Groovy remove beginning of path - java

I'm trying to delete the beginning of a path that has '\' and ' ' in it. I seem to be getting the some issues saying escape character issue at character 3.
Example:
SomePath: C:\Users\ADMINISTRATOR\App Play\blah\blah
SomePath.replaceFirst('C:\\Users\\ADMINISTRATOR\\App Play\\', '');
Path should be blah\blah
I've tried:
SomePath.replaceFirst("C:\Users\ADMINISTRATOR\App Play\", "");
SomePath.replaceFirst("C:\\Users\\ADMINISTRATOR\\App Play\\", "");
SomePath.replaceFirst("C:\\\\Users\\\\ADMINISTRATOR\\\\App Play\\\\", "");
SomePath.replaceAll("C:\Users\ADMINISTRATOR\App Play\", "");
SomePath.replaceAll("C:\\Users\\ADMINISTRATOR\\App Play\\", "");
SomePath.replaceAll("C:\\\\Users\\\\ADMINISTRATOR\\\\App Play\\\\", "");

Just gave it a try... the examples with four backslashes work for me:
def somePath = "C:\\Users\\ADMINISTRATOR\\App Play\\blah\\blah"
println somePath
somePath.replaceFirst("C:\\\\Users\\\\ADMINISTRATOR\\\\App Play\\\\", "");
The problem is that the string needs one escaping \ and since the replaceFirst uses a regexp, the regexp-engine needs another \ to escape the \. The result are four backslashes.
Btw: you can use string operations to get your path, but you could also try file operations like this:
def root= new File("C:\\Users\\ADMINISTRATOR\\App Play\\")
def full= new File("C:\\Users\\ADMINISTRATOR\\App Play\\blah\\blah")
def relPath = root.toPath().relativize( full.toPath() ).toFile()
println relPath
(taken from https://gist.github.com/ysb33r/5804364)

You can tackle this problem differently. You could tokenize your input path using \ as a delimiter and then you could pick the last 2 elements (blah and blah) or skip first 4 elements (C:, Users, ADMINISTRATOR, App Play). It depends which assumption is easier to deduct for you. Consider following example:
def somePath = 'C:\\Users\\ADMINISTRATOR\\App Play\\blah\\blah'
// Build a new path by accepting the last 2 parts of the initial path
assert 'blah\\blah' == somePath.tokenize('\\')[-2..-1].join('\\')
// Build a new path by skipping the first 4 parts from initial path
assert 'blah\\blah' == somePath.tokenize('\\').drop(4).join('\\')
First option works better if you want only two last parts from the initial path. Second option works better if you can expect final path like blah\blah\blahhhh because you don't know how many nested children initial path contains and you want to start building a new path right after \App Play\ .

Related

Why can't Nextflow handle this awk phrase?

