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toledano
Asked: 2020-03-11 11:26:58 +0800 CST 2020-03-11 11:26:58 +0800 CST

Delete text after ==

  • 772

requirements file contains all the installed Python packages, so that file can be used elsewhere and rebuild the original programming environment.

A requirements file looks like this:

alabaster==0.7.9
arrow==0.8.0
awesome-slugify==1.6.5
Babel==2.3.4
binaryornot==0.4.0
blessings==1.6

What I want is to remove the part that indicates the version, in the case of the first line alabaster==0.7.9, remove the part ==0.7.9and leave only alabaster.

I understand that finding a match creates two groups, but I can't get it to work. I am trying it on using as follows.

  1. When I order the first group:

    $ awk -F"==" '{print $1}' base.txt

    I get this:

    alabaster==0.7.9
arrow==0.8.0
awesome-slugify==1.6.5

    that is, the file is repeated.

  2. When I order the second group with

    $ awk -F"==" '{print $2}' base.txt

    I only get 50 blank lines.

ADDITION:

Now I search with this pattern (\w+)(==.)with which I make two match groups, I am interested in the first one. But if the package is called python-mimeparsethere is no match anymore . You should be able to add hyphens in case some package is called paquete_pythonorpaquete-python .

Addendum 2

This expression (.+)(==)(.+)finds three groups, the first is the package (which is what I'm looking for) and the third is the version. Now I just need to know how to use it in awk.

third edition

I posted an answer that solves the problem in Python, but the idea is that the solution is applied with some other tool like awk, gawk, sedor even perl.

There are several options in this SOen post , but I haven't been able to use my search pattern on any of them. I get no errors, but no output either.

Some considerations:

  • I am looking to get only the package name , not the version
  • There is no package installed, so there is nothing to update
  • The solution can use another tool, like sedorgrep
regex

4 Answers

  1. A. VALUES ON THE LEFT OF ==

    Option 1.

    Capture everything before ==

    ^.*?(?=\=\=)/gm

    Option 2.

    Make a match without capturing the group from ==

    Thank you @fedorqui

    ^.*?(?:==)/gm

    DEMO

    Result

    alabaster
arrow
awesome-slugify
Babel
binaryornot
blessings

    B. VALUES TO THE RIGHT OF ==

    =.*

    DEMO

    Result

    ==0.7.9
==0.8.0
==1.6.5
==2.3.4
==0.4.0
==1.6
    • 3
  2. Best Answer

    The solution awk -F'==' '{print $1}' archivouses a field separator ( FS ) with multicharacters. This is valid as long as you are using a version of awkPOSIX-compliant. For example, on Solaris it won't work.

    So the question is: how to make it work?

    So let's simplify: the file consists of lines of the form módulo==versión. Therefore, what we can do is delete =and everything that follows it:

    $ cut -d'=' -f1 fichero
alabaster
arrow
awesome-slugify
Babel
binaryornot
blessings

    This is saying: separate the line based on =as separator ( -d=) and print the first resulting field ( -f1).

    It can be a bit fragile, so you can also choose to use sed:

    sed 's/=.*//' fichero

    This does the same thing: removes from the first symbol =. However it allows to extend the command to something more complex like:

    $ sed '/==/s/=.*//' fichero
alabaster
arrow
awesome-slugify
Babel
binaryornot
blessings

    Which performs this substitution only on lines containing ==. And if you push me, you can get to say:

    sed -n '/==/s/=.*//p' fichero

    To print only these lines ( -ndisables printing by default and pprints the current line).

    If you really want match()to use awk, use:

    $ awk 'match($0, /^(.+)==(.+)/, res) {print res[1]}' fichero
alabaster
arrow
awesome-slugify
Babel
binaryornot
blessings

    As you can see, the syntax is match(línea, patrón, matriz de resultados). Therefore, it is a matter of capturing the ones that interest us: in this case only the first one, so in fact we could limit ourselves to saying match($0, /^(.+)==/, res), without the need to capture the rest.

    In short: awkit doesn't seem like the best solution here because depending on which environments you may have problems with the multicharacter field separator. Make your life easy using sedin this case: there is no need to use such complex regular expressions when a sedsimple one already gives you everything you need.

    • 2

  3. Try this command in Bash:

    cat requirements.txt | grep -oP "\w+[-_]{0,1}\w+"

    requirements.txtwould be the pip requirements file.

    The important thing is the regular expression to use and the one I put includes the requirement of the hyphen or underscore separator; I updated the example from @A. Cedano so you can see it live here .

    If you need to save the output to a file (you probably do), you can obviously use output redirection; namely:

    cat requirements.txt | grep -oP "\w+[-_]{0,1}\w+" > salida.txt

    I hope it helps you, greetings.

    • 1

  4. The alternative in Python is as follows:

    import re


r = re.compile('(?P<paquete>.+)(==)(?P<version>.+)'
for l in open('base.txt').readlines():
    print (r.search(l).group('paquete'))
    • The first line imports the module rethat handles regular expressions.
    • Since we are going to apply the same search to all the lines, we create a search object patternor pattern with the expression we are looking for:
      • The first group has a name paqueteand is formed with any character and any quantity.
      • The second group is just a separator, made up of the equals signs.
      • The third group is named versionand is made up of the rest of the characters after the second group.
    • We go through the requirements file, line by line,
    • And we pass it as an argument to the search (which uses the previously compiled object) and only the result of the group paquete(that is, the match ) is printed.
    • 1