I would like to take the number after the - sign and put is as column 2 in my matrix. I know how to grep the string but not how to print it after the text string.
in:
1-967764 GGCTGGTCCGATGGTAGTGGGTTATCAGAACT
3-425354 GCATTGGTGGTTCAGTGGTAGAATTCTCGCC
4-376323 GGCTGGTCCGATGGTAGTGGGTTATCAGAAC
5-221398 GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT
6-180339 TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT
out:
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
awk -F'[[:space:]-]+' '{print $3,$2}' file
Seems like a simple substitution should do the job:
sed -E 's/[0-9]+-([0-9]+)[[:space:]]*(.*)/\2 \1/' file
Capture the parts you're interested in and use them in the replacement.
Alternatively, using awk:
awk 'sub(/^[0-9]+-/, "") { print $2, $1 }' file
Remove the leading digits and - from the start of the line. When this is successful, sub returns true, so the action is performed, printing the second field, followed by the first.
Using regex ( +|-) as field separator:
$ awk -F"( +|-)" '{print $3,$2}' file
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
here is another awk
$ awk 'split($1,a,"-") {print $2,a[2]}' file
awk '{sub(/.-/,"");print $2,$1}' file
GGCTGGTCCGATGGTAGTGGGTTATCAGAACT 967764
GCATTGGTGGTTCAGTGGTAGAATTCTCGCC 425354
GGCTGGTCCGATGGTAGTGGGTTATCAGAAC 376323
GGAAGAGCACACGTCTGAACTCCAGTCACGTGAAAATCTCGTATGCCGTCT 221398
TCCCTGGTGGTCTAGTGGTTAGGATTCGGCGCT 180339
Related
I have the following sentence within a file
FQDN=joe.blogs.com.
How can I print the string "joe"
I have tried using -->> awk -F"=" '{print $2}' file
but this returns joe.blogs.com as "=" is the delimiter.
Is it possible to use 2 delimiters on the same line?
You might use regular expression as FS. Let file.txt content be
FQDN=joe.blogs.com.
then
awk 'BEGIN{FS="[=.]"}{print $2}' file.txt
output
joe
In case you are ok with sed, could you please try following.
sed 's/.*=\([^.]*\)\..*/\1/' Input_file
With GNU grep and using its -oP flag we could try following too.
grep -oP '(.*=)\K([^.]*)' Input_file
You could use GNU grep:
grep -oP '(?<=FQDN=)[^.]+' file
^ all characters up to a '.'
^ lookbehind for 'FQDN='
^ only print match and Perl style regex
Or with Perl:
perl -lne 'print $1 if /(?<=FQDN=)([^.]+)/' file
With awk I would probably do:
awk 'BEGIN{FS="[.=]"} /FQDN=/{print $2}' file
why not keeping it simple and pipe awk?
awk -F"=" '{print $2}' | awk -F"." '{print $1}'
can I use two field delimiters on one line?
No. You may do further string manipulation as post processing, or you could use a regex as field delimiter.
Another option is to use awk's split function:
awk -F= '{ split($2,map,".");print map[1] }' file
Split the second = separated field into the array map using "." as the delimiter. Print the first index of the array.
Having a very large file where two lines shown below and having two fields name and revision having colon delimiter. I need to print only the second column.
sam:7.[0:6]
Ram:8.[6:6]_rev[2:4] h_ack[2:6]
vincent:58
I tried this code:
#!/bin/bash
awk -F: '{print $2}'
7.[0
8.[6
58
Output should be:
7.[0:6]
8.[6:6]_rev[2:4] h_ack[2:6]
58
What went wrong in my code.
The problem in your awk expression is that you are splitting on all :.
Instead, you want to split only on the first : from the start.
$ awk -F'^[^:]+:' '{print $2}' file
The regex pattern matches the start of the string ^, any character other than a :, and finally a :.
If you specify field separator as :, it's normal behavior of awk to output this, ex:
7.[0, because you need the other columns after $2.
cut here, better suits the requirement:
cut -d: -f2- file
Could you please try following.
awk '
match($0,/:.*/){
print substr($0,RSTART+1,RLENGTH-1)
}
' Input_file
I have a file like this
name|age
Bob|30
Tom|50
Cindy|10
I want the first row to have a different seperator, "^".
awk 'NR==1 { gsub("|","^")1}1' f
But I keep getting
^n^a^m^e^|^a^g^e^
Bob|30
Tom|50
Cindy|10
Desired output is
name^age
Bob|30
Tom|50
Cindy|10
Your code with gsub("|","^") doesn't have special meta character | (used for alternation in regex) escaped hence it will match every position in input.
