I have the following line of CoffeeScript:
names = (mail.folder for mail in #data when mail.service_name is service.name).unique()
This line is too long, so it won't pass linting by CoffeeLint.
I'm trying to break it, but I always get indentation errors by CoffeeLint.
What is the proper way to indent this?
This is the most readable version of that without getting overly lengthy:
names =
(for mail in #data when mail.service_name is service.name
mail.folder).unique()
You can't split list comprehensions over multiple lines, but a normal for loop can also return a value, so using one of those solves the problem. If you're willing to grant an extra line, there's no need for the awkward parentheses around the loop:
names =
for mail in #data when mail.service_name is service.name
mail.folder
names = names.unique()
Finally, the indentation of the for line is up to you; I find my first version more readable but this is also valid:
names =
(for mail in #data when mail.service_name is service.name
mail.folder).unique()
This compiles fine:
names = (mail.folder for mail in #data \
when mail.service_name is service.name).unique()
You can also invert the for ... when and the expression:
names = (for mail in #data when mail.service_name is service.name
mail.folder).unique()
Apparently splitting list comprehension over multiple lines is not allowed:
https://stackoverflow.com/a/8553292
Related
I am trying to select the below value from database:
Reporting that one of #its many problems had been the recent# extended
sales slump in women's apparel, the seven-store retailer said it would
start a three-month liquidation sale in all of its stores.~(A) its
many problems had been the recent~(B) its many problems has been the
recently~(C) its many problems is the recently~(D) their many problems
is the recent~(E) their many problems had been the recent~
i am selecting this value in variable $ques and then selecting a text as below:
$ques=~s/^(.*?)\#(.*?)\#(.*?)$/$2/;
Now, while replacing the ~ character in the string by
$3=~s/~/\n/g; ---->line 171
and running the script, I am getting one error as:
Modification of a read-only value attempted at main.pl line 171
I want to replace all the ~ character with '\n' and print the final value. Please suggest how to do it.
*I have researched this on net, but got confused that how to handle these read only variables.
You've already got a good explanation of the problem from José Castro. But there's another solution if you're using a recent-ish version of Perl (Update: having checked more carefully, I find that means 5.14+). The /r argument to the substitution operator will copy your string, make the substitution on the copy and then return that altered value.
So you could write:
my $new_value = $3 =~ s/~/\n/rg;
It sounds like what you really want in this case is split rather than regular expression capture groups:
my #parts = split(/#/, $ques);
$parts[2] =~ s/~/\n/g;
It makes the intent of your code clearer since you are, in fact, splitting on # symbols.
Just like you say, the special variables $1, $2, etc., are read-only, and that means that you can't perform that substitution on them.
Performing the substitution on $ques will do what you need:
$ques =~ s/~/\n/g;
print $ques;
Do note that in the earlier substitution that you're performing on $ques you're getting rid of all the ~ characters.
I am still learning perl and have all most got a program written. My question, as simple as it may be, is if I want to hardcode a string to a field would the below do that? Thank you :).
$out[45]="VUS";
In the other lines I use the below to define the values that are passed into the `$[out], but the one in question is hardcoded and the others come from a split.
my #vals = split/\t/; # this splits the line at tabs
my #mutations=split/,/,$vals[9]; # splits on comma to create an array of mutations
my ($gene,$transcript,$exon,$coding,$aa);
for (#mutations)
{
($gene,$transcript,$exon,$coding,$aa) = split/\:/; # this takes col AB and splits it at colons
grep {$transcript eq $_} keys %nms or next;
}
my #out=($.,#colsleft,$_,#colsright);
$out[2]=$gene;
$out[3]=$nms{$transcript};
$out[4]=$transcript;
$out[15]=$coding;
$out[17]=$aa;
Your line of code: $out[45]="VUS"; is correct in that it is defining that 46th element of the array #out to the string, "VUS". I am trying to understand from your code, however why you would want to do that? Usually, it is better practice to not hardcode if at all possible. You want to make it your goal to make your program as dynamic as possible.
I am trying to extract a part of a string and put it into a new variable. The string I am looking at is:
maker-scaffold_26653|ref0016423-snap-gene-0.1
(inside a $gene_name variable)
and the thing I want to match is:
scaffold_26653|ref0016423
I'm using the following piece of code:
my $gene_name;
my $scaffold_name;
if ($gene_name =~ m/scaffold_[0-9]+\|ref[0-9]+/) {
$scaffold_name = $1;
print "$scaffold_name\n";
}
I'm getting the following error when trying to execute:
Use of uninitialized value $scaffold_name in concatenation (.) or string
I know that the pattern is right, because if I use $' instead of $1 I get
-snap-gene-0.1
I'm at a bit of a loss: why will $1 not work here?
