the real world app are similar to the
original
benchmark
Third: Premature optimization is a terrible thing.
Premature? Could you elaborate?
Just as an FYI, you don't need exists in your code at all. It is just a waste
of time in your example. Should be beter writen as:
print hi if $n{11};
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Re: hash slice/exists vs. grep benchmark weirdness...
Just as an FYI, you don't need exists in your code at all. It is just a
waste of time in your example. Should be beter writen as:
print hi if $n{11};
What if it's value is 0, '', or undef? It would exist but your test
would miss it :)
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RE: hash slice/exists vs. grep benchmark weirdness...
[EMAIL PROTECTED] suggested:
Just as an FYI, you don't need exists in your code at all.
It is just a waste of time in your example. Should be beter writen as:
print hi if $n{11};
Bad idea, if you consider this:
$n{'oops'} = 0;
print hi if $n{'oops'};
print ho if exists $n{'oops'};
HTH,
Thomas
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RE: hash slice/exists vs. grep benchmark weirdness...
You have to stop spending so much time playing with all this bogus benchmarking :) It not bogus :) Its an example to find the best method for a project. If you prefer I'll ask the question I was trying to answer with the benchmark: Assuming you have an array of 0-15 elements is it quicker/more efficient to: 1) create a hash slice and use exists when checking for specific ones or 2) grep a regex out of the array when checking for specific ones If you're going to be looking for something in an array over and over again, perhaps you shouldn't be using an array as your data structure. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: hash slice/exists vs. grep benchmark weirdness...
- Original Message -
From: JupiterHost.Net [EMAIL PROTECTED]
Date: Tuesday, January 25, 2005 11:37 am
Subject: Re: hash slice/exists vs. grep benchmark weirdness...
Just as an FYI, you don't need exists in your code at all. It
is just a waste of time in your example. Should be beter writen as:
print hi if $n{11};
What if it's value is 0, '', or undef? It would exist but your
test
would miss it :)
In that case defined would be a beter function to use.
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Re: hash slice/exists vs. grep benchmark weirdness...
You have to stop spending so much time playing with all this bogus benchmarking :) It not bogus :) Its an example to find the best method for a project. And I'll reiterate more clearly. Benchmarking is *not* the tool to find the best method. The BEST method for a project is that method which is self evident and maintainable. ONLY AFTER you determine that you have missed your performance envelope, do you go back through your code using rakes and shovels and implements of destruction to find the slow parts. Early Optimization Is The Root of All Evil. If you prefer I'll ask the question I was trying to answer with the benchmark: Assuming you have an array of 0-15 elements is it quicker/more efficient to: 1) create a hash slice and use exists when checking for specific ones or 2) grep a regex out of the array when checking for specific ones Both. Each one will 'win' in terms of total execution time for some set of conditions, none of which have been expressed in the problem. (For example, will the keys always be small integers? How complex will they be? This will have a tremendous impact on the performance of the match operator.) And, in fact, you yourself may not know a priori what the conditions will be - which is why I say (again because you really need to hear this): Premature Optimization Is The Root of All Evil. Write your code so you can maintain it. The microseconds will take care of themselves. The programmers' time is orders of magnitude more costly than the computers', and your value is (hopefully!) increasing, while the cost of computational oomph is (thankfully) decreasing. Now, as to this specific question -- use whichever idiom you are more comfortable with. Furthermore, that is a tiny implementation detail -- in the Big Picture, it won't matter a gnats eyebrow. Finally, there were serious errors in your methodology in your original benchmark. It turns out the printing dominated the total execution time, such that the variance between the two different selection methods were pushed down into the noise. When I benchmarked using lists of 10 elements searching for five random members in the list, the exists-in-hash version was about 10 times faster for *JUST THE SELECTION OPERATION* than the grep-the-list version. -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- Lawrence Statton - [EMAIL PROTECTED] s/aba/c/g Computer software consists of only two components: ones and zeros, in roughly equal proportions. All that is required is to sort them into the correct order. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: hash slice/exists vs. grep benchmark weirdness...
print hi if $n{11};
What if it's value is 0, '', or undef? It would exist but your
test
would miss it :)
In that case defined would be a beter function to use.
ok, but I'm still not interested in the value of the key I;'m interested
inthe key, so why not exists since its made for hashes :)
I simply asked what if to illustrate why if $n{11} is not better
than exists $n{11} in this case.
