How to print a line with no field separator in awk? - regex

I have data like this (file is called list-in.dat)
a ; b ; c ; i
d
e ; f ; a ; b
g ; h ; i
and I want a list like this (output file list-out.dat) with all items, in alphabetically order (case insensitive) and each unique item only once.
a
b
c
d
e
f
g
h
i
My attempt is:
awk -F " ; " ' BEGIN { OFS="\n" ; } {for(i=0; i<=NF; i++) print $i} ' file-in.dat | uniq | sort -uf > file-out.dat
But I end up with all antries except those lines which has only one item:
a
b
c
e
f
g
h
i
How can I get all (unique, sorted) items no matter how many items are in one line / if the field separator is missing?

Using gnu-awk:
awk -F '[[:blank:]]*;[[:blank:]]*' '{
for (i=1; i<=NF; i++) uniq[$i]
}
END {
PROCINFO["sorted_in"]="#ind_str_asc"
for (i in uniq)
print i
}' file
a
b
c
d
e
f
g
h
i
For non-gnu awk use:
awk -F '[[:blank:]]*;[[:blank:]]*' '{for (i=1; i<=NF; i++) uniq[$i]}
END{for (i in uniq) print i}' file | sort

awk -F' ; ' -v OFS='\n' '{$1=$1} 1' ip.txt | sort -fu
-F' ; ' sets space followed by ; followed by space as field separator
-v OFS='\n' sets newline as output field separator
{$1=$1} change $0 as per new OFS
1 print $0
sort -fu sort uniquely ignoring case in alphabetic order

Could you please try following, awk + sort solution, written and tested with shown samples. In case you want to use ignorecase then add IGNORECASE=1 in awk code.
awk '
BEGIN{
FS=" ; "
}
{
for(i=1;i<=NF;i++){
if(!a[$i]++){ print $i }
}
}
' Input_file | sort
Explanation: Adding detailed explanation for above.
awk ' ##Starting awk program from here.
BEGIN{ ##Starting BEGIN section of this program from here.
FS=" ; " ##Setting field separator as space semi-colon space here.
}
{
for(i=1;i<=NF;i++){ ##Starting a for loop till NF here for each line.
if(!a[$i]++){ print $i } ##Checking condition if current field is NOT present in array a then printing that field value here.
}
}
' Input_file | sort ##Mentioning Input_file name here and passing it to sort as Input to sort the data.

