insert anonymous hash into anonymous hash for counting in a loop
I'm trying to count starts and stops of some services i keep track of in logs.
I'm not going to past here entire code, but my way of doing hash is this:
I'm passing those starts and stops into anonymous hash .
First I'm creating anonymous hash filled with keys and values (in my case $knot is a key an zeros are values). Next im replaqcing values with another hash.
My code looks like this:
foreach $knot (@knots)
chomp $knot;
$variable = $variable."$knot;0;";
$Services = split(/;/,$variable);
my $data =
Starts=>'0',
Stops=>'0',
;
foreach my $key (keys %$Services)
$Services->$key = $data;
print Dumper $Services;
Printing shows:
$VAR1 = {
' knot1' =>
'Stops' => '0',
'Starts' => '0'
,
' knot2' => $VAR1->' knot1',
' knot3' => $VAR1->' knot1',
' knot4' => $VAR1->' knot1',
' knot5' => $VAR1->' knot1',
and so on. Is there a better way of doing this? My way if i'm correct is badly written because changing knot1 starts/stops changes every other knot values.
perl hash
add a comment |
I'm trying to count starts and stops of some services i keep track of in logs.
I'm not going to past here entire code, but my way of doing hash is this:
I'm passing those starts and stops into anonymous hash .
First I'm creating anonymous hash filled with keys and values (in my case $knot is a key an zeros are values). Next im replaqcing values with another hash.
My code looks like this:
foreach $knot (@knots)
chomp $knot;
$variable = $variable."$knot;0;";
$Services = split(/;/,$variable);
my $data =
Starts=>'0',
Stops=>'0',
;
foreach my $key (keys %$Services)
$Services->$key = $data;
print Dumper $Services;
Printing shows:
$VAR1 = {
' knot1' =>
'Stops' => '0',
'Starts' => '0'
,
' knot2' => $VAR1->' knot1',
' knot3' => $VAR1->' knot1',
' knot4' => $VAR1->' knot1',
' knot5' => $VAR1->' knot1',
and so on. Is there a better way of doing this? My way if i'm correct is badly written because changing knot1 starts/stops changes every other knot values.
perl hash
3
You are only ever assigning the same reference$data. The easiest approach to make them different between each other is to movemy $data = ...;into the loop.
– Corion
Nov 14 '18 at 13:01
add a comment |
I'm trying to count starts and stops of some services i keep track of in logs.
I'm not going to past here entire code, but my way of doing hash is this:
I'm passing those starts and stops into anonymous hash .
First I'm creating anonymous hash filled with keys and values (in my case $knot is a key an zeros are values). Next im replaqcing values with another hash.
My code looks like this:
foreach $knot (@knots)
chomp $knot;
$variable = $variable."$knot;0;";
$Services = split(/;/,$variable);
my $data =
Starts=>'0',
Stops=>'0',
;
foreach my $key (keys %$Services)
$Services->$key = $data;
print Dumper $Services;
Printing shows:
$VAR1 = {
' knot1' =>
'Stops' => '0',
'Starts' => '0'
,
' knot2' => $VAR1->' knot1',
' knot3' => $VAR1->' knot1',
' knot4' => $VAR1->' knot1',
' knot5' => $VAR1->' knot1',
and so on. Is there a better way of doing this? My way if i'm correct is badly written because changing knot1 starts/stops changes every other knot values.
perl hash
I'm trying to count starts and stops of some services i keep track of in logs.
I'm not going to past here entire code, but my way of doing hash is this:
I'm passing those starts and stops into anonymous hash .
First I'm creating anonymous hash filled with keys and values (in my case $knot is a key an zeros are values). Next im replaqcing values with another hash.
My code looks like this:
foreach $knot (@knots)
chomp $knot;
$variable = $variable."$knot;0;";
$Services = split(/;/,$variable);
my $data =
Starts=>'0',
Stops=>'0',
;
foreach my $key (keys %$Services)
$Services->$key = $data;
print Dumper $Services;
Printing shows:
$VAR1 = {
' knot1' =>
'Stops' => '0',
'Starts' => '0'
,
' knot2' => $VAR1->' knot1',
' knot3' => $VAR1->' knot1',
' knot4' => $VAR1->' knot1',
' knot5' => $VAR1->' knot1',
and so on. Is there a better way of doing this? My way if i'm correct is badly written because changing knot1 starts/stops changes every other knot values.
perl hash
perl hash
edited Nov 28 '18 at 16:28
Boris Däppen
658517
658517
asked Nov 14 '18 at 12:28
Marcin KMarcin K
31
31
3
You are only ever assigning the same reference$data. The easiest approach to make them different between each other is to movemy $data = ...;into the loop.
