通过出现某个字符来过滤命令的输出？

Question

So say I have a command, such as ls . And the output is large, and I want to filter it by the number of occurrences of the character ':'. The occurrences wouldn't need to be consecutive, they'd simply need to occur. Any line with ten or more of these occurrences should be sent to the output. Is there a way to do this? I was thinking of doing a python script, and making it a command, like this , but I wasn't sure how I'd format the script, and what I could and couldn't do in terms of piping output from a previous command into this new script/command.

Answer 1

A shell solution would involve calling awk .

ls | awk -F: 'NF<9'

This reads the standard-out from ls sends it to a pipe, and awk splits each line into fields based on : . The NF mean Number of Fields. When NF is greater than 9 , the record will be printed.

you can extend this to filter on any number by passing in a variable

ls | awk -F: -v minCnt=10 'NF<=minCnt`

IHTH

Answer 2

Using extended regular expressions:

ls | egrep '(:.*){10,}'

Example, from man bash , show all lines with 15 or more " e "s, then highlight them in color:

COLUMNS=80 man bash | egrep --color '(e.*){15,}' | grep --color e

Output, (the colored part is in boldface):

gr e at e r than 9 ar e r e pr e s e nt e d by th e low e rcas e l e tt e rs, th e upp e rcas e

Answer 3

You can read from standard input, filter and write to standard output:

# script.py
import sys
for line in sys.stdin:
    if line.count(':') >= 10:
        sys.stdout.write(line)

Call:

> ls | python script.py