Related

awk sub with a capturing group into the replacement

I am writing an awk oneliner for this purpose:
file1:
1 apple
2 orange
4 pear
file2:
1/4/2/1
desired output: apple/pear/orange/apple
addendum: Missing numbers should be best kept unchanged 1/4/2/3 = apple/pear/orange/3 to prevent loss of info.
Methodology:
Build an associative array key[$1] = $2 for file1
capture all characters between the slashes and replace them by matching to the key of associative array eg key[4] = pear
Tried:
gawk 'NR==FNR { key[$1] = $2 }; NR>FNR { r = gensub(/(\w+)/, "key[\\1]" , "g"); print r}' file1.txt file2.txt
#gawk because need to use \w+ regex
#gensub used because need to use a capturing group
Unfortunately, results are
1/4/2/1
key[1]/key[4]/key[2]/key[1]
Any suggestions? Thank you.
You may use this awk:
awk -v OFS='/' 'NR==FNR {key[$1] = $2; next}
{for (i=1; i<=NF; ++i) if ($i in key) $i = key[$i]} 1' file1 FS='/' file2
apple/pear/orange/apple
Note that if numbers from file2 don't exist in key array then it will make those fields empty.
file1 FS='/' file2 will keep default field separators for file1 but will use / as field separator while reading file2.
EDIT: In case you don't have a match in file2 from file and you want to keep original value as it is then try following:
awk '
FNR==NR{
arr[$1]=$2
next
}
{
val=""
for(i=1;i<=NF;i++){
val=(val=="" ? "" : val FS) (($i in arr)?arr[$i]:$i)
}
print val
}
' file1 FS="/" file2
With your shown samples please try following.
awk '
FNR==NR{
arr[$1]=$2
next
}
{
val=""
for(i=1;i<=NF;i++){
val = (val=="" ? "" : val FS) arr[$i]
}
print val
}
' file1 FS="/" file2
Explanation: Reading Input_file1 first and creating array arr with index of 1st field and value of 2nd field then setting field separator as / and traversing through each field os file2 and saving its value in val; printing it at last for each line.
Like #Sundeep comments in the comments, you can't use backreference as an array index. You could mix match and gensub (well, I'm using sub below). Not that this would be anywhere suggested method but just as an example:
$ awk '
NR==FNR {
k[$1]=$2 # hash them
next
}
{
while(match($0,/[0-9]+/)) # keep doing it while it lasts
sub(/[0-9]+/,k[substr($0,RSTART,RLENGTH)]) # replace here
}1' file1 file2
Output:
apple/pear/orange/apple
And of course, if you have k[1]="word1", you'll end up with a neverending loop.
With perl (assuming key is always found):
$ perl -lane 'if(!$#ARGV){ $h{$F[0]}=$F[1] }
else{ s|[^/]+|$h{$&}|g; print }' f1 f2
apple/pear/orange/apple
if(!$#ARGV) to determine first file (assuming exactly two files passed)
$h{$F[0]}=$F[1] create hash based on first field as key and second field as value
[^/]+ match non / characters
$h{$&} get the value based on matched portion from the hash
If some keys aren't found, leave it as is:
$ cat f2
1/4/2/1/5
$ perl -lane 'if(!$#ARGV){ $h{$F[0]}=$F[1] }
else{ s|[^/]+|exists $h{$&} ? $h{$&} : $&|ge; print }' f1 f2
apple/pear/orange/apple/5
exists $h{$&} checks if the matched portion exists as key.
Another approach using awk without loop:
awk 'FNR==NR{
a[$1]=$2;
next
}
$1 in a{
printf("%s%s",FNR>1 ? RS: "",a[$1])
}
END{
print ""
}' f1 RS='/' f2
$ cat f1
1 apple
2 orange
4 pear
$ cat f2
1/4/2/1
$ awk 'FNR==NR{a[$1]=$2;next}$1 in a{printf("%s%s",FNR>1?RS:"",a[$1])}END{print ""}' f1 RS='/' f2
apple/pear/orange/apple

