extracting numbers whose lengths are fixed from text file
Multi tool use
I have a text file. This file includes characters and numbers as follows:
ANKR00TUR_R_20183240000_01D_30S_MO.rnx: 2018 11 20 00 00 0.0000000 GPS TIME OF FIRST OBS
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
rinex_quantity:grep "TIME OF FIRST OBS" * > time_of_first_epochs
I need to extract only 4 digits numbers and store them into another file as follows:
2018
2014
2014
I applied the following code but it extracts all 4 digit numbers:
grep -Po "d{4}" data
2018
3240
2018
0000
3350
2014
0000
1204
2014
0000
text-processing
asked Jan 14 at 10:14
deepblue_86deepblue_86
5851023
add a comment |
I have a text file. This file includes characters and numbers as follows:
ANKR00TUR_R_20183240000_01D_30S_MO.rnx: 2018 11 20 00 00 0.0000000 GPS TIME OF FIRST OBS
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
rinex_quantity:grep "TIME OF FIRST OBS" * > time_of_first_epochs
I need to extract only 4 digits numbers and store them into another file as follows:
2018
2014
2014
I applied the following code but it extracts all 4 digit numbers:
grep -Po "d{4}" data
2018
3240
2018
0000
3350
2014
0000
1204
2014
0000
text-processing
asked Jan 14 at 10:14
deepblue_86deepblue_86
5851023
you need to extract the digit number after the colon?
– AtomiX84
Jan 14 at 10:17
add a comment |
I have a text file. This file includes characters and numbers as follows:
ANKR00TUR_R_20183240000_01D_30S_MO.rnx: 2018 11 20 00 00 0.0000000 GPS TIME OF FIRST OBS
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
rinex_quantity:grep "TIME OF FIRST OBS" * > time_of_first_epochs
I need to extract only 4 digits numbers and store them into another file as follows:
2018
2014
2014
I applied the following code but it extracts all 4 digit numbers:
grep -Po "d{4}" data
2018
3240
2018
0000
3350
2014
0000
1204
2014
0000
text-processing
asked Jan 14 at 10:14
deepblue_86deepblue_86
5851023
I have a text file. This file includes characters and numbers as follows:
ANKR00TUR_R_20183240000_01D_30S_MO.rnx: 2018 11 20 00 00 0.0000000 GPS TIME OF FIRST OBS
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
rinex_quantity:grep "TIME OF FIRST OBS" * > time_of_first_epochs
I need to extract only 4 digits numbers and store them into another file as follows:
2018
2014
2014
I applied the following code but it extracts all 4 digit numbers:
grep -Po "d{4}" data
2018
3240
2018
0000
3350
2014
0000
1204
2014
0000
text-processing
text-processing
asked Jan 14 at 10:14
deepblue_86deepblue_86
5851023
asked Jan 14 at 10:14
deepblue_86deepblue_86
5851023
asked Jan 14 at 10:14
deepblue_86deepblue_86
5851023
asked Jan 14 at 10:14
deepblue_86deepblue_86
5851023
asked Jan 14 at 10:14
deepblue_86deepblue_86
5851023
5851023
you need to extract the digit number after the colon?
– AtomiX84
Jan 14 at 10:17
add a comment |
you need to extract the digit number after the colon?
– AtomiX84
Jan 14 at 10:17
you need to extract the digit number after the colon?
– AtomiX84
Jan 14 at 10:17
you need to extract the digit number after the colon?
– AtomiX84
Jan 14 at 10:17
add a comment |
2 Answers
2
active
oldest
votes
Your grep command was almost correct, you just have to anchor the pattern to match only if there is a word boundary before or after it.
Word boundaries are zero-length patterns that match between a word-character (letters, digits, underscore) and a non-word charater (e.g. spaces, other punctuation, line end, and everything else).
In grep, you can either do this by surrounding your pattern with b, or by using the -w switch to enable word matching:
$ grep -Po 'bd{4}b' data
2018
2014
2014
$ grep -Pow 'd{4}' data
2018
2014
2014
answered Jan 14 at 10:45
Byte CommanderByte Commander
64.1k27176295
add a comment |
with miller (http://johnkerl.org/miller/doc) is
mlr --implicit-csv-header --pprint cut -f 2 then label year input
As output you will have
year
2014
2014
Mi input is
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
I have simply extracted the second column with cut
answered Jan 14 at 10:47
aborrusoaborruso
1714
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
Your grep command was almost correct, you just have to anchor the pattern to match only if there is a word boundary before or after it.
