I know 2 things about awk:
1.
PAT='aGeneName'
awk -v var="$PAT" '$3 ~ var {print $0}' file.txt # will print the line where 3rd field includes the variable $PAT
2.
awk '$3 ~ /^aGeneName/' file.txt # will print the line where 3rd field starts with string "aGeneName"
But what I want is the combination of these two: I want to print the line where the 3rd field starts with the variable $PAT, something like
PAT='aGeneName'
awk -v var="$PAT" '$3 ~ /^var/ {print $0}' file.txt # but this is wrong, since variable can't be put into //
One way is like this:
PAT='aGeneName'
awk -v var="$PAT" '$3 ~ "^" var {print $0}' file.txt
And the {print $0} can be saved here, it's implied.
Another way, when the pattern var is a simple string, no RegEX character inside:
PAT='aGeneName'
awk -v var="$PAT" 'index($3, var)==1' file.txt
Related
This command works. It outputs the field separator (in this case, a comma):
$ echo "hi,ho"|awk -F, '/hi/{print $0}'
hi,ho
This command has strange output (it is missing the comma):
$ echo "hi,ho"|awk -F, '/hi/{$2="low";print $0}'
hi low
Setting the OFS (output field separator) variable to a comma fixes this case, but it really does not explain this behaviour.
Can I tell awk to keep the OFS?
When you modify the line ($0) awk re-constructs all columns and puts the value of OFS between them which by default is space. You modified the value of $2 which means you forced awk to re-evaluate$0.
When you print the line as is using $0 in your first case, since you did not modify any fields, awk did not re-evaluated each field and hence the field separator is preserved.
In order to preserve the field separator, you can specify that using:
BEGIN block:
$ echo "hi,ho" | awk 'BEGIN{FS=OFS=","}/hi/{$2="low";print $0}'
hi,low
Using -v option:
$ echo "hi,ho" | awk -F, -v OFS="," '/hi/{$2="low";print $0}'
hi,low
Defining at the end of awk:
$ echo "hi,ho" | awk -F, '/hi/{$2="low";print $0}' OFS=","
hi,low
You first example does not change anything, so all is printed out as the input.
In second example, it change the line and it will use the default OFS, that is (one space)
So to overcome this:
echo "hi,ho"|awk -F, '/hi/{$2="low";print $0}' OFS=","
hi,low
In your BEGIN action, set OFS = FS.
I would like to modify a file by including the size of following line using awk.
My file is like this:
>AAAS:1220136:1220159:-:0::NW_015494524.1:1220136-1220159(-)
ATGTCGATGCTCGATC
>AAAS::1215902:1215986:-:1::NW_015494524.1:1215902-1215986(-)
ATGCGATGCTAGCTAGCTCGAT
>AAAS:1215614:1215701:-:1::NW_015494524.1:1215614-1215701(-)
ATGCCGCGACGCAGCACCCGACGCGCAG
I am using awk to modify it to have the following format:
>Assembly_AAAS_1_16
ATGTCGATGCTCGATC
>Assembly_AAAS_2_22
ATGCGATGCTAGCTAGCTCGAT
>Assembly_AAAS_3_28
ATGCCGCGACGCAGCACCCGACGCGCAG
I have used awk to modify the first part.
awk -F":" -v i=1 '/>/{print ">Assembly_" $1 "_" val i "_";i++;next} {print length($0)} 1' infile | sed -e "s/_>/_/g" > outfile
I can use print length($0) but how to print it in the same line?
Thanks
EDIT2: Since OP has changed the sample data again so adding this code now.
awk -v val="Assembly_AAAS_" '/>/{++i;val=">"val i "_";next} {sub(/ +$/,"");print val length($0) ORS $0}' Input_file
OR
awk -v val="Assembly_AAAS_" '/>/{++i;val=">"val i "_";next} {print val length($1) ORS $0;}' Input_file
Above will remove spaces from last of the lines of Input_file, in case you don't need it then remove sub(/ +$/,""); part from above code please.
EDIT: As per OP changed solution now.
awk -v i=1 -v val=">Assembly_GeneName1_" -v val1="_sizeline" '/>/{value="\047" val i val1;i++;next} {print value length($0) ORS $0}' Input_file
OR
awk -v i=1 -v val=">Assembly_GeneName1_" -v val1="_sizeline" '
/>/{ value="\047" val i val1;
i++;
next}
{
print value length($0) ORS $0
}
' Input_file
Following awk may help you on same.
awk -v i="" -v j=2 '/>/{print "\047>Assembly_GeneName1_"++i"_sizeline"j;j+=2;next} 1' Input_file
Solution 2nd:
awk -v i=1 -v j=2 -v val=">Assembly_GeneName1_" -v val1="_sizeline" '/>/{print "\047" val i val1 j;j+=2;i++;next} 1' Input_file
What you are dealing with is a beautiful example of records which are not lines. awk is a record parser and by default, a record is defined to be a line. With awk you can define a record to be a block of text using the record separator RS.
RS : The first character of the string value of RS shall be the input record separator; a <newline> by default. If RS contains more
than one character, the results are unspecified. If RS is null, then
records are separated by sequences consisting of a <newline> plus one
or more blank lines, leading or trailing blank lines shall not result
in empty records at the beginning or end of the input, and a <newline>
shall always be a field separator, no matter what the value of FS is.
