find string and print first and last characters of line
up vote
1
down vote
favorite
I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.
For example, cat file1.txt
1234567 New line with irrelevant info x end line
2345678 irrelevant line
3456789 New line with different irrelevant info y end line
4567890 irrelevant line
5678901 New line with yet more irrelevant info z end line
And my output would be:
1234567 x
3456789 y
5678901 z
sed grep cut
asked Dec 1 at 21:09
user2535719
62
add a comment |
up vote
1
down vote
favorite
I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.
For example, cat file1.txt
1234567 New line with irrelevant info x end line
2345678 irrelevant line
3456789 New line with different irrelevant info y end line
4567890 irrelevant line
5678901 New line with yet more irrelevant info z end line
And my output would be:
1234567 x
3456789 y
5678901 z
sed grep cut
asked Dec 1 at 21:09
user2535719
62
1
Expected output seems wrong .... second line starting with2345678is irrelevant.
– George Vasiliou
Dec 1 at 21:36
1
... as is the third.
– RudiC
Dec 1 at 21:45
thanks, typo on my part. fixed now.
– user2535719
Dec 1 at 21:56
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.
For example, cat file1.txt
1234567 New line with irrelevant info x end line
2345678 irrelevant line
3456789 New line with different irrelevant info y end line
4567890 irrelevant line
5678901 New line with yet more irrelevant info z end line
And my output would be:
1234567 x
3456789 y
5678901 z
sed grep cut
asked Dec 1 at 21:09
user2535719
62
I have files with hundreds of lines of varying length. I want to find each line with the string "New" and print the first 7 characters and the 10th from the last character.
For example, cat file1.txt
1234567 New line with irrelevant info x end line
2345678 irrelevant line
3456789 New line with different irrelevant info y end line
4567890 irrelevant line
5678901 New line with yet more irrelevant info z end line
And my output would be:
1234567 x
3456789 y
5678901 z
sed grep cut
sed grep cut
asked Dec 1 at 21:09
user2535719
62
asked Dec 1 at 21:09
user2535719
62
edited Dec 1 at 21:55
asked Dec 1 at 21:09
user2535719
62
asked Dec 1 at 21:09
user2535719
62
asked Dec 1 at 21:09
user2535719
62
62
1
Expected output seems wrong .... second line starting with2345678is irrelevant.
– George Vasiliou
Dec 1 at 21:36
1
... as is the third.
– RudiC
Dec 1 at 21:45
thanks, typo on my part. fixed now.
– user2535719
Dec 1 at 21:56
add a comment |
1
Expected output seems wrong .... second line starting with2345678is irrelevant.
– George Vasiliou
Dec 1 at 21:36
1
... as is the third.
– RudiC
Dec 1 at 21:45
thanks, typo on my part. fixed now.
– user2535719
Dec 1 at 21:56
1
1
Expected output seems wrong .... second line starting with
2345678 is irrelevant.– George Vasiliou
Dec 1 at 21:36
Expected output seems wrong .... second line starting with
2345678 is irrelevant.– George Vasiliou
Dec 1 at 21:36
1
1
... as is the third.
– RudiC
Dec 1 at 21:45
... as is the third.
– RudiC
Dec 1 at 21:45
thanks, typo on my part. fixed now.
– user2535719
Dec 1 at 21:56
thanks, typo on my part. fixed now.
– user2535719
Dec 1 at 21:56
add a comment |
3 Answers
3
active
oldest
votes
up vote
1
down vote
Choose one you like:
awk solution:
awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt
sed solution:
sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt
Sample output (for both approaches):
1234567 x
3456789 y
5678901 z
answered Dec 1 at 21:37
RomanPerekhrest
22.8k12246
add a comment |
up vote
1
down vote
POSIXly:
Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):
awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'
or assuming the lines contain at least 17 characters (the lines that don't will be skipped):
sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'
answered Dec 1 at 21:48
Stéphane Chazelas
297k54562908
1
What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
– George Vasiliou
Dec 1 at 21:53
@GeorgeVasiliou, he edited theawksolution in after I had started writing mine and I hadn't seen his edit before I posted mine. Hissedone is GNU specific, my intention was to offer a standard solution.
– Stéphane Chazelas
Dec 1 at 21:56
ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
– George Vasiliou
Dec 1 at 22:00
add a comment |
up vote
0
down vote
This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.
awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
1234567 x
3456789 y
5678901 z
More solutions with grep / sed will follow.
answered Dec 1 at 21:33
George Vasiliou
5,59531028
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Choose one you like:
awk solution:
awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt
sed solution:
sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt
Sample output (for both approaches):
1234567 x
3456789 y
5678901 z
answered Dec 1 at 21:37
RomanPerekhrest
22.8k12246
add a comment |
up vote
1
down vote
Choose one you like:
awk solution:
awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt
sed solution:
sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt
Sample output (for both approaches):
1234567 x
3456789 y
5678901 z
answered Dec 1 at 21:37
RomanPerekhrest
22.8k12246
add a comment |
up vote
1
down vote
up vote
1
down vote
Choose one you like:
awk solution:
awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt
sed solution:
sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt
Sample output (for both approaches):
1234567 x
3456789 y
5678901 z
answered Dec 1 at 21:37
RomanPerekhrest
22.8k12246
Choose one you like:
awk solution:
awk '/New/{ print substr($0, 1, 7), substr($0, length-9, 1) }' file1.txt
sed solution:
sed -rn '/New/ s/^(.{7}).*(.).{9}$/1 2/p' file1.txt
Sample output (for both approaches):
1234567 x
3456789 y
5678901 z
answered Dec 1 at 21:37
RomanPerekhrest
22.8k12246
edited Dec 1 at 21:46
answered Dec 1 at 21:37
RomanPerekhrest
22.8k12246
answered Dec 1 at 21:37
RomanPerekhrest
22.8k12246
answered Dec 1 at 21:37
RomanPerekhrest
22.8k12246
22.8k12246
add a comment |
add a comment |
up vote
1
down vote
POSIXly:
Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):
awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'
or assuming the lines contain at least 17 characters (the lines that don't will be skipped):
sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'
answered Dec 1 at 21:48
Stéphane Chazelas
297k54562908
1
What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
– George Vasiliou
Dec 1 at 21:53
@GeorgeVasiliou, he edited theawksolution in after I had started writing mine and I hadn't seen his edit before I posted mine. Hissedone is GNU specific, my intention was to offer a standard solution.
