I have a comma-delimited file to which I want to append a string in specific columns. I am trying to do something like this, but couldn't do it until now.
re1,1,a1e,a2e,AGT
re2,2,a1w,a2w,AGT
re3,3,a1t,a2t,ACGTCA
re12,4,b1e,b2e,ACGTACT
And I want to append 'some_string' to columns 3 and 4:
re1,1,some_stringa1e,some_stringa2e,AGT
re2,2,some_stringa1w,some_stringa2w,AGT
re3,3,some_stringa1t,some_stringa2t,ACGTCA
re12,4,some_stringb1e,some_stringb2e,ACGTACT
I was trying something similar to the suggestion solution, but to no avail:
awk -v OFS=$'\,' '{ $3="some_string" $3; print}' $lookup_file
Also, I would like my string to be added to both columns. How would you do this with awk or bash?
Thanks a lot in advance
You can do that with (almost) what you have:
pax> echo 're1,1,a1e,a2e,AGT
re2,2,a1w,a2w,AGT
re3,3,a1t,a2t,ACGTCA
re12,4,b1e,b2e,ACGTACT' | awk 'BEGIN{FS=OFS=","}{$3 = "pre3:"$3; $4 = "pre4:"$4; print}'
re1,1,pre3:a1e,pre4:a2e,AGT
re2,2,pre3:a1w,pre4:a2w,AGT
re3,3,pre3:a1t,pre4:a2t,ACGTCA
re12,4,pre3:b1e,pre4:b2e,ACGTACT
The begin block sets the input and output field separators, the two assignments massage fields 3 and 4, and the print outputs the modified line.
You need to set FS to comma, not just OFS. There's a shortcut for setting FS, it's the -F option.
awk -F, -v OFS=',' '{ $3="some_string" $3; $4 = "some_string" $4; print}' "$lookup_file"
awk's default action is to concatenate, so you can simply place strings next to each other and they'll be treated as one. 1 means true, so with no {action} it will assume "print". You can use Bash's Brace Expansion to assign multiple variables after the script.
awk '{$3 = "three" $3; $4 = "four" $4} 1' {O,}FS=,
Related
I have semicolon-separated columns, and I would like to add some characters to a specific column.
aaa;111;bbb
ccc;222;ddd
eee;333;fff
to the second column I want to add '#', so the output should be;
aaa;#111;bbb
ccc;#222;ddd
eee;#333;fff
I tried
awk -F';' -OFS=';' '{ $2 = "#" $2}1' file
It adds the character but removes all semicolons with space.
You could use sed to do your job:
# replaces just the first occurrence of ';', note the absence of `g` that
# would have made it a global replacement
sed 's/;/;#/' file > file.out
or, to do it in place:
sed -i 's/;/;#/' file
Or, use awk:
awk -F';' '{$2 = "#"$2}1' OFS=';' file
All the above commands result in the same output for your example file:
aaa;#111;bbb
ccc;#222;ddd
eee;#333;fff
#atb: Try:
1st:
awk -F";" '{print $1 FS "#" $2 FS $3}' Input_file
Above will work only when your Input_file has 3 fields only.
2nd:
awk -F";" -vfield=2 '{$field="#"$field} 1' OFS=";" Input_file
Above code you could put any field number and could make it as per your request.
Here I am making field separator as ";" and then taking a variable named field which will have the field number in it and then that concatenating "#" in it's value and 1 is for making condition TRUE and not making and action so by default print action will happen of current line.
You just misunderstood how to set variables. Change -OFS to -v OFS:
awk -F';' -v OFS=';' '{ $2 = "#" $2 }1' file
but in reality you should set them both to the same value at one time:
awk 'BEGIN{FS=OFS=";"} { $2 = "#" $2 }1' file
I have a file which has the following form:
#id|firstName|lastName|gender|birthday|creationDate|locationIP|browserUsed
111|Arkas|Sarkas|male|1995-09-11|2010-03-17T13:32:10.447+0000|192.248.2.123|Midori
Every field is separated with "|". I am writing a shell script and my goal is to remove the "-" from the fifth field (birthday), in order to make comparisons as if they were numbers.
For example i want the fifth field to be like |19950911|
The only solution I have reached so far, deletes all the "-" from each line which is not what I want using sed.
i would be extremely grateful if you show me a solution to my problem using awk.
If this is a homework writing the complete script will be a disservice. Some hints: the function you should be using is gsub in awk. The fifth field is $5 and you can set the field separator by -F'|' or in BEGIN block as FS="|"
Also, line numbers are in NR variable, to skip first line for example, you can add a condition NR>1
An awk one liner:
awk 'BEGIN { FS="|" } { gsub("-","",$5); print }' infile.txt
To keep "|" as output separator, it is better to define OFS value as "|" :
... | awk 'BEGIN { FS="|"; OFS="|"} {gsub("-","",$5); print $0 }'
I have a line like:
one:two:three:four:five:six seven:eight
and I want to use awk to get $1 to be one and $2 to be two:three:four:five:six seven:eight
I know I can get it by doing sed before. That is to change the first occurrence of : with sed then awk it using the new delimiter.
However replacing the delimiter with a new one would not help me since I can not guarantee that the new delimiter will not already be somewhere in the text.
I want to know if there is an option to get awk to behave this way
So something like:
awk -F: '{print $1,$2}'
will print:
one two:three:four:five:six seven:eight
I will also want to do some manipulations on $1 and $2 so I don't want just to substitute the first occurrence of :.