If you want to use a value from the matching you have to make () arround the character in regex
To expand on Jens' answer, () in a regex signifies an anonymous capture group. The content matched in a capture group is stored in $1-9+ from left to right, so for example,
/(..):(..):(..)/
on an HH:MM:SS time string will store hours, minutes, and seconds in $1, $2, $3 respectively. Naturally this begins to become unwieldy and is not self-documenting, so you can assign the results to a list instead:
my ($hours, $mins, $secs) = $time =~ m/(..):(..):(..)/;
So your example could bypass the use of $ variables by doing direct assignment:
my ($scaffold_name) = $gene_name =~ m/(scaffold_[0-9]+[|]ref[0-9]+)/;
# $scaffold_name now contains 'scaffold_26653|ref0016423'
You can even get rid of the ugly =~ binding by using for as a topicalizer:
my $scaffold_name;
for ($gene_name) {
($scaffold_name) = m/(scaffold_\d+[|]ref\d+)/;
print $scaffold_name;
}
If things start to get more complex, I prefer to use named capture groups (introduced in Perl v5.10.0):
$gene_name =~ m{
(?<scaffold_name> # ?<name> creates a named capture group
scaffold_\d+? # 'scaffold' and its trailing digits
[|] # Literal pipe symbol
ref\d+ # 'ref' and its trailing digits
)
}xms; # The x flag lets us write more readable regexes
print $+{scaffold_name}, "\n";
The results of named capture groups are stored in the magic hash %+. Access is done just like any other hash lookup, with the capture groups as the keys. %+ is locally scoped in the same way the $ are, so it can be used as a drop-in replacement for them in most situations.
It's overkill for this particular example, but as regexes start to get larger and more complicated, this saves you the trouble of either having to scroll all the way back up and count anonymous capture groups from left to right to find which of those darn $ variables is holding the capture you wanted, or scan across a long list assignment to find where to add a new variable to hold a capture that got inserted in the middle.
My personal rule of thumb is to assign the results of anonymous captured to descriptively named lexically scoped variables for 3 or less captures, then switch to using named captures, comments, and indentation in regexes when more are necessary.
How do I transform my data to an array with Perl?
Here is my data:
my $data =
"203.174.38.128203.174.38.129203.174.38.1" .
"30203.174.38.131203.174.38.132203.174.38" .
".133203.174.38.134173.174.38.135203.174." .
"38.136203.174.38.137203.174.38.142";
And I want to transform it to be array like this
my #array= (
"203.174.38.128",
"203.174.38.129",
"203.174.38.130",
"203.174.38.131",
"203.174.38.132",
"203.174.38.133",
"203.174.38.134",
"173.174.38.135",
"203.174.38.136",
"203.174.38.137",
"203.174.38.142"
);
Anyone know how to do that with Perl?
If the first part of IP logged is always 203, it's kinda easy:
my #arr = split /(?<=\d)(?=203\.)/, $data;
In the example given it's not, but the first part is always 3-digit, and the second part is always 174, so it's enough to do...
my #arr = split /(?<=\d)(?=\d{3}\.174\.)/, $data;
... to get the correct result.
But please understand that it's close to impossible to give a more generic (and bulletproof) solution here - when these 'marker' parts are... too dynamic. For example, take this string...
11.11.11.22222.11.11.11
The question is, where to split it? Should it be 11.11.11.22; 222.11.11.11? Or 11.11.11.222; 22.11.11.11? Both are quite valid IPs, if you ask me. And it could get even worse, with trying to split '2222' part (can be '2; 222', '22; 22' and even '222; 2').
You can, for example, make a rule: "split each sequence of > 3 digits followed by a dot sign so that the second part of this split would always start from 3 digits":
my #arr = split /(?<=\d)(?=\d{3}\.)/, $data;
... but this will obviously fail to work properly in the ambiguous cases mentioned earlier IF there are IPs with two- or even one-digit first octet in your datastring.
