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I have a large file (>10000 lines) that contains one word per line, with a newline character after each word. The words contain no spaces.

I'd like to list (or even better, output to a new file) any words that start and/or end with a number, then I'd like to remove these from the original file. But I don't want to remove words that just contain numbers.

For example, if I had the contents

789
hello
1hello
112121hello3323
he11o
hello9
88888

Then the strings 1hello, 112121hello3323, hello9 would get output and then removed from the file.

How can I do this?

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3 Answers 3

Reset to default
2

GNU grep

grep -vP '^\d+\D|\D\d+$'

produces

789
hello
he11o
88888
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2
  • 1
    or POSIXly, grep -vE '^[[:digit:]]+[^[:digit:]]|[^[:digit:]][[:digit:]]+$'
    – iruvar
    Feb 9, 2016 at 13:37
  • @1_CR Yeah i was going to add that but wasn't 100% that -E was completely posix compliant.
    – 123
    Feb 9, 2016 at 13:48
1

To actually edit the source file and create a new file with the discards is a bit trickier. I would do this

$ cat file
789
hello
1hello
112121hello3323
he11o
hello9
88888

$ perl -i -lne 'if (/^\d+\D|\D\d+$/) {warn "$_\n"} else {print}' file 2>file_nums

$ cat file
789
hello
he11o
88888

$ cat file_nums
1hello
112121hello3323
hello9

The matched lines are output on stderr, which is then redirected to a separate file. perl's -i flag takes care of saving the changes in-place.

The one-liner can be even trickier:

perl -i -lne 'print {/^\d+\D|\D\d+$/ ? STDERR : ARGVOUT} $_' file 2>file_nums
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1
  • 1
    Note that this will also write any warnings or errors from perl into file_nums.
    – 123
    Feb 9, 2016 at 14:53
1

An awk solution:

awk '$0!~/.*[[:alpha:]][[:digit:]]+$/ && $0!~/^[[:digit:]]+[[:alpha:]]+/' words.txt
789
hello
he11o
88888
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