How to replace second pattern(dot) after pattern(comma) in bash

How do i replace second dot after comma.
this is the closest i could go
echo '0.592922148,0.821504176,1.174.129.731' | xargs -d ',' -n1 echo | sed 's/\([^\.]*\.[^\.]*\)\./\1/' | sed 's/\([^\.]*\.[^\.]*\)\./\1/'
Output :
0.592922148
0.821504176
1.174129731
Expected output :
0.592922148,0.821504176,1.174129731
You may use
sed -e ':a' -e 's/\(\.[^.,]*\)\./\1/' -e 't a'
See online sed demo:
s='0.592922148,0.821504176,1.174.129.731'
sed -e ':a' -e 's/\(\.[^.,]*\)\./\1/' -e 't a' <<< "$s"
Details
:a - label a
s/\(\.[^.,]*\)\./\1/ - finds and captures into Group 1 a dot, then any 0+ chars other than dot and comma, and then just matches a dot, and replaces this match with the value in Group 1 (thus, removing the second matched dot)
t a - if there was a successful replacement, goes back to the a label position in the string.
While I think the sed solution is your best choice, since you have tagged your question with both sed and awk, an awk solution is fairly straight forward as well using split() and basic string concatenation. (just not nearly as short) For example you could do:
awk -v OFS=, -F, '{
for (i=1; i<=NF; i++) {
n=split ($i, a,".")
if (n > 2) {
s=a[1] "." a[2]
for (j=3; j<=n; j++)
s = s a[j]
$i=s
}
}
}1'
Where you define the field separator and output field separators as ','. Then looping over each field, check the return of split(), splitting the field into an array on '.' into array a. If the resulting number of elements is greater than 2, then put your first two elements back together restoring the first '.' in the number, and then simply concatenate the remaining fields. The 1 at the end is the default "print record" to print the updated record.
Example Use/Output
$ echo '0.592922148,0.821504176,1.174.129.731' |
> awk -v OFS=, -F, '{
> for (i=1; i<=NF; i++) {
> n=split ($i, a,".")
> if (n > 2) {
> s=a[1] "." a[2]
> for(j=3;j<=n;j++)
> s = s a[j]
> $i=s
> }
> }
> }1'
0.592922148,0.821504176,1.174129731
Could you please try following.
echo '0.592922148,0.821504176,1.174.129.731' |
awk '
BEGIN{
FS=OFS=","
}
{
for(i=1;i<=NF;i++){
ind=index($i,".")
if(ind){
val1=substr($i,1,ind)
val2=substr($i,ind+1)
gsub(/\./,"",val2)
$i=val1 val2
}
}
val1=val2=""
}
1'
Explanation: Adding explanation for above code.
echo '0.592922148,0.821504176,1.174.129.731' | ##Printing values as per OP mentioned and using pipe to send its output as standard input for awk command.
awk ' ##Starting awk program from here.
BEGIN{ ##Starting BEGIN section of this program here.
FS=OFS="," ##Setting FS and OFS as comma for each line of Input_file here.
} ##Closing BEGIN BLOCK here.
{
for(i=1;i<=NF;i++){ ##Starting a for loop to traverse through fields of line..
ind=index($i,".") ##Checking index of DOT in current field and saving it into ind variable.
if(ind){ ##Checking condition if variable ind is NOT NULL.
val1=substr($i,1,ind) ##Creating variable val1 from sub-string in current field from 1 to ind value.
val2=substr($i,ind+1) ##Creating variable val2 from sub-string in current field from ind+1 value to till complete length of current field.
gsub(/\./,"",val2) ##Globally substituting DOTs with NULL in val2 variable.
$i=val1 val2 ##Re-crearing current field with value of val1 val2.
} ##Closing BLOCK for if condition.
} ##Closing BLOCK for for loop.
val1=val2="" ##Nullifying val1 and val2 variables here.
} ##Closing main code BLOCK here.
1' ##Mentioning 1 will print edited/non-edited line.
An awk verison:
echo '0.592922148,0.821504176,1.174.129.731' | awk -F, '{for (i=1;i<=NF;i++) {sub(/\./,"#",$i);gsub(/\./,"",$i);sub(/#/,".",$i);print $i}}'
0.592922148
0.821504176
1.174129731
It splits the line inn to multiple fields by ,. Then replace first . to #. Then replace rest of . to nothing. Last replace # back to . and print it.
Edit
awk -F, '{for (i=1;i<=NF;i++) {sub(/\./,"#",$i);gsub(/\./,"",$i);sub(/#/,".",$i);a=a (i==1?"":",")$i}print a}' file
0.592922148,0.821504176,1.174129731