– Corion
Nov 14 '18 at 13:01
add a comment |
3
You are only ever assigning the same reference$data. The easiest approach to make them different between each other is to movemy $data = ...;into the loop.
– Corion
Nov 14 '18 at 13:01
3
3
You are only ever assigning the same reference
$data. The easiest approach to make them different between each other is to move my $data = ...; into the loop.– Corion
Nov 14 '18 at 13:01
You are only ever assigning the same reference
$data. The easiest approach to make them different between each other is to move my $data = ...; into the loop.– Corion
Nov 14 '18 at 13:01
add a comment |
1 Answer
1
active
oldest
votes
Counting is very simple in Perl, thanks to Autovivification. You can just create anonymous data structures on the fly, like so:
use Data::Dumper;
my %hash = ();
$hashapplegreen++;
$hashapplered ++;
$hashpearyellow++;
$hashapplegreen++;
$hashapplered ++;
$hashapplegreen++;
print Dumper(%hash);
This will produce the desired structure for counting:
$VAR1 =
'apple' =>
'green' => 3,
'red' => 2
,
'pear' =>
'yellow' => 1
;
This also works in loops using variables (here using a reference to a hash):
my $hash_ref = ;
for my $fruit (qw( apple pear apple peach apple pear ))
$hash_ref->$fruit++;
print Dumper($hash_ref);
resulting in:
$VAR1 =
'peach' => 1,
'pear' => 2,
'apple' => 3
;
1
my %hash = ();is equivalent tomy %hash; %hash = ();, which is redundant becausemyalready creates the hash empty.
– melpomene
Nov 16 '18 at 14:06
Thank you, this was helpful , i have made loop with added your code and it works like a charm :)
– Marcin K
Nov 22 '18 at 11:06
@MarcinK nice! Glad I could be of help. Could you please accept the answer as your valid answer by clicking the checkmark at the left side of this answer. This will help you, me and all the other stack users. thx
– Boris Däppen
Nov 26 '18 at 12:17
@Boris Däppen Done :)
– Marcin K
Nov 27 '18 at 14:28
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
Counting is very simple in Perl, thanks to Autovivification. You can just create anonymous data structures on the fly, like so:
use Data::Dumper;
my %hash = ();
$hashapplegreen++;
$hashapplered ++;
$hashpearyellow++;
$hashapplegreen++;
$hashapplered ++;
$hashapplegreen++;
print Dumper(%hash);
This will produce the desired structure for counting:
$VAR1 =
'apple' =>
'green' => 3,
'red' => 2
,
'pear' =>
'yellow' => 1
;
This also works in loops using variables (here using a reference to a hash):
my $hash_ref = ;
for my $fruit (qw( apple pear apple peach apple pear ))
$hash_ref->$fruit++;
print Dumper($hash_ref);
resulting in:
$VAR1 =
'peach' => 1,
'pear' => 2,
'apple' => 3
;
1
my %hash = ();is equivalent tomy %hash; %hash = ();, which is redundant becausemyalready creates the hash empty.
– melpomene
Nov 16 '18 at 14:06
Thank you, this was helpful , i have made loop with added your code and it works like a charm :)
– Marcin K
Nov 22 '18 at 11:06
@MarcinK nice! Glad I could be of help. Could you please accept the answer as your valid answer by clicking the checkmark at the left side of this answer. This will help you, me and all the other stack users. thx
– Boris Däppen
Nov 26 '18 at 12:17
@Boris Däppen Done :)
– Marcin K
Nov 27 '18 at 14:28
add a comment |
Counting is very simple in Perl, thanks to Autovivification. You can just create anonymous data structures on the fly, like so:
use Data::Dumper;
my %hash = ();
$hashapplegreen++;
$hashapplered ++;
$hashpearyellow++;
$hashapplegreen++;
$hashapplered ++;
$hashapplegreen++;
print Dumper(%hash);
This will produce the desired structure for counting:
$VAR1 =
'apple' =>
'green' => 3,
'red' => 2
,
'pear' =>
'yellow' => 1
;
This also works in loops using variables (here using a reference to a hash):
my $hash_ref = ;
for my $fruit (qw( apple pear apple peach apple pear ))
$hash_ref->$fruit++;
print Dumper($hash_ref);
resulting in:
$VAR1 =
'peach' => 1,
'pear' => 2,
'apple' => 3
;
1
my %hash = ();is equivalent tomy %hash; %hash = ();, which is redundant becausemyalready creates the hash empty.