– Stéphane Chazelas
Dec 1 at 21:56
ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
– George Vasiliou
Dec 1 at 22:00
add a comment |
up vote
1
down vote
POSIXly:
Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):
awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'
or assuming the lines contain at least 17 characters (the lines that don't will be skipped):
sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'
answered Dec 1 at 21:48
Stéphane Chazelas
297k54562908
1
What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
– George Vasiliou
Dec 1 at 21:53
@GeorgeVasiliou, he edited theawksolution in after I had started writing mine and I hadn't seen his edit before I posted mine. Hissedone is GNU specific, my intention was to offer a standard solution.
– Stéphane Chazelas
Dec 1 at 21:56
ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
– George Vasiliou
Dec 1 at 22:00
add a comment |
up vote
1
down vote
up vote
1
down vote
POSIXly:
Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):
awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'
or assuming the lines contain at least 17 characters (the lines that don't will be skipped):
sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'
answered Dec 1 at 21:48
Stéphane Chazelas
297k54562908
POSIXly:
Assuming the lines contain at least 10 characters (if not, the behaviour is unspecified for the second substr(), you can add a && length >= 10 or && length >= 17 after /New/ to skip the lines that have fewer than 10 or 17 characters):
awk '/New/ {print substr($0, 1, 7), substr($0, length - 9, 1)}'
or assuming the lines contain at least 17 characters (the lines that don't will be skipped):
sed -n '/New/ s/^(.{7}).*(.).{9}$/1 2/p'
answered Dec 1 at 21:48
Stéphane Chazelas
297k54562908
answered Dec 1 at 21:48
Stéphane Chazelas
297k54562908
answered Dec 1 at 21:48
Stéphane Chazelas
297k54562908
answered Dec 1 at 21:48
Stéphane Chazelas
297k54562908
297k54562908
1
What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
– George Vasiliou
Dec 1 at 21:53
@GeorgeVasiliou, he edited theawksolution in after I had started writing mine and I hadn't seen his edit before I posted mine. Hissedone is GNU specific, my intention was to offer a standard solution.
– Stéphane Chazelas
Dec 1 at 21:56
ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
– George Vasiliou
Dec 1 at 22:00
add a comment |
1
What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
– George Vasiliou
Dec 1 at 21:53
@GeorgeVasiliou, he edited theawksolution in after I had started writing mine and I hadn't seen his edit before I posted mine. Hissedone is GNU specific, my intention was to offer a standard solution.
– Stéphane Chazelas
Dec 1 at 21:56
ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
– George Vasiliou
Dec 1 at 22:00
1
1
What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
– George Vasiliou
Dec 1 at 21:53
What is the difference with the answer of RomanPerekhrest ? Your awk solution is identical and your sed solution is also identical, it just uses basic regex instead of extended regex....
– George Vasiliou
Dec 1 at 21:53
@GeorgeVasiliou, he edited the
awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.– Stéphane Chazelas
Dec 1 at 21:56
@GeorgeVasiliou, he edited the
awk solution in after I had started writing mine and I hadn't seen his edit before I posted mine. His sed one is GNU specific, my intention was to offer a standard solution.– Stéphane Chazelas
Dec 1 at 21:56
ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
– George Vasiliou
Dec 1 at 22:00
ok then. Fair answer. Has happened also to me. If i had more time my answer would also be like Roman's answer, but this guy was fast! :-)
– George Vasiliou
Dec 1 at 22:00
add a comment |
up vote
0
down vote
This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.
awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
1234567 x
3456789 y
5678901 z
More solutions with grep / sed will follow.
answered Dec 1 at 21:33
George Vasiliou
5,59531028
add a comment |
up vote
0
down vote
This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.
awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
1234567 x
3456789 y
5678901 z
More solutions with grep / sed will follow.
answered Dec 1 at 21:33
George Vasiliou
5,59531028
add a comment |
up vote
0
down vote
up vote
0
down vote
This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.
awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
1234567 x
3456789 y
5678901 z
More solutions with grep / sed will follow.
answered Dec 1 at 21:33
George Vasiliou
5,59531028
This is one "brutal" gawk solution that does the job, using null as field separator FS and output field separator OFS , meaning each char of inputfile is considered to be a field for awk.
awk '/New/{print $1,$2,$3,$4,$5,$6,$7," ",$(NF-9)}' FS="" OFS="" file1
1234567 x
3456789 y
5678901 z
More solutions with grep / sed will follow.
answered Dec 1 at 21:33
George Vasiliou
5,59531028
answered Dec 1 at 21:33
George Vasiliou
5,59531028
answered Dec 1 at 21:33
George Vasiliou
5,59531028
answered Dec 1 at 21:33
George Vasiliou
5,59531028
5,59531028
add a comment |
add a comment |
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1
Expected output seems wrong .... second line starting with
2345678is irrelevant.– George Vasiliou
Dec 1 at 21:36
1
... as is the third.
– RudiC
Dec 1 at 21:45
thanks, typo on my part. fixed now.
– user2535719
Dec 1 at 21:56