So the goal is to define the record to be
AAAS:1220136:1220159:-:0::NW_015494524.1:1220136-1220159(-)
ATGTCGATGCTCGATC
And this can be done by defining the RS="\n<". Furthremore we will use \n as a field separator FS. This way you can get the requested length as length($2) and the count by using the record count NR.
A simple awk script is then:
awk 'BEGIN{RS="\n<"; FS=OFS="\n"}
{$1=">Assembly_AAAS_"NR"_"length($2)}
{print $1,$2}' <file>
This will do exactly what you want.
note: we use print $1,$2 and not print $0 as the last record might have 3 fields (if the last char of the file is a newline). This would imply that you would have an extra empty line at the end of your file.
If you want to pick the AAAS string out of $1 you can use substr($1,1,match($1,":")-1) to pick it up. This results in this:
awk 'BEGIN{RS="\n<"; FS=OFS="\n"}
{$1=">Assembly_"substr($1,1,match($1,":")-1)"_"NR"_"length($2)}
{print $1,$2}' <file>
Finally, be aware that the above solution only works if there are no spaces in $2, if you want to change that, you can do this :
awk 'BEGIN{RS="\n<"; FS=OFS="\n"}
{ gsub(/[[:blank:]]/,"",$2);
$1=">Assembly_"substr($1,1,match($1,":")-1)"_"NR"_"length($2)
}
{ print $1,$2 }' <file>
I know that initialising FS in BEGIN is the correct practice but what if i need different field seperators for different lines(lines containing a particular pattern)? eg: my awk script is
{if($0 ~ /.*youtube.*/){FS="=";print $2}}
This code is not processing the first line.How to fix this?
You can use split. Eks get the middle date from third field green
echo "on,cat ,blue|green|red,more" | awk -F, '{split($3,a,"|");print a[2]}'
green
And you BEGIN block is not only where you can set the Field Separator:
echo "on,two,three" | awk -F, '{print $2}'
echo "on,two,three" | awk '{print $2}' FS=,
echo "on,two,three" | awk 'BEGIN{FS=","} {print $2}'
echo "on,two,three" | awk -v FS=, '{print $2}'
All these will print two
But they may have some different impact in when they can be used.
awk -F, 'BEGIN{print FS}'
,
and this does not work and gives no output.
awk 'BEGIN{print FS}' FS=,
Back to your problem:
This:
awk '{if($0 ~ /.*youtube.*/){FS="=";print $2}}' file
should be:
awk '{if($0 ~ /.*youtube.*/){split($0,a,"=");print a[2]}}' file
You do not need to test for any characters before and after regex, so:
awk '{if($0 ~ /youtube/){split($0,a,"=");print a[2]}}' file
And this could even more be simplified:
awk '/youtube/ {split($0,a,"=");print a[2]}' file
If data is like this:
cat file
youtube=thisisyoutube1 //starts here
youtube=thisisyoutube2
youtube=thisisyoutube3
youtube=thisisyoutube4
yautube=thisisnottobeprinted
Then do like this:
awk -F= '/youtube/ {split($2,a," ");print a[1]}' file
thisisyoutube1
thisisyoutube2
thisisyoutube3
thisisyoutube4
This command works. It outputs the field separator (in this case, a comma):
$ echo "hi,ho"|awk -F, '/hi/{print $0}'
hi,ho
This command has strange output (it is missing the comma):
$ echo "hi,ho"|awk -F, '/hi/{$2="low";print $0}'
hi low
Setting the OFS (output field separator) variable to a comma fixes this case, but it really does not explain this behaviour.
Can I tell awk to keep the OFS?
When you modify the line ($0) awk re-constructs all columns and puts the value of OFS between them which by default is space. You modified the value of $2 which means you forced awk to re-evaluate$0.
When you print the line as is using $0 in your first case, since you did not modify any fields, awk did not re-evaluated each field and hence the field separator is preserved.
In order to preserve the field separator, you can specify that using:
BEGIN block:
$ echo "hi,ho" | awk 'BEGIN{FS=OFS=","}/hi/{$2="low";print $0}'
hi,low
Using -v option:
$ echo "hi,ho" | awk -F, -v OFS="," '/hi/{$2="low";print $0}'
hi,low
Defining at the end of awk:
$ echo "hi,ho" | awk -F, '/hi/{$2="low";print $0}' OFS=","
hi,low
You first example does not change anything, so all is printed out as the input.
In second example, it change the line and it will use the default OFS, that is (one space)
So to overcome this:
echo "hi,ho"|awk -F, '/hi/{$2="low";print $0}' OFS=","
hi,low
In your BEGIN action, set OFS = FS.
Configuration.xml has "mysearchstring" at position (line) 23.
If I use the following statement, it returns me line 23
awk '/"mysearchstring"/{print NR}' Configuration.xml
But if I use an assigned variable, it returns me nothing
str="mySearchString";awk '/$str/{print NR}' Configuration.xml
Can someone tell me what is incorrect in the second statement?
You need to pass the variable to awk with -v and then use the ~ comparison:
awk -v myvar="$str" '$0 ~ myvar {print NR}' Configuration.xml
Example
$ cat a
hello
how
are
you
$ awk '/e/ {print NR}' a <---- hardcoded
1
3
$ awk -v myvar="e" '$0~myvar {print NR}' a <---- through variable
1
3
You can use the command-line option -v to pass variables to awk:
awk -v searchstr="$str" '$0 ~ searchstr { print NR }'
Or you can pass variable like this, after code:
awk '$0~s {print NR}' s="$str" file