How to get the first 7 characters of the lines which contain...??

songor · 06-23-2014, 04:09 AM

Hi All

I hope someone can help. Not enough in the subject. But my full question is:
How to get the first 7 characters of the lines which contain character 'D' in position 120 from a file ???

eg
2222222 jason... ...DXXY (D is position 120)
2222223 peter... ...DXXY
2222224 john.... ...OXXY

The output should be:
2222222
2222223

But not 2222224 because position 120 is 'O', not 'D'

Is it possible to do it in sed ?? Please help

Many many thanks

NevemTeve · 06-23-2014, 04:45 AM

First part:

Code:

grep '^.{119}D'

Second part:

Code:

cut -b1-7

Now merge these together.

danielbmartin · 06-23-2014, 06:55 AM

For ease in testing I changed the "D in position 120" to "D in position 20."

With this InFile ...

Code:

yes,yes,90123456ABCDEFGHIJK
reject,89012345678D0DD000DDD
oui,oui,90123456abcDefghijklmn

... this awk ...

Code:

awk -F "" '{if ($20=="D") print substr($0,1,7)}' $InFile >$OutFile

... produced this OutFile ...

Code:

yes,yes
oui,oui

Daniel B. Martin

danielbmartin · 06-23-2014, 05:26 PM

Quote:

Originally Posted by songor

Is it possible to do it in sed ??

Yes, but it's not pretty.

For ease in testing I changed the "D in position 120" to "D in position 20."

With this InFile ...

Code:

yes,yes,90123456ABCDEFGHIJK
reject,89012345678D0DD000DDD
oui,oui,90123456abcDefghijklmn

... this sed ...

Code:

sed -e '/^.\{19\}D/!d' -e 's/\(^.\{7\}\)\(.*\)/\1/' $InFile >$OutFile

... produced this OutFile ...

Code:

yes,yes
oui,oui

This part: -e '/^.\{19\}D/!d' keeps lines with D in position 20, and then ...
this part: -e 's/\(^.\{7\}\)\(.*\)/\1/' preserves the first 7 characters of those lines.

In my opinion the awk solution (see previous post) is easier to write, easier to read.

Daniel B. Martin

songor · 06-25-2014, 01:24 AM

Quote:

Originally Posted by danielbmartin

For ease in testing I changed the "D in position 120" to "D in position 20."

With this InFile ...

Code:

yes,yes,90123456ABCDEFGHIJK
reject,89012345678D0DD000DDD
oui,oui,90123456abcDefghijklmn

... this awk ...

Code:

awk -F "" '{if ($20=="D") print substr($0,1,7)}' $InFile >$OutFile

... produced this OutFile ...

Code:

yes,yes
oui,oui

Daniel B. Martin

Wow, work great! Many thanks

danielbmartin · 06-25-2014, 07:00 PM

Quote:

Originally Posted by songor

Is it possible to do it in sed ??

For ease in testing I changed the "D in position 120" to "D in position 20."

With this InFile ...

Code:

yes,yes,90123456ABCDEFGHIJK
reject,89012345678D0DD000DDD
oui,oui,90123456abcDefghijklmn

... this sed ...

Code:

sed -n 's/\(^.\{7\}\)\(.\{12\}D.*\)/\1/p' $InFile >$OutFile

... produced this OutFile ...

Code:

yes,yes
oui,oui

In my opinion the awk solution (see previous post) is easier to write, easier to read.

Daniel B. Martin

firstfire · 06-28-2014, 01:19 AM

Hi.

Extended regular expressions (-r option) allow one to omit most of backslashes (but are a GNU extension):

Code:

$ cat /tmp/infile
yes,yes,90123456ABCDEFGHIJK
reject,89012345678D0DD000DDD
oui,oui,90123456abcDefghijklmn

$ sed -nr 's/(^.{7})(.{12}D.*)/\1/p' /tmp/infile
yes,yes
oui,oui

danielbmartin · 06-28-2014, 09:09 AM

Quote:

Originally Posted by firstfire

Code:

sed -nr 's/(^.{7})(.{12}D.*)/\1/p' /tmp/infile

Thank you for this improvement. Your code is fewer keystrokes and easier to read. Nice!

Daniel B. Martin

firstfire · 06-28-2014, 05:35 PM

Hi.

I just noticed that last pair of parentheses can be omitted too:

Code:

$ sed -nr 's/(^.{7}).{12}D.*/\1/p' /tmp/infile
yes,yes
oui,oui