Printing the actual field delimiter value not the regular expression - regex

Given the following input:
check1;check2
check1;;check2
check1,check2
and the awk command:
awk -F';+|,' '{print $1 FS $2}'
FS should contain the selected delimiter?
How can you print the delimiter which is selected i.e. either of ;, ;; or , not the regular expression that the describes the delimiters.
If the input is check1;check2 then the output should be check1;check2.

If you're using GNU Awk (gawk) you can use the 4th argument of split():
gawk '{split($0, a, /;+|,/, seps); print a[1] seps[1] a[2]}' file
Output:
check1;check2
check1;;check2
check1,check2
Using it within a loop is also easy to handle:
gawk '{nf = split($0, a, /;+|,/, seps); for (i = 1; i <= nf; ++i) printf "%s%s", a[i], seps[i]; print ""}' file
22011,25029;;3331,25275
6740,16516;;27292,1217
13480,31488;;7947,18804
328,30623;;12470,6883
If you only need the fields you would only have to touch a. Separators would be separated in seps and the indices of those are aligned with a.

I don't think awk stores the matched delimiter anywhere. If you use GNU awk, you can do it yourself:
gawk '{match($0, /([^;,]*)(;+|,)(.*)/, a); print a[1], a[2], a[3]}'

GNU awk has this feature for records not fields so you could also do something like this:
$ awk '{printf "%s%s",$0,RT}' RS=';+|,|\n' file
check1;check2
check1;;check2
check1,check2
Where RT is the value match by RS for the given record which you can see by:
$ awk '{printf "%s",RT}' RS=';+|,|\n' file
;
;;
,

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

Removing multiple delimiters between outside delimiters on each line

Using awk or sed in a bash script, I need to remove comma separated delimiters that are located between an inner and outer delimiter. The problem is that wrong values ends up in the wrong columns, where only 3 columns are desired.
For example, I want to turn this:
2020/11/04,Test Account,569.00
2020/11/05,Test,Account,250.00
2020/11/05,More,Test,Accounts,225.00
Into this:
2020/11/04,Test Account,569.00
2020/11/05,Test Account,250.00
2020/11/05,More Test Accounts,225.00
I've tried to use a few things, testing regex:
But I cannot find a solution to only select the commas in order to remove.
awk -F, '{ printf "%s,",$1;for (i=2;i<=NF-2;i++) { printf "%s ",$i };printf "%s,%s\n",$(NF-1),$NF }' file
Using awk, print the first comma delimited field and then loop through the rest of the field up to the last but 2 field printing the field followed by a space. Then for the last 2 fields print the last but one field, a comma and then the last field.
With GNU awk for the 3rd arg to match():
$ awk -v OFS=, '{
match($0,/([^,]*),(.*),([^,]*)/,a)
gsub(/,/," ",a[2])
print a[1], a[2], a[3]
}' file
2020/11/04,Test Account,569.00
2020/11/05,Test Account,250.00
2020/11/05,More Test Accounts,225.00
or with any awk:
$ awk '
BEGIN { FS=OFS="," }
{
n = split($0,a)
gsub(/^[^,]*,|,[^,]*$/,"")
gsub(/,/," ")
print a[1], $0, a[n]
}
' file
2020/11/04,Test Account,569.00
2020/11/05,Test Account,250.00
2020/11/05,More Test Accounts,225.00
Use this Perl one-liner:
perl -F',' -lane 'print join ",", $F[0], "#F[1 .. ($#F-1)]", $F[-1];' in.csv
The Perl one-liner uses these command line flags:
-e : Tells Perl to look for code in-line, instead of in a file.
-n : Loop over the input one line at a time, assigning it to $_ by default.
-l : Strip the input line separator ("\n" on *NIX by default) before executing the code in-line, and append it when printing.
-a : Split $_ into array #F on whitespace or on the regex specified in -F option.
-F',' : Split into #F on comma, rather than on whitespace.
$F[0] : first element of the array #F (= first comma-delimited value).
$F[-1] : last element of #F.
#F[1 .. ($#F-1)] : elements of #F between the second from the start and the second from the end, inclusive.
"#F[1 .. ($#F-1)]" : the above elements, joined on blanks into a string.
join ",", ... : join the LIST "..." on a comma, and return the resulting string.
SEE ALSO:
perldoc perlrun: how to execute the Perl interpreter: command line switches
perl -pe 's{,\K.*(?=,)}{$& =~ y/,/ /r}e' file
sed -e ':a' -e 's/\(,[^,]*\),\([^,]*,\)/\1 \2/; t a' file
awk '{$1=$1","; $NF=","$NF; gsub(/ *, */,","); print}' FS=, file
awk '{for (i=2; i<=NF; ++i) $i=(i>2 && i<NF ? " " : ",") $i} 1' FS=, OFS= file
awk doesn't support look arounds, we could have it by using match function of awk; using that could you please try following, written and tested with shown samples in GNU awk.
awk '
match($0,/,.*,/){
val=substr($0,RSTART+1,RLENGTH-2)
gsub(/,/," ",val)
print substr($0,1,RSTART) val substr($0,RSTART+RLENGTH-1)
}
' Input_file
Yet another perl
$ perl -pe 's/(?:^[^,]*,|,[^,]*$)(*SKIP)(*F)|,/ /g' ip.txt
2020/11/04,Test Account,569.00
2020/11/05,Test Account,250.00
2020/11/05,More Test Accounts,225.00
(?:^[^,]*,|,[^,]*$) matches first/last field along with the comma character
(*SKIP)(*F) this would prevent modification of preceding regexp
|, provide , as alternate regexp to be matched for modification
With sed (assuming \n is supported by the implementation, otherwise, you'll have to find a character that cannot be present in the input)
sed -E 's/,/\n/; s/,([^,]*)$/\n\1/; y/,/ /; y/\n/,/'
s/,/\n/; s/,([^,]*)$/\n\1/ replace first and last comma with newline character
y/,/ / replace all comma with space
y/\n/,/ change newlines back to comma
A similar answer to Timur's, in awk
awk '
BEGIN { FS = OFS = "," }
function join(start, stop, sep, str, i) {
str = $start
for (i = start + 1; i <= stop; i++) {
str = str sep $i
}
return str
}
{ print $1, join(2, NF-1, " "), $NF }
' file.csv
It's a shame awk doesn't ship with a join function builtin