Background:
Using a csv as input, I want to combine the first two columns into a new one (separated by an underscore) and add that new column to the end of a new csv.
Input:
column1,column2,column3
1,2,3
a,b,c
Desired output:
column1,column2,column3,column1_column2
1,2,3,1_2
a,b,c,a_b
The below awk phrase works from the command line:
awk 'BEGIN{FS=OFS=","} {print \$0, (NR>1 ? \$1"_"\$2 : "column1_column2")}' file.csv > full_template.csv
However, when placed within a nextflow script (below) it gives an error.
#!/usr/bin/env nextflow
params.input = '/file/location/here/file.csv'
process unique {
input:
path input from params.input
output:
path 'full_template.csv' into template
"""
awk 'BEGIN{FS=OFS=","} {print \$0, (NR>1 ? \$1"_"\$2 : "combined_header")}' $input > full_template.csv
"""
}
Here is the error:
N E X T F L O W ~ version 21.10.0
Launching `file.nf` [awesome_pike] - revision: 1b63d4b438
class groovyx.gpars.dataflow.expression.DataflowInvocationExpression cannot be cast to class java.nio.file.FileSystem (groovyx.gpars.dataflow.expression.Dclass groovyx.gpars.dataflow.expression.DataflowInvocationExpression cannot be cast to class java.nio.file.FileSystem (groovyx.gpars.dataflow.expression.DataflowInvocationExpression is in unnamed module of loader 'app'; java.nio.file.FileSystem is in module java.base of loader 'bootstrap')
I'm not sure what is causing this, and any help would be appreciated.
Thanks!
Edit:
Yes it seems this was not the source of the error (sorry!). I'm trying to use splitCsv on the resulting csv and this appears to be what's causing the error. Like so:
Channel
.fromPath(template)
.splitCsv(header:true, sep:',')
.map{ row -> tuple(row.column1, file(row.column2), file(row.column3)) }
.set { split }
I expect my issue is it's not acceptable to use .fromPath on a channel, but I can't figure out how else to do it.
Edit 2:
So this was a stupid mistake. I simply needed to add the .splitCsv option directly after the input line where I invoked the channel. Hardly elegant, but appears to be working great now.
process blah {
input:
what_you_want from template.splitCsv(header:true, sep:',').map{ row -> tuple(row.column1, file(row.column2), file(row.column3)) }
I was unable to reproduce the error you're seeing with your example code and Nextflow version. In fact, I get the expected output. This shouldn't be much of a surprise though, because you have correctly escaped the special dollar variables in your AWK command. The cause of the error is likely somewhere else in your code.
If escaping the special characters gets tedious, another way is to use a shell block instead:
It is an alternative to the Script definition with an important
difference, it uses the exclamation mark ! character as the variable
placeholder for Nextflow variables in place of the usual dollar
character.
The example becomes:
params.input_csv = '/file/location/here/file.csv'
input_csv = file( params.input_csv)
process unique {
input:
path input_csv
output:
path 'full_template.csv' into template
shell:
'''
awk 'BEGIN { FS=OFS="," } { print $0, (NR>1 ? $1 "_" $2 : "combined_header") }' \\
"!{input_csv}" > "full_template.csv"
'''
}
template.view { it.text }
Results:
$ nextflow run file.nf
N E X T F L O W ~ version 20.10.0
Launching `file.nf` [wise_hamilton] - revision: b71ff1eb03
executor > local (1)
[76/ddbb87] process > unique [100%] 1 of 1 ✔
column1,column2,column3,combined_header
1,2,3,1_2
a,b,c,a_b

Scala RegEx String extractors behaving inconsistently

I have two regular expression extractors.
One for .java files and the other is for .scala files
val JavaFileRegEx =
"""\S*
\s+
//
\s{1}
([^\.java]+)
\.java
""".replaceAll("(\\s)", "").r
val ScalaFileRegEx =
"""\S*
\s+
//
\s{1}
([^\.scala]+)
\.scala
""".replaceAll("(\\s)", "").r
I want to use these extractors above to extract a java file name and a scala file name from the example code below.
val string1 = " // Tester.java"
val string2 = " // Hello.scala"
string1 match {
case JavaFileRegEx(fileName1) => println(" Java file: " + fileName1)
case other => println(other + "--NO_MATCH")
}
string2 match {
case ScalaFileRegEx(fileName2) => println(" Scala file: " + fileName2)
case other => println(other + "--NO_MATCH")
}
I get this output indicating that the .java file matched but the .scala file did not.
Java file: Tester
// Hello.scala--NO_MATCH
How is it that the Java file matched but the .scala file did not?
NOTE
[] denotes character class. It matches only a single character.
[^] denotes match anything except the characters present in the character class.
In your first regex
\S*\s+//\s{1}([^\.java]+)\.java
\S* matches nothing as there is space in starting
\s+ matches the space which is in starting
// matches // literally
\s{1} matches next space
You are using [^\.java] which says match anything except . or j or a or v or a which can be written as [^.jav].
So, the left string now to be tested is
Tester.java
(Un)luckily any character from Tester does not matches . or j or a or v until we encounter a .. So Tester is matched and then java is also matched.
In your second regex
\S*\s+//\s{1}([^\.scala]+)\.scala
\S* matches nothing as there is space in starting
\s+ matches the space which is in starting
// matches // literally
\s{1} matches next space
Now, you are using [^\.scala] which says that match anything except . or s or c or a or l or a which can be written as [^.scla].
You have now
Hello.scala
but (un)luckily Hello here contains l which is not allowed according to character class and the regex fails.
How to correct it?
I will modify only a bit of your regex
\S*\s+//\s{1}([^.]*)\.java
<-->
This says that match anything except .
You can also use \w here instead if [^.]
Regex Demo
\S*\s+//\s{1}([^.]*)\.scala
Regex Demo
There is no need of {1} in \s{1}. You can simply write it as \s and it will match exactly one space like
\S*\s+//\s([^.]*)\.java