So the original question remains:
Which is faster:
1) slice the array into a has and do exists $hsh{key}
or
2) just grep a regex on the array
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Re: hash slice/exists vs. grep benchmark weirdness...
Bakken, Luke wrote: You have to stop spending so much time playing with all this bogus benchmarking :) It not bogus :) Its an example to find the best method for a project. If you prefer I'll ask the question I was trying to answer with the benchmark: Assuming you have an array of 0-15 elements is it quicker/more efficient to: 1) create a hash slice and use exists when checking for specific ones or 2) grep a regex out of the array when checking for specific ones If you're going to be looking for something in an array over and over again, perhaps you shouldn't be using an array as your data structure. Hence, my original question :) I have an array not matter what. So should I slice it into a hash and do exists or skip the slice and grep a regex on the array each time :) -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: hash slice/exists vs. grep benchmark weirdness...
Thanks for your input :) Finally, there were serious errors in your methodology in your Serious? I thought in the Big Picture, it won't matter a gnats eyebrow. :) original benchmark. It turns out the printing dominated the total That is why I made both identical except for the difference I'm concerned with. execution time, such that the variance between the two different selection methods were pushed down into the noise. When I benchmarked using lists of 10 elements searching for five random members in the list, the exists-in-hash version was about 10 times faster for *JUST THE SELECTION OPERATION* than the grep-the-list version. Excellent, what code did you benchmark? -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: hash slice/exists vs. grep benchmark weirdness...
Thanks for your input :) No problem -- I really do enjoy this. However, I do have some actual WORK that I need to do today :( Finally, there were serious errors in your methodology in your Serious? I thought in the Big Picture, it won't matter a gnats eyebrow. :) And I stand by that ... Such is the fallacy of the benchmark: hah! Technique Foo was ten times faster than Technique Bar! But, if one takes 5 microseconds and the other takes 50, they're both going to be blown out of the water by the first page-fault that crosses their path, or waiting for an I/O operation to complete, or someone moving the mouse causing the screen to refresh, or a million other things that will make that 45 microsecond difference pale in comparison. original benchmark. It turns out the printing dominated the total That is why I made both identical except for the difference I'm concerned with. Still poor methodology -- A thought-experiment. I am testing two different implementations -- now $DEITY knows that one of them will take 10 milliseconds to execute, and the other an entire second. (I, however, not being omniscient don't know that -- hence my benchmarking test). I put them into a test-crib with a second function that takes precisely 24-hours to complete. So, one of them takes 86400.01 seconds, and the other takes 86401.00 seconds. Now, assume that the 24-hour extraneous function really might vary randomly plus or minus five minutes. Suddenly that variance in the spurious function will totally swamp the variance I am trying to measure between the two implementations. In your benchmark, you did precisely that -- by putting a function that takes a relatively very-very-long and randomly-variable amount of time (printing) inside the function you are trying to test, you are basically destroying the exact number you are trying to compute, and have merely computed the deviation of the print operator. -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- Lawrence Statton - [EMAIL PROTECTED] s/aba/c/g Computer software consists of only two components: ones and zeros, in roughly equal proportions. All that is required is to sort them into the correct order. -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
Re: hash slice/exists vs. grep benchmark weirdness...
JupiterHost.Net wrote:
In benchmarking some code I've come across something I did not expect:
slice:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
my %n;@[EMAIL PROTECTED] = @k;
print hi if exists $n{1};
print hi if exists $n{3};
print hi if exists $n{5};
print hi if exists $n{7};
print hi if exists $n{9};
print hi if exists $n{11};
grep:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
print hi if grep /^1$/, @k;
print hi if grep /^3$/, @k;
print hi if grep /^5$/, @k;
print hi if grep /^7$/, @k;
print hi if grep /^9$/, @k;
print hi if grep /^11$/, @k;
Benchmark: timing 500 iterations of grep, slice...
grep: 3.65945 wallclock secs ( 2.33 usr + 0.04 sys = 2.37 CPU) @
2109704.64/s (n=500)
slice: 2.37966 wallclock secs ( 2.52 usr + -0.01 sys = 2.51 CPU) @
1992031.87/s (n=500)
Rate slice grep
slice 1992032/s-- -6%
grep 2109705/s6%--
I would've thought the slice and then use exists would have been
faster then greping the entire array each time and using regexes when
you check it. but its consistently faster by an average 6-10%
Any ideas why that might be?