Without any substitutions
echo "one:two:three:four:five" | awk -F: '{ st = index($0,":");print $1 " " substr($0,st+1)}'
The index command finds the first occurance of the ":" in the whole string, so in this case the variable st would be set to 4. I then use substr function to grab all the rest of the string from starting from position st+1, if no end number supplied it'll go to the end of the string. The output being
one two:three:four:five
If you want to do further processing you could always set the string to a variable for further processing.
rem = substr($0,st+1)
Note this was tested on Solaris AWK but I can't see any reason why this shouldn't work on other flavours.
Some like this?
echo "one:two:three:four:five:six" | awk '{sub(/:/," ")}1'
one two:three:four:five:six
This replaces the first : to space.
You can then later get it into $1, $2
echo "one:two:three:four:five:six" | awk '{sub(/:/," ")}1' | awk '{print $1,$2}'
one two:three:four:five:six
Or in same awk, so even with substitution, you get $1 and $2 the way you like
echo "one:two:three:four:five:six" | awk '{sub(/:/," ");$1=$1;print $1,$2}'
one two:three:four:five:six
EDIT:
Using a different separator you can get first one as filed $1 and rest in $2 like this:
echo "one:two:three:four:five:six seven:eight" | awk -F\| '{sub(/:/,"|");$1=$1;print "$1="$1 "\n$2="$2}'
$1=one
$2=two:three:four:five:six seven:eight
Unique separator
echo "one:two:three:four:five:six seven:eight" | awk -F"#;#." '{sub(/:/,"#;#.");$1=$1;print "$1="$1 "\n$2="$2}'
$1=one
$2=two:three:four:five:six seven:eight
The closest you can get with is with GNU awk's FPAT:
$ awk '{print $1}' FPAT='(^[^:]+)|(:.*)' file
one
$ awk '{print $2}' FPAT='(^[^:]+)|(:.*)' file
:two:three:four:five:six seven:eight
But $2 will include the leading delimiter but you could use substr to fix that:
$ awk '{print substr($2,2)}' FPAT='(^[^:]+)|(:.*)' file
two:three:four:five:six seven:eight
So putting it all together:
$ awk '{print $1, substr($2,2)}' FPAT='(^[^:]+)|(:.*)' file
one two:three:four:five:six seven:eight
Storing the results of the substr back in $2 will allow further processing on $2 without the leading delimiter:
$ awk '{$2=substr($2,2); print $1,$2}' FPAT='(^[^:]+)|(:.*)' file
one two:three:four:five:six seven:eight
A solution that should work with mawk 1.3.3:
awk '{n=index($0,":");s=$0;$1=substr(s,1,n-1);$2=substr(s,n+1);print $1}' FS='\0'
one
awk '{n=index($0,":");s=$0;$1=substr(s,1,n-1);$2=substr(s,n+1);print $2}' FS='\0'
two:three:four five:six:seven
awk '{n=index($0,":");s=$0;$1=substr(s,1,n-1);$2=substr(s,n+1);print $1,$2}' FS='\0'
one two:three:four five:six:seven
Just throwing this on here as a solution I came up with where I wanted to split the first two columns on : but keep the rest of the line intact.
Comments inline.
echo "a:b:c:d::e" | \
awk '{
split($0,f,":"); # split $0 into array of fields `f`
sub(/^([^:]+:){2}/,"",$0); # remove first two "fields" from `$0`
print f[1],f[2],$0 # print first two elements of `f` and edited `$0`
}'
Returns:
a b c:d::e
In my input I didn't have to worry about the first two fields containing escaped :, if that was a requirement, this solution wouldn't work as expected.
Amended to match the original requirements:
echo "a:b:c:d::e" | \
awk '{
split($0,f,":");
sub(/^([^:]+:)/,"",$0);
print f[1],$0
}'
Returns:
a b:c:d::e
I have a file that contains the following information
organic_apple;2;organic_apple_212_212
organic_tomato;3;organic_tomato_24_29
fruit_juice;5;fruit_juice_15_15
So i want a file that contains the output
organic_apple;2;organic_apple_212
organic_tomato;3;organic_tomato_24_29
fruit_juice;5;fruit_juice_15
compare the last two fields, if they are the same display it once , if not , display them both
I'm writing in unix bash using solaris
Regardless of the number of underscores, compare the last two:
awk 'BEGIN{FS=OFS="_"}$NF==$(NF-1){--NF;$1=$1}1' test.in
Try this :
awk -vOFS=_ -F_ '{if ($2 == $3) print $1, $2; else print $1, $2, $3}' file.txt
This script removes the last field, if it is equal to the one before last:
awk -F "_" '$NF==$(NF-1){$NF=""}1' file
I have a string containing this (field separator is the percentage sign), stored in a variable called data
201%jkhjfhn%kfhngjm%mkdfhgjdfg%mkdfhgjdfhg%mkdhfgjdhfg%kdfhgjgh%kdfgjhgfh%mkfgnhmkgfnh%k,gnhjkgfn%jkdfhngjdfng
I'm trying to print out that string replacing the percentage sign with a pipe but it seems harden than i thought:
echo ${data} | awk -F"%" 'BEGIN {OFS="|"} {print $0}'
I know I'm very close to it just not close enough.
I see that code as:
1 echo the variable value into a awk session
2 set field separator as "%"
3 set as output field separator "|"
4 print the line
Try this :
echo "$data" | awk -F"%" 'BEGIN {OFS="|"} {$1=$1; print $0}'
From awk manual
Finally, there are times when it is convenient to force awk to rebuild the entire
record, using the current value of the fields and OFS. To do this, use the seemingly
innocuous assignment:
$1 = $1 # force record to be reconstituted
print $0 # or whatever else with $0
Another lightweight way using only tr if you search an alternative for awk :
tr '%' '|' <<< "$data"
Sputnick gave you the awk solution, but you don't actually need awk at all, just use your shell:
echo ${data//%/|}