Is there a way to use Perl's range operator .. to use letters AND numbers?
For example, have:
for my $i ('X'..'9') {
print "$i ";
}
output X Y Z 1 2 3 4 5 6 7 8 9
Not unless for my $i ('X' .. 'Z', 1 .. 9) counts. "Z" will never increment to 1.
Related
I am new to Perl. An exercise, where I am to create a numeric ruler from which, I size columns for data at 20 characters-width, is proving a little difficult to complete. So far, I have,
printf “%10d” x 5, (1..6);
#ruler = (1..10) x 7;
Print #ruler, “\n”;
It should look something like,
1 2 3 4
1234567890123456789012345678901234567890
What I get for the top row of numbers is an error, ‘Redundant argument in printf at <script.pl> line #; the bottom row produces numbers from 1 to 10, as it ought with the range operator, but I would like it to produce 1 to 9 with a zero on the end. I did think to start the range from 0, but I haven’t figured out how to remove the first index and only the first index.
I would be grateful for your guidance with both issues.
The warning is due to the fact that you pass 6 numbers to printf, but the format only requires 5.
To me,
1 2 3 4
1234567890123456789012345678901234567890
reads as
11, 12, 13, ..., 19, 10, 21, 22, 23, ...
Why does it start with 11? Why is 10 between 19 and 21?
The following makes more sense:
1 2 3 4
01234567890123456789012345678901234567890 0-based
and
1 2 3 4
1234567890123456789012345678901234567890 1-based
I'm not going to give the solution outright.
If you want the numbers 1 to 9 and 0, that would be 1..9, 0.
%10d will add padding on the left. %-10d will add padding on the right.
Nothing says you can't prefix the output with something that doesn't repeat, like a zero or a space.
Provided desired output starts count from 11 instead 1 -- it doesn't look right.
Perhaps OP intended to start count from 1 until some $max value with placing a digit representing tens above main counter.
Please study following code sample for compliance with your requirements.
use strict;
use warnings;
use feature 'say';
my $max = shift || 45;
rule($max);
sub rule {
my $max = shift;
my($a,$b);
$a .= ' ' x 9 . $_ for 1..$max/10;
$b .= $_ % 10 for 1..$max;
say $a . "\n" . $b;
}
Output
1 2 3 4
123456789012345678901234567890123456789012345
Original OP's code requires slight modification to achieve desired output
use strict;
use warnings;
use feature 'say';
my $max = shift || 45;
printf "%10d" x int($max/10) . "\n", (1..$max/10);
print $_ % 10 for 1..$max;
print "\n";
I have 2 arrays, one is the root array, which contains huge mount of elements. Another is the tested array, all of which elements is a subset of root array. I want to construct a new array, with size equal to root arrays, and its element value at a specific position represents the count of that element in the tested array.
Below codes works well when 2 array size is small. But the practical problem I meet is that, the root arrays has about 15000 elements, and there are about 14000 tested arrays.
I wonder to have better algorithm. Do you guys have some suggestions?
my #root=qw(1 2 3 4 5 6 7 8 10);
my #aa=qw(1 1 2 3);
my #count;
foreach my $eleroot(#root){
my $mathnum=0;
my ($i) = grep { $root[$_] ~~ $eleroot } 0 .. $#root;
foreach my $eleaa(#aa){
if ($eleroot==$eleaa){
$mathnum++;
}
}
$count[$i]=$mathnum;
}
print #count;
A better algorithm would be to use a hash to keep counts. For your sample arrays, it would look like this. (And will run considerably faster than your solution).
#!/usr/bin/perl
use strict;
use warnings;
my #root=qw(1 2 3 4 5 6 7 8 10);
my #aa=qw(1 1 2 3);
print join("\t", #root), "\n";
my %seen;
for my $data (#aa) {
$seen{$data}++;
}
print join("\t", map {$_ // '0'} #seen{#root}), "\n";
Output is:
1 2 3 4 5 6 7 8 10
2 1 1 0 0 0 0 0 0
#seen{#root} is a hash slice keyed by the #root array. If no item was found for any of the #root elements, the map supplies a zero.
The map {$_ // '0'} portion is saying pass to join a count if $_ has a count otherwise pass a zero.
I have code to print out the first 9 squared numbers:
#!/usr/local/bin/perl
for($i=1;$i<10 ;$i++ )
{
printf $i^2 . "\n";
}
but for some reason this just outputs 30167451011. How do I properly square a number?
To square you have to use $i**2
#!/usr/local/bin/perl
for ( my $i = 1; $i < 10; $i++ ) {
print $i**2 . "\n";
}
This will output:
1
4
9
16
25
36
49
64
81
To explain what happened in the original code, you need to know 3 things: first, ^ is the XOR operator in perl, because it is the XOR operator in C.
1 ^ 2 = 3
2 ^ 2 = 0
3 ^ 2 = 1
4 ^ 2 = 6
...
Second, the ^ operator has lower precedence than the string concatenation operator . so $i^2 . "\n" is equivalent to $i ^ (2 . "\n")
Third, perl converts between strings and numbers as necessary. The . operator requires strings on both sides, so the 2 is converted to "2" and concatenated with the "\n" to become the string "2\n".
Then the ^ operator requires numbers on both sides, so the string "2\n" is converted to a number - by taking the leading number-looking portion and throwing away the rest. So the result of $i ^ 2 . "\n" is ultimately the same as $i ^ 2. Your "\n" didn't have any effect at all, so all the results are printed with nothing between them. 3, 0, 1, 6, ... became 3016...