Well, let's see:
#!/usr/bin/perl
use warnings;
use strict;
use Benchmark 'cmpthese';
my @k = qw( 1 2 3 4 5 6 );
cmpthese( -10, {
exists = sub {
my $count = 0;
my %hash;
@hash{ @k } = ();
for my $num ( 1, 3, 5, 7, 9, 11 ) {
$count++ if exists $hash{ $num };
}
return $count;
},
hash= sub {
my $count = 0;
my %hash = map { $_ = 1 } @k;
for my $num ( 1, 3, 5, 7, 9, 11 ) {
$count++ if $hash{ $num };
}
return $count;
},
grep_r = sub {
my $count = 0;
my @array = @k;
for my $num ( 1, 3, 5, 7, 9, 11 ) {
$count++ if grep /^$num$/, @array;
}
return $count;
},
grep_e = sub {
my $count = 0;
my @array = @k;
for my $num ( 1, 3, 5, 7, 9, 11 ) {
$count++ if grep $_ == $num, @array;
}
return $count;
},
} );
Produces the result (on my computer):
Rate grep_r grep_e hash exists
grep_r 4838/s -- -81% -83% -90%
grep_e 25923/s 436% -- -10% -49%
hash 28848/s 496%11% -- -43%
exists 50524/s 944%95%75% --
So in my test exists() is definitely faster.
John
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Re: hash slice/exists vs. grep benchmark weirdness...
#!/usr/bin/perl
use warnings;
use strict;
use Benchmark 'cmpthese';
my @k = qw( 1 2 3 4 5 6 );
cmpthese( -10, {
exists = sub {
my $count = 0;
my %hash;
@hash{ @k } = ();
for my $num ( 1, 3, 5, 7, 9, 11 ) {
$count++ if exists $hash{ $num };
}
return $count;
},
hash= sub {
my $count = 0;
my %hash = map { $_ = 1 } @k;
for my $num ( 1, 3, 5, 7, 9, 11 ) {
$count++ if $hash{ $num };
}
return $count;
},
grep_r = sub {
my $count = 0;
my @array = @k;
for my $num ( 1, 3, 5, 7, 9, 11 ) {
$count++ if grep /^$num$/, @array;
}
return $count;
},
grep_e = sub {
my $count = 0;
my @array = @k;
for my $num ( 1, 3, 5, 7, 9, 11 ) {
$count++ if grep $_ == $num, @array;
}
return $count;
},
} );
Produces the result (on my computer):
Rate grep_r grep_e hash exists
grep_r 4838/s -- -81% -83% -90%
grep_e 25923/s 436% -- -10% -49%
hash 28848/s 496%11% -- -43%
exists 50524/s 944%95%75% --
So in my test exists() is definitely faster.
Awesome, definately soemthing for me to look into, thanks a bunch John!
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Re: hash slice/exists vs. grep benchmark weirdness...
Lawrence Statton wrote: [snip] Gotcha, thx So then can you suggest a method of benchmarking these 2 methods that would be more accurate? I believe John's solution was excellent at illustrating th deficiencies in the way I was doing it and supplying a solution and answering my question all at once ;p I appreciate your input, I'll be more careful of the way I benchmark next time :) -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED] http://learn.perl.org/ http://learn.perl.org/first-response
hash slice/exists vs. grep benchmark weirdness...