You may use this awk without involving any regex:
awk 'BEGIN{FS=OFS="|"} FNR==1{OFS="^"; $1=$1; OFS=FS} 1' f
name^age
Bob|30
Tom|50
Cindy|10
Details:
FS="|": Sets FS as |
OFS="^": Sets OFS as ^
$1=$1: Forces awk to reformat each of the fields using OFS
You can also use sed:
sed '1 s/|/^/' ip.txt
1 address for the command, which is first line here
| is not special, because by default sed uses BRE, see this Q&A for BRE vs ERE differences
use s/|/^/g if you can have multiple matches
Like this :
awk -F'|' 'NR==1{print $1,$2;next}1' OFS='^' file
or a mix between anubhava response and mine:
awk -F'|' 'NR==1{$1=$1}1' OFS='^' file
Could you please try following.
awk 'FNR==1{sub(/\|/,"^")} 1' Input_file
Use gsub in place of sub in case of multiple occurrences needs to be changed.
awk 'FNR==1{gsub(/\|/,"^")} 1' Input_file
Using awk, I want to print all lines that have a string in the first column that starts with 22_
I tried the following, but obviously * does not work as a wildcard in awk:
awk '$1=="22_*" {print $0}' input > output
Is this possible in awk?
Let's start with a test file:
$ cat >file
22_something keep
23_other omit
To keep only lines that start with 22_:
$ awk '/^22_/' file
22_something keep
Alternatively, if you prefer to reference the first field explicitly, we could use:
$ awk '$1 ~ /^22_/' file
22_something keep
Note that we don't have to write {print $0} after the condition because that is exactly the default action that awk associates with a condition.
At the start of a regular expressions, ^ matches the beginning of a line. Thus, if you want 22_ to occur at the start of a line or the start of a field, you want to write ^22_.
In the condition $1 ~ /^22_/, note that the operator is ~. That operator tells awk to check if the preceding string, $1, matches the regular expression ^22_.
Chosen answer does not answer how to use a wildcard in awk, which is achieved using .* (instead of *):
awk '$1=="22_.*" {print $0}' input > output
I have lines like this:
Volume.Free_IBM_LUN59_28D: 2072083693568
I would like to get only IBM_LUN59_28D from this line using awk.
Thanks
You can use sub to do substitutions on each input line, as per the following transcript:
pax> echo 'Volume.Free_IBM_LUN59_28D: 2072083693568' | awk '
...> {
...> sub (".*Free_", "");
...> sub (":.*", "");
...> print
...> }'
IBM_LUN59_28D
That command crosses multiple lines for readability but, if you're operating on a file and not too concerned about readability, you can just use the compressed version:
awk '{sub(".*Free_","");sub(":.*","");print}' inputFile
If you're amenable to non-awk solutions, you could also use sed:
sed -e 's/.*Free_//' -e 's/:.*//' inputFile
Note that both those solutions rely on your (somewhat sparse) test data. If your definition of "like" includes preceding textual segments other than Free_ or subsequent characters other than :, some more work may be needed.
For example, if you wanted the string between the first _ and the first :, you could use:
awk '{sub("[^_]*_","");sub(":.*","");print}'
With sed:
sed 's/[^_]*_\(.*\):.*/\1/'
Search for sequence of non _ characters followed by _ (this will match Volume.Free_), then another sequence of characters (this will match IBM_LUN59_28D, we group this for future use), followed by : and any char sequence. Substitute with the saved pattern (\1). That's it.
Sample:
$ echo "Volume.Free_IBM_LUN59_28D: 2072083693568" | sed 's/[^_]*_\(.*\):.*/\1/'
IBM_LUN59_28D
Here is one awk
awk -F"Free_" 'NF>1{split($2,a,":");print a[1]}'
Eks:
echo "Volume.Free_IBM_LUN59_28D: 2072083693568" | awk -F"Free_" 'NF>1{split($2,a,":");print a[1]}'
IBM_LUN59_28D
It divides the line by Free_.
If line then have more than one field NF>1 then:
Split second field bye : and print first part a[1]
With awk:
echo "$val" | awk -F: '{print $1}' | awk -F. '{print $2}' | awk '{print substr($0,6)}'
where the given string is in $val.