Grep and Cut Command and divide string

Question

I have a grep command that gives the following string:

20121121001100 18 0 16 2 18

but I would like to modify this string to get

20121121 001 18 0 16 2 18

the above value are being extracted by the following:

for i in `ls -1 file.txt | sort`; do echo $i`
grep datetime $i | wc -l ``
grep abc $i | wc -l ``
grep def $i | wc -l ``
grep ghi $i | wc -l ``
grep jkl $i | wc -l ` ; done | cut -c9-500

cut -c9-500 is used because the original string is in the form of

datetime20121121001100 18 0 16 2 18

and cut -c9-500 returns

20121121001100 18 0 16 2 18

Can someone please help me to get

20121121 001 18 0 16 2 18

(ie remove the last 3 digits from the date part)

Zsolt Botykai · Accepted Answer · 2012-11-22T15:06:34.240

1

Most of what you want/do can be accomplished with awk. But for the minimum you want:

for i in `ls -1 file.txt | sort`; do echo $i`
grep datetime $i | wc -l ``
grep abc $i | wc -l ``
grep def $i | wc -l ``
grep ghi $i | wc -l ``
grep jkl $i | wc -l ` ; done | cut -c9-500 | awk '{print substr($0,1,11) substr($0,15) }'

awk is very capable at text processing.

Edit: I'm not sure of what are you doing, but, basicly this does (almost) the same:

awk 'FILENAME != oldfilename {oldfilename = FILENAME; dt = 0 ; a = 0; d = 0; g = 0; j = 0}
     /datetime/ {dt++}
     /abc/ {a++}
     /def/ {d++}
     /ghi/ {g++}
     /j/ {j++}
     END {print FILENAME, dt, a, d, g, j}' *

And it's faster, fewer processes, etc... Basically awk process the file, counts the occurences of the specified strings, and when it finishes the file (after the last line) prints the report.

Changed specs:

for i in `ls -1 file.txt | sort`; do echo $i`
grep datetime $i | wc -l ``
grep abc $i | wc -l ``
grep def $i | wc -l ``
grep ghi $i | wc -l ``
grep jkl $i | wc -l ` ; done | cut -c9-500 | awk '{print substr($0,1,8) " " substr($0,9,4) substr($0,15) }'

edited Nov 22 '12 at 15:06

answered Nov 22 '12 at 14:36

Zsolt Botykai

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thanks .... re"after the last line will print report" will the use a large amount of memory? – mouthpiec Nov 22 '12 at 14:47
Nope, as you can see it will count in 5 variables (which are reset for every file), so most probably modern machines (after ummm 1960) can deal with it. – Zsolt Botykai Nov 22 '12 at 14:51
thanks .... you first suggestion worked fine! can you please arrange your command to get "20121121 0010" instead of "20121121001 " ? – mouthpiec Nov 22 '12 at 14:59
yes ... but now the spec changed a bit :) ... is it possible to get "20121121 0010"? – mouthpiec Nov 22 '12 at 15:01
Yes it is. see this page: http://mattgemmell.com/2008/12/08/what-have-you-tried/ (and I'm editing the post...)? – Zsolt Botykai Nov 22 '12 at 15:04

score 0 · Answer 2 · answered Nov 22 '12 at 14:38

0

Pipe to sed:

echo "20121121001100 18 0 16 2 18" | sed -r 's/^([0-9]+)[0-9][0-9][0-9] (.*)$/\1 \2/'

gives

20121121001 18 0 16 2 18

answered Nov 22 '12 at 14:38

August Karlstrom

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can you add the above to my grep command cause i cannot understand how to implement it. – mouthpiec Nov 22 '12 at 14:54
@mouthpiec: Just pipe the output from your cut command to the sed command above. – August Karlstrom Nov 22 '12 at 15:31

Grep and Cut Command and divide string

2 Answers2