If you write a regex that will match any valid value for one of the numbers in the quartet then you can just search for them all and recombine them in sets of four. This
/2[0-5][0-5]|1\d\d|[1-9]\d|\d/
matches 200-255 or 100-199 or 10-99 or 0-9, and a program to use it is shown below.
There is no way to know which option to take if there is more than one way to split the string, and this solution assigns the longest value to the first of the two ip addresses. For instance, 1.1.1.1234.1.1.1 will split as 1.1.1.123 and 4.1.1.1
use strict;
use warnings;
use feature 'say';
my $data =
"203.174.38.128203.174.38.129203.174.38.1" .
"30203.174.38.131203.174.38.132203.174.38" .
".133203.174.38.134173.174.38.135203.174." .
"38.136203.174.38.137203.174.38.142";
my $byte = qr/2[0-5][0-5]|1\d\d|\d\d|\d/;
my #bytes = $data =~ /($byte)/g;
my #addresses;
push #addresses, join('.', splice(#bytes, 0, 4)) while #bytes;
say for #addresses;
output
203.174.38.128
203.174.38.129
203.174.38.130
203.174.38.131
203.174.38.132
203.174.38.133
203.174.38.134
173.174.38.135
203.174.38.136
203.174.38.137
203.174.38.142
Using your sample, it looks like you have 3 digits for the first and last node. That would prompt using this pattern:
/(\d{3}\.\d{1,3}\.\d{1,3}\.\d{3})/
Add that with a /g switch and it will pull every one.
However, if you have a larger and divergent set of data than what you show for your sample, somebody should have separated the ips before dumping them into this string. If they are separate data points, they should have some separation.
I am using Net::Whois::Raw to query a list of domains from a text file and then parse through this to output relevant information for each domain.
It was all going well until I hit Nominet results as the information I require is never on the same line as that which I am pattern matching.
For instance:
Name servers:
ns.mistral.co.uk 195.184.229.229
So what I need to do is pattern match for "Name servers:" and then display the next line or lines but I just can't manage it.
I have read through all of the answers on here but they either don't seem to work in my case or confuse me even further as I am a simple bear.
The code I am using is as follows:
while ($record = <DOMAINS>) {
$domaininfo = whois($record);
if ($domaininfo=~ m/Name servers:(.*?)\n/){
print "Nameserver: $1\n";
}
}
I have tried an example of Stackoverflow where
<DOMAINS>;
will take the next line but this didn't work for me and I assume it is because we have already read the contents of this into $domaininfo.
EDIT: Forgot to say thanks!
how rude.
So, the $domaininfo string contains your domain?
What you probably need is the m parameter at the end of your regular expression. This treats your string as a multilined string (which is what it is). Then, you can match on the \n character. This works for me:
my $domaininfo =<<DATA;
Name servers:
ns.mistral.co.uk 195.184.229.229
DATA
$domaininfo =~ m/Name servers:\n(\S+)\s+(\S+)/m;
print "Server name = $1\n";
print "IP Address = $2\n";
Now, I can match the \n at the end of the Name servers: line and capture the name and IP address which is on the next line.
This might have to be munged a bit to get it to work in your situation.
This is half a question and perhaps half an answer (the question's in here as I am not yet allowed to write comments...). Okay, here we go:
Name servers:
ns.mistral.co.uk 195.184.229.229
Is this what an entry in the file you're parsing looks like? What will follow immediately afterwards - more domain names and IP addresses? And will there be blank lines in between?
Anyway, I think your problem may (in part?) be related to your reading the file line by line. Once you get to the IP address line, the info about 'Name servers:' having been present will be gone. Multiline matching will not help if you're looking at your file line by line. Thus I'd recommend switching to paragraph mode:
{
local $/ = ''; # one paragraph instead of one line constitutes a record
while ($record = <DOMAINS>) {
# $record will now contain all consecutive lines that were NOT separated
# by blank lines; once there are >= 1 blank lines $record will have a
# new value
# do stuff, e.g. pattern matching
}
}
But then you said
I have tried an example of Stackoverflow where
<DOMAINS>;
will take the next line but this didn't work for me and I assume it is because we have already read the contents of this into $domaininfo.
so maybe you've already tried what I have just suggested? An alternative would be to just add another variable ($indicator or whatever) which you'll set to 1 once 'Name servers:' has been read, and as long as it's equal to 1 all following lines will be treated as containing the data you need. Whether this is feasible, however, depends on you always knowing what else your data file contains.
I hope something in here has been helpful to you. If there are any questions, please ask :)