Split on Regular Expression per Path

If I have this:
thisisgibberish 1234 /hello/world/
more gibberish 43/7 /good/timing/
just onemore 8888 /thanks/mate
what would the regular expression inside the Java String.split() method be to obtain the paths per line?
ie.
[0]: /hello/world/
[1]: /good/timing/
[2]: /thanks/mate
Doing
myString.split("\/[a-zA-Z]")
causes the splits to occur to every /h, /w, /g, /t, and /m.
How would I go about writing a regular expression to split it only once per line while only capturing the paths?
Thanks in advance.
Why split ? I think running a match here is better, try the following expression:
(?<=\s)(/[a-zA-Z/])+
Regex101 Demo
This uses split() :
String[] split = myString.split(myString.substring(0, myString.lastIndexOf(" ")));
OR
myString.split(myString.substring(0, myString.lastIndexOf(" ")))[1]; //works for current inputs
You must first remove the leading junk, then split on the intervening junk:
String[] paths = str.replaceAll("^.*? (?=/[a-zA-Z])", "")
.split("(?m)((?<=[a-zA-Z]/|[a-zA-Z])\\s|^).*? (?=/[a-zA-Z])");
One important point here is the use of (?m), which is a switch that turns on "dot matches newline", which is required to split across the newlines.
Here's some test code:
String str = "thisisgibberish 1234 /hello/world/\nmore gibberish 43/7 /good/timing/\njust onemore 8888 /thanks/mate";
String[] paths = str.replaceAll("^.*? (?=/[a-zA-Z])", "")
.split("(?m)((?<=[a-zA-Z]/|[a-zA-Z])\\s|^).*? (?=/[a-zA-Z])");
System.out.println( Arrays.toString( paths));
Output (achieving requirements):
[/hello/world/, /good/timing/, /thanks/mate]

A custom tokenizer for Java

I am developing an application in which I need to process text files containing emails. I need all the tokens from the text and the following is the definition of token:
Alphanumeric
Case-sensitive (case to be preserved)
'!' and '$' are to be considered as constituent characters. Ex: FREE!!, $50 are tokens
'.' (dot) and ',' comma are to be considered as constituent characters if they occur between numbers. For ex:
192.168.1.1, $24,500
are tokens.
and so on..
Please suggest me some open-source tokenizers for Java which are easy to customize to suit my needs. Will simply using StringTokenizer and regex be enough? I have to perform stopping also and that's why I was looking for an open source tokenizer which will also perform some extra things like stopping, stemming.
A few comments up front:
From StringTokenizer javadoc:
StringTokenizer is a legacy class that is retained for compatibility
reasons although its use is discouraged in new code. It is recommended
that anyone seeking this functionality use the split method of String
or the java.util.regex package instead.
Always use Google first - the first result as of now is JTopas. I did not use it, but it looks it could work for this
As for regex, it really depends on your requirements. Given the above, this might work:
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Mkt {
public static void main(String[] args) {
Pattern p = Pattern.compile("([$\\d.,]+)|([\\w\\d!$]+)");
String str = "--- FREE!! $50 192.168.1.1 $24,500";
System.out.println("input: " + str);
Matcher m = p.matcher(str);
while(m.find()) {
System.out.println("token: " + m.group());
}
}
}
Here's a sample run:
$ javac Mkt.java && java Mkt
input: --- FREE!! $50 192.168.1.1 $24,500
token: FREE!!
token: $50
token: 192.168.1.1
token: $24,500
Now, you might need to tweak the regex, for example:
You gave $24,500 as an example. Should this work for $24,500abc or $24,500EUR?
You mentioned 192.168.1.1 should be included. Should it also include 192,168.1,1 (given . and , are to be included)?
and I guess there are other things to consider.
Hope this helps to get you started.

How to append a backslash in a string in java

I want to add a '\' character to every string in a list of strings... I m doing something like this but it adds 2 backslashes instead.
feedbackMsgs.add(behaviorName+"\\"+fbCode);
result is like: "abc\\def"
how to make sure a single backslash is added??
I've just run a program with the following -
String s = "test" + "\\" + "test2";
System.out.println(s);
And it prints out the following -
test\test2
Are you sure there is no \ in the behaviourName or fbCode variables?
Looks like either your behaviourName ends with a \ or fbCode starts with one.
Try to Log/print behaviorName fbCode and find it yourself !
System.out.println(behaviorName);
System.out.println(fbCode);

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