Word boundaries are zero-length patterns that match between a word-character (letters, digits, underscore) and a non-word charater (e.g. spaces, other punctuation, line end, and everything else).
In grep, you can either do this by surrounding your pattern with b, or by using the -w switch to enable word matching:
$ grep -Po 'bd{4}b' data
2018
2014
2014
$ grep -Pow 'd{4}' data
2018
2014
2014
answered Jan 14 at 10:45
Byte CommanderByte Commander
64.1k27176295
add a comment |
Your grep command was almost correct, you just have to anchor the pattern to match only if there is a word boundary before or after it.
Word boundaries are zero-length patterns that match between a word-character (letters, digits, underscore) and a non-word charater (e.g. spaces, other punctuation, line end, and everything else).
In grep, you can either do this by surrounding your pattern with b, or by using the -w switch to enable word matching:
$ grep -Po 'bd{4}b' data
2018
2014
2014
$ grep -Pow 'd{4}' data
2018
2014
2014
answered Jan 14 at 10:45
Byte CommanderByte Commander
64.1k27176295
add a comment |
Your grep command was almost correct, you just have to anchor the pattern to match only if there is a word boundary before or after it.
Word boundaries are zero-length patterns that match between a word-character (letters, digits, underscore) and a non-word charater (e.g. spaces, other punctuation, line end, and everything else).
In grep, you can either do this by surrounding your pattern with b, or by using the -w switch to enable word matching:
$ grep -Po 'bd{4}b' data
2018
2014
2014
$ grep -Pow 'd{4}' data
2018
2014
2014
answered Jan 14 at 10:45
Byte CommanderByte Commander
64.1k27176295
Your grep command was almost correct, you just have to anchor the pattern to match only if there is a word boundary before or after it.
Word boundaries are zero-length patterns that match between a word-character (letters, digits, underscore) and a non-word charater (e.g. spaces, other punctuation, line end, and everything else).
In grep, you can either do this by surrounding your pattern with b, or by using the -w switch to enable word matching:
$ grep -Po 'bd{4}b' data
2018
2014
2014
$ grep -Pow 'd{4}' data
2018
2014
2014
answered Jan 14 at 10:45
Byte CommanderByte Commander
64.1k27176295
answered Jan 14 at 10:45
Byte CommanderByte Commander
64.1k27176295
answered Jan 14 at 10:45
Byte CommanderByte Commander
64.1k27176295
answered Jan 14 at 10:45
Byte CommanderByte Commander
64.1k27176295
64.1k27176295
add a comment |
add a comment |
with miller (http://johnkerl.org/miller/doc) is
mlr --implicit-csv-header --pprint cut -f 2 then label year input
As output you will have
year
2014
2014
Mi input is
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
I have simply extracted the second column with cut
answered Jan 14 at 10:47
aborrusoaborruso
1714
add a comment |
with miller (http://johnkerl.org/miller/doc) is
mlr --implicit-csv-header --pprint cut -f 2 then label year input
As output you will have
year
2014
2014
Mi input is
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
I have simply extracted the second column with cut
answered Jan 14 at 10:47
aborrusoaborruso
1714
add a comment |
with miller (http://johnkerl.org/miller/doc) is
mlr --implicit-csv-header --pprint cut -f 2 then label year input
As output you will have
year
2014
2014
Mi input is
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
I have simply extracted the second column with cut
answered Jan 14 at 10:47
aborrusoaborruso
1714
with miller (http://johnkerl.org/miller/doc) is
mlr --implicit-csv-header --pprint cut -f 2 then label year input
As output you will have
year
2014
2014
Mi input is
brmu3350.14o: 2014 12 1 0 0 0.0000000 GPS TIME OF FIRST OBS
KNY12040.14o: 2014 7 23 0 0 0.0000000 GPS TIME OF FIRST OBS
I have simply extracted the second column with cut
answered Jan 14 at 10:47
aborrusoaborruso
1714
answered Jan 14 at 10:47
aborrusoaborruso
1714
answered Jan 14 at 10:47
aborrusoaborruso
1714
answered Jan 14 at 10:47
aborrusoaborruso
1714
1714
add a comment |
add a comment |
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KiForG
you need to extract the digit number after the colon?
– AtomiX84
Jan 14 at 10:17