In benchmarking some code I've come across something I did not expect:
slice:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
my %n;@[EMAIL PROTECTED] = @k;
print hi if exists $n{1};
print hi if exists $n{3};
print hi if exists $n{5};
print hi if exists $n{7};
print hi if exists $n{9};
print hi if exists $n{11};
grep:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
print hi if grep /^1$/, @k;
print hi if grep /^3$/, @k;
print hi if grep /^5$/, @k;
print hi if grep /^7$/, @k;
print hi if grep /^9$/, @k;
print hi if grep /^11$/, @k;
Benchmark: timing 500 iterations of grep, slice...
grep: 3.65945 wallclock secs ( 2.33 usr + 0.04 sys = 2.37 CPU)
@ 2109704.64/s (n=500)
slice: 2.37966 wallclock secs ( 2.52 usr + -0.01 sys = 2.51 CPU)
@ 1992031.87/s (n=500)
Rate slice grep
slice 1992032/s-- -6%
grep 2109705/s6%--
I would've thought the slice and then use exists would have been
faster then greping the entire array each time and using regexes when
you check it. but its consistently faster by an average 6-10%
Any ideas why that might be?
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Re: hash slice/exists vs. grep benchmark weirdness...
In benchmarking some code I've come across something I did not expect:
You have to stop spending so much time playing with all this bogus
benchmarking :)
slice:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
my %n;@[EMAIL PROTECTED] = @k;
print hi if exists $n{1};
print hi if exists $n{3};
print hi if exists $n{5};
print hi if exists $n{7};
print hi if exists $n{9};
print hi if exists $n{11};
grep:
use strict;
use warnings;
my @k=qw(1 2 3 4 5 6);
print hi if grep /^1$/, @k;
print hi if grep /^3$/, @k;
print hi if grep /^5$/, @k;
print hi if grep /^7$/, @k;
print hi if grep /^9$/, @k;
print hi if grep /^11$/, @k;
Benchmark: timing 500 iterations of grep, slice...
grep: 3.65945 wallclock secs ( 2.33 usr + 0.04 sys = 2.37 CPU)
@ 2109704.64/s (n=500)
slice: 2.37966 wallclock secs ( 2.52 usr + -0.01 sys = 2.51 CPU)
@ 1992031.87/s (n=500)
Rate slice grep
slice 1992032/s-- -6%
grep 2109705/s6%--
I would've thought the slice and then use exists would have been
faster then greping the entire array each time and using regexes when
you check it. but its consistently faster by an average 6-10%
Any ideas why that might be?
Well, first -- you're creating the hash inside the benchmark loop,
which is not particularly light-weight.
Second: You're using pathologically small lists. Try running it with
10,000 elements instead of ten.
Third: Premature optimization is a terrible thing.
-- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- -- --
Lawrence Statton - [EMAIL PROTECTED] s/aba/c/g
Computer software consists of only two components: ones and
zeros, in roughly equal proportions. All that is required is to
sort them into the correct order.
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create hash slice from hash
I am attempting to create a hash slice from a hash. The hash is: %hash =(test1 = test10, test2 = test12 , test3 = test13) I want the slice to include only the keys test1 and test3. How can I accomplish this? -- To unsubscribe, e-mail: [EMAIL PROTECTED] For additional commands, e-mail: [EMAIL PROTECTED]
Re: create hash slice from hash
--- Lisa Neclos [EMAIL PROTECTED] wrote:
I am attempting to create a hash slice from a hash. The hash is:
%hash =(test1 = test10,
test2 = test12 ,
test3 = test13)
I want the slice to include only the keys test1 and test3. How can I
accomplish this?
Lisa,
my @results = @hash{ qw/ test1 test3 / };
Cheers,
Curtis Ovid Poe
=
Senior Programmer
Onsite! Technology (http://www.onsitetech.com/)
Ovid on http://www.perlmonks.org/
__
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Make a great connection at Yahoo! Personals.
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Re: create hash slice from hash
LN I am attempting to create a hash slice from a hash. The hash is:
LN %hash =(test1 = test10,
LNtest2 = test12 ,
LN test3 = test13)
LN I want the slice to include only the keys test1 and test3. How can I
LN accomplish this?
you could use map:
my %new_hash = map { $_ = $hash{$_} } qw(test1 test2);
map is a really useful function...
hth,
daniel
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