^ is the Bitwise Xor operator
To square a number, you want the Exponentiation operator **
for my $i ( 1 .. 9 ) {
print $i**2, "\n";
}
Outputs:
1
4
9
16
25
36
49
64
81
foreach my $i (1..9){
say $i**2;
}
It can be achieve by doing this way also:
for (1..9) {
print $_*$_,"\n"
}
Output:
1
4
9
16
25
36
49
64
81
You can use the code below to get the squares.
print $_**2,"\n" for 1..10;
Here is my code sample:
for $i(1..100){
if ($i%15==0){$s="Divisible by 15"}
elsif($i%5==0){$s="Divisible by 5"}
else {$i%3==0 ? $s="Divisible by 3" : $s=$i};
print $s."\n"}
This displays partial correct results, if a number if divisible by 3, it displays the "number" and not "Divisible by 3".
Example of output:
1
2
3
4
Divisible by 5
6
7
8
9
Divisible by 5
PS: I have to write this code in the minimum no. of characters possible. (The reason why the code is so sluggish)
The fourth line is parsed as
(((($i % 3) == 0) ? ($s = 'Divisible by 3') : $s) = $i)
meaning that even when $i is divisible by 3, the assignment looks like
($s = 'Divisible by 3') = $i
Some fixes:
$i%3==0 ? ($s="Divisible by 3") : ($s=$i)
$s = ($i%3==0 ? "Divisible by 3" : $i)
Pro-tip: B::Deparse is very helpful for figuring this stuff out. I ran the command
perl -MO=Deparse,-p -e '$i%3==0 ? $s="Divisible by 3" : $s=$i'
to see exactly what the problem was.
The ternary ?: has higher precedence than assignment. Use parens to sort that out:
for my $i (1 .. 100) {
my $s = "Divisible by ";
$i%15==0 ? ($s .= 15)
: $i%5==0 ? ($s .= 5)
: $i%3==0 ? ($s .= 3)
: ($s =$i);
print "$s\n";
}
Output:
1
2
Divisible by 3
4
Divisible by 5
Divisible by 3
7
8
Divisible by 3
Divisible by 5
If you want a version with less characters,
for $i(1..100){print(($i%3&&$i%5?$i:"Divisible by ".($i%5?3:$i%15?5:15))."\n");}
Parenthesis after the print function means call the print function, not group the expression. Caused by the fact that parenthesis are optional for that function.
I've a problem in making a Perl program for matching the numbers in two documents. Let's say there are documents A and B.
So I want to have the numbers which is present in document A and absent in the document B.
Example 1:
DocA: 1 2 3 5 6 8 9 10 11 12 13
DocB: 1 2 3 6 7 8 9 10 11
output:
5 12 13
EDITED:
#a=qw( 1 2 3 5 6 8 9 10 11 12 13);
#b=qw( 1 2 3 4 5 6 7 8 9 10 11);
#new=();
#new1=();
for($i=0;$i<=$#a;$i++)
{
for($j=0;$j<=$#b;$j++)
{
if($a[$i] ne $b[$j])
{
push(#new,$b[$j]);
}
}
}
You could use the CPAN module Array::Utils. The following will do what you need:
use Array::Utils qw(:all);
my #a = qw( 1 2 3 5 6 8 9 10 11 12 13);
my #b = qw( 1 2 3 4 5 6 7 8 9 10 11);
my #diff = array_minus(#a, #b);`
By the way, the reason your program does not work is because you made a logic error. You are adding a value to #new EVERY TIME the value does not match. So, in the first iteration of the loop you compare the a value to the b value. Even though the value is equal to the first element of #b, it is not equal to the other ten elements of #b and hence all of these elements are added to #new. I have rewritten your loop. Note the more Perlish loops instead of the C-loop you used in your code.
my #a = qw( 1 2 3 5 6 8 9 10 11 12 13);
my #b = qw( 1 2 3 4 6 7 8 9 10 11);
my #new = ();
for my $a_value (#a) {
my $b_not_in_a = 1;
INNER: for my $b_value (#b)
{ if($a_value == $b_value) {
$b_not_in_a = 0;
last INNER; }
}
if ($b_not_in_a)
{
push(#new,$a_value);
}
}
Consider Algorithm::Diff ?
You could use a hash. Read DocA and initialize a hash using the read numbers as keys:
open(INPUT, "DocA");
while (<INPUT>)
{
chomp;
$myhash{$_} = 1;
}
Then read DocB and foreach number, check if already defined in your hash:
open(INPUT, "DocB");
while (<INPUT>)
{
chomp;
if (not defined $myhash{$_})
{
print "$_\n";
}
}
This code asumes you have a number per line. If your files are formatted differently, you will need to adapt it.
This code will work even if your numbers aren't ordered.
The lists in your example are sorted. I am assuming that they are, and that you're not allowed to use modules since it is homework. Also, since it is homework, I won't give the answer, but some hints in the right direction.
If you would do this by hand, and you are only allowed to look at the front of each row, how would you do it? If the head of A is a number smaller than B, what does that mean? If it is equal, what does that mean? If it is larger, what does that mean?
Now you know how you can handle one situation, from that you can create some kind of step to reduce the problem. Now define when you need to stop, and what the possible leftovers of the lists are at that point, and how you can get your answer from the values you collected in the step, and the remainder after you stop.
Some examples of extreme cases:
#a = qw();
#b = qw(1 2 3);
#a = qw (1 2 3);
#b = qw (4 5 6);
#a = qw(1 3 5);
#b = qw(2 4 6)
Good luck!