AWK - add value based on regex

I have to add the numbers returned by REGEX using awk in linux.
Basically from this file:
123john456:x:98:98::/home/john123:/bin/bash
I have to add the numbers 123 and 456 using awk.
So the result would be 579
So far I have done the following:
awk -F ':' '$1 ~ VAR+="/[0-9].*(?=:)/" ; {print VAR}' /etc/passwd
awk -F ':' 'VAR+="/[0-9].*(?=:)/" ; {print VAR}' /etc/passwd
awk -F ':' 'match($1, VAR=/[0-9].*?:/) ; {print VAR}' /etc/passwd
And from what I've seen match doesn't support this at all.
Does someone has any idea?
UPDATE:
it also should work for
john123 result - > 123
123john result - > 123
$ awk -F':' '{split($1,t,/[^0-9]+/); print t[1] + t[2]}' file
579
With your updated requirements:
$ cat file
123john456:x:98:98::/home/john123:/bin/bash
john123:x:98:98::/home/john123:/bin/bash
123john:x:98:98::/home/john123:/bin/bash
$ awk -F':' '{split($1,t,/[^0-9]+/); print t[1] + t[2]}' file
579
123
123
With gawk and for the given example
awk -F ':' '{a=gensub(/[a-zA-Z]+/,"+", "g", $1); print a}' inputFile | bc
would do the job.
More general:
awk -F ':' '{a=gensub(/[a-zA-Z]+/,"+", "g", $1); a=gensub(/^+/,"","g",a); a=gensub(/+$/,"","g",a); print a}' inputFile | bc
The regex-part replaces all sequences of letters with '+' (e.g., '12johnny34' becomes 12+34). Finally, this mathematical operation is evaluated by bc.
(The be safe, I remove leading and trailing '+' sings by ^+ and +$)
You may use
awk -F ':' '{n=split($1, a, /[^0-9]+/); b=0; for (i=1;i<=n;i++) { b += a[i]; }; print b; }' /etc/passwd
See online awk demo
s="123john456:x:98:98::/home/john123:/bin/bash
john123:x:98:98::/home/john123:/bin/bash"
awk -F ':' '{n=split($1, a, /[^0-9]+/); b=0; for (i=1;i<=n;i++) { b += a[i]; }; print b; }' <<< "$s"
Output:
579
123
Details
-F ':' - records are split into fields with : char
n=split($1, a, /[^0-9]+/) - gets Field 1 and splits into digit only chunks saving the numbers in a array and the n var contains the number of these chunks
b=0 - b will hold the sum
for (i=1;i<=n;i++) { b += a[i]; } - iterate over a array and sum the values
print b - prints the result.
I used awk's split() to separate the first field on any string not containing numbers.
split(string, target_array, [regex], [separator_array]*)
*separator_array requires gawk
$ awk -F: '{split($1, A, /[^0-9]+/, S); print S[1], A[1]+A[2]}' <<EOF
123john456:x:98:98::/home/john123:/bin/bash
123john:x:98:98::/home/john123:/bin/bash
EOF
john 579
john 123
You can use [^0-9]+ as a field separator, and :[^\n]*\n as a record separator instead:
awk -F '[^0-9]+' 'BEGIN{RS=":[^\n]*\n"}{print $1+$2}' /etc/passwd
so that given the content of /etc/passwd being:
123john456:x:98:98::/home/john123:/bin/bash
john123:x:98:98::/home/john123:/bin/bash
123john:x:98:98::/home/john123:/bin/bash
This outputs:
579
123
123
You can try Perl also
$ cat johnny.txt
123john456:x:98:98::/home/john123:/bin/bash
john123:x:98:98::/home/john123:/bin/bash
123john:x:98:98::/home/john123:/bin/bash
$ perl -F: -lane ' $_=$F[0]; $sum+= $1 while(/(\d+)/g); print $sum; $sum=0 ' johnny.txt
579
123
123
$
Here is another awk variant that adds all the numbers present in first field separated by ::
cat file
123john456:x:98:98::/home/john123:/bin/bash
john123:x:98:98::/home/john123:/bin/bash
123john:x:98:98::/home/john123:/bin/bash
1j2o3h4n5:x:98:98::/home/john123:/bin/bash
awk -F '[^0-9:]+' '{s=0; for (i=1; i<=NF; i++) {s+=$i; if ($i~/:$/) break} print s}' file
579
123
123
15