– melpomene
Nov 16 '18 at 14:06
Thank you, this was helpful , i have made loop with added your code and it works like a charm :)
– Marcin K
Nov 22 '18 at 11:06
@MarcinK nice! Glad I could be of help. Could you please accept the answer as your valid answer by clicking the checkmark at the left side of this answer. This will help you, me and all the other stack users. thx
– Boris Däppen
Nov 26 '18 at 12:17
@Boris Däppen Done :)
– Marcin K
Nov 27 '18 at 14:28
add a comment |
Counting is very simple in Perl, thanks to Autovivification. You can just create anonymous data structures on the fly, like so:
use Data::Dumper;
my %hash = ();
$hashapplegreen++;
$hashapplered ++;
$hashpearyellow++;
$hashapplegreen++;
$hashapplered ++;
$hashapplegreen++;
print Dumper(%hash);
This will produce the desired structure for counting:
$VAR1 =
'apple' =>
'green' => 3,
'red' => 2
,
'pear' =>
'yellow' => 1
;
This also works in loops using variables (here using a reference to a hash):
my $hash_ref = ;
for my $fruit (qw( apple pear apple peach apple pear ))
$hash_ref->$fruit++;
print Dumper($hash_ref);
resulting in:
$VAR1 =
'peach' => 1,
'pear' => 2,
'apple' => 3
;
Counting is very simple in Perl, thanks to Autovivification. You can just create anonymous data structures on the fly, like so:
use Data::Dumper;
my %hash = ();
$hashapplegreen++;
$hashapplered ++;
$hashpearyellow++;
$hashapplegreen++;
$hashapplered ++;
$hashapplegreen++;
print Dumper(%hash);
This will produce the desired structure for counting:
$VAR1 =
'apple' =>
'green' => 3,
'red' => 2
,
'pear' =>
'yellow' => 1
;
This also works in loops using variables (here using a reference to a hash):
my $hash_ref = ;
for my $fruit (qw( apple pear apple peach apple pear ))
$hash_ref->$fruit++;
print Dumper($hash_ref);
resulting in:
$VAR1 =
'peach' => 1,
'pear' => 2,
'apple' => 3
;
edited Nov 16 '18 at 13:07
answered Nov 16 '18 at 7:52
Boris DäppenBoris Däppen
658517
658517
1
my %hash = ();is equivalent tomy %hash; %hash = ();, which is redundant becausemyalready creates the hash empty.
– melpomene
Nov 16 '18 at 14:06
Thank you, this was helpful , i have made loop with added your code and it works like a charm :)
– Marcin K
Nov 22 '18 at 11:06
@MarcinK nice! Glad I could be of help. Could you please accept the answer as your valid answer by clicking the checkmark at the left side of this answer. This will help you, me and all the other stack users. thx
– Boris Däppen
Nov 26 '18 at 12:17
@Boris Däppen Done :)
– Marcin K
Nov 27 '18 at 14:28
add a comment |
1
my %hash = ();is equivalent tomy %hash; %hash = ();, which is redundant becausemyalready creates the hash empty.
– melpomene
Nov 16 '18 at 14:06
Thank you, this was helpful , i have made loop with added your code and it works like a charm :)
– Marcin K
Nov 22 '18 at 11:06
@MarcinK nice! Glad I could be of help. Could you please accept the answer as your valid answer by clicking the checkmark at the left side of this answer. This will help you, me and all the other stack users. thx
– Boris Däppen
Nov 26 '18 at 12:17
@Boris Däppen Done :)
– Marcin K
Nov 27 '18 at 14:28
1
1
my %hash = (); is equivalent to my %hash; %hash = ();, which is redundant because my already creates the hash empty.– melpomene
Nov 16 '18 at 14:06
my %hash = (); is equivalent to my %hash; %hash = ();, which is redundant because my already creates the hash empty.– melpomene
Nov 16 '18 at 14:06
Thank you, this was helpful , i have made loop with added your code and it works like a charm :)
– Marcin K
Nov 22 '18 at 11:06
Thank you, this was helpful , i have made loop with added your code and it works like a charm :)
– Marcin K
Nov 22 '18 at 11:06
@MarcinK nice! Glad I could be of help. Could you please accept the answer as your valid answer by clicking the checkmark at the left side of this answer. This will help you, me and all the other stack users. thx
– Boris Däppen
Nov 26 '18 at 12:17
@MarcinK nice! Glad I could be of help. Could you please accept the answer as your valid answer by clicking the checkmark at the left side of this answer. This will help you, me and all the other stack users. thx
– Boris Däppen
Nov 26 '18 at 12:17
@Boris Däppen Done :)
– Marcin K
Nov 27 '18 at 14:28
@Boris Däppen Done :)
– Marcin K
Nov 27 '18 at 14:28
add a comment |
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3
You are only ever assigning the same reference
$data. The easiest approach to make them different between each other is to movemy $data = ...;into the loop.– Corion
Nov 14 '18 at 13:01