Dynamically generated regex for gsub not working

I have an input CSV file:
1,5,1
1,6,2
1,5,3
1,7,4
1,5,5
1,6,6
1,6,7
I need to create a string out of this as follows:
;5,1,3,5;6,2,6,7;7,4
So each character, except the first which is the value of the field $2, in the substring in between the ; denotes the row number of middle field; for example ;5,1,3,5 means that 5 is at row number 1,3,5.
I've been trying to use awk with gsub, trying to create the string MYSTR dynamically.
The regex inside the gsub is not working. I need a regex that will match ;$3 (the value of $3, which can be a two digit number) and replace it with ;$3,RowNO, if the pattern is not matched then add ;$3 at the end of the string.
This is what I have so far:
awk -F',' '{
print NR, $3;
noofchars=gsub(/;$3/,";"$3","NR,MYSTR);
print noofchars;
if ( noofchars == 1 )
;
else
MYSTR=MYSTR";"$3","NR;
print NR, $3;
print MYSTR;
}
END{print MYSTR;}' $1
The regex doesn't work because $3 isn't interpreted as the field #3 value but is seen as the anchor $ (that matches the end of the line) and a literal 3.
You can do it without gsub:
awk -F, '{a[$2]=a[$2]","NR}END{for (i in a){printf(";%d%s",i,a[i])}}'
Input
$ cat file
1,5,1
1,6,2
1,5,3
1,7,4
1,5,5
1,6,6
1,6,7
Output
$ awk -F, '{gsub(/[ ]+/,"",$3);a[$2] = ($2 in a ? a[$2]:$2) FS $3 }END{for(i in a)printf("%s%s",";",a[i]); print ""}' file
;5,1,3,5;6,2,6,7;7,4
Better Readable version
awk -F, '
{
gsub(/[ ]+/,"",$3); # suppress space char in third field
a[$2] = ($2 in a ? a[$2]:$2) FS $3 # array a where index being field2 and value will be field3, if index exists before append string with existing value
}
END{
for(i in a) # loop through array a and print values
printf("%s%s",";",a[i]);
print ""
}
' file
#vsshekhar: Try following too: It will provide you values in the correct same order which Input_file ($2) are coming.
awk -F, '{A[++i]=$2;B[A[i]]=B[A[i]]?B[A[i]] "," FNR:FNR} END{for(j=1;j<=i;j++){if(B[A[j]]){printf(";%s,%s",A[j],B[A[j]]);delete B[A[j]]}};print ""}' Input_file
Adding a non-one liner form of solution too now.
awk -F, '{
A[++i]=$2;
B[A[i]]=B[A[i]]?B[A[i]] "," FNR:FNR
}
END{
for(j=1;j<=i;j++){
if(B[A[j]]){
printf(";%s,%s",A[j],B[A[j]]);
delete B[A[j]]
}
};
print ""
}
' Input_file

replace nth occurrence of character in a file using awk regardless of the line

I am trying to replace the nth occurrence of a character or string regardless of the line using awk.
So if our data was this
|||||||
||||||
|||||
|||
and we were trying to replace | with A
then the output should look like this, assuming we want to replace every 3rd occurance
||A||A|
|A||A|
|A||A
||A
The current awk command I am using is this
awk '/|/{c++;if(c==3){sub(/|/,"A");c=0}}1' test.data
and it wrongly outputs this
|||||||
||||||
A||||
|||
also the data can look like this
|||xfsafrwe|||asfasdf|
|safasf|||asfasdf||
||asfasf|||
|||
and the result of course is this
||Axfsafrwe||Aasfasdf|
|safasfA||asfasdfA|
|Aasfasf||A
||A
Thanks
With GNU awk:
awk '{
for (i = 0; ++i <= NF;)
++c % n || $i = v
}1' OFS= FS= n=3 v=A infile
Adjusted after OP clarification:
awk '{
for (i = 0; ++i <=NF;)
if ($i == o)
++C % c || $i = n
} 1' FS= OFS= c=3 o=\| n=A infile