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

How to reverse all the words in a file with bash in Ubuntu?

I would like to reverse the complete text from the file.
Say if the file contains:
com.e.h/float
I want to get output as:
float/h.e.com
I have tried the command:
rev file.txt
but I have got all the reverse output: taolf/h.e.moc
Is there a way I can get the desired output. Do let me know. Thank you.
Here is teh link of teh sample file: Sample Text
You can use sed and tac:
str=$(echo 'com.e.h/float' | sed -E 's/(\W+)/\n\1\n/g' | tac | tr -d '\n')
echo "$str"
float/h.e.com
Using sed we insert \n before and after all non-word characters.
Using tac we reverse the output lines.
Using tr we strip all new lines.
If you have gnu-awk then you can do all this in a single awk command using 4 argument split function call that populates split strings and delimiters separately:
awk '{
s = ""
split($0, arr, /\W+/, seps)
for (i=length(arr); i>=1; i--)
s = s seps[i] arr[i]
print s
}' file
For non-gnu awk, you can use:
awk '{
r = $0
i = 0
while (match(r, /[^a-zA-Z0-9_]+/)) {
a[++i] = substr(r, RSTART, RLENGTH) substr(r, 0, RSTART-1)
r = substr(r, RSTART+RLENGTH)
}
s = r
for (j=i; j>=1; j--)
s = s a[j]
print s
}' file
Is it possible to use Perl?
perl -nlE 'say reverse(split("([/.])",$_))' f
This one-liner reverses all the lines of f, according to PO's criteria.
If prefer a less parentesis version:
perl -nlE 'say reverse split "([/.])"' f
For portability, this can be done using any awk (not just GNU) using substrings:
$ awk '{
while (match($0,/[[:alnum:]]+/)) {
s=substr($0,RLENGTH+1,1) substr($0,1,RLENGTH) s;
$0=substr($0,RLENGTH+2)
} print s
}' <<<"com.e.h/float"
This steps through the string grabbing alphanumeric strings plus the following character, reversing the order of those two captured pieces, and prepending them to an output string.
Using GNU awk's split, splitting from separators . and /, define more if you wish.
$ cat program.awk
{
for(n=split($0,a,"[./]",s); n>=1; n--) # split to a and s, use n from split
printf "%s%s", a[n], (n==1?ORS:s[(n-1)]) # printf it pretty
}
Run it:
$ echo com.e.h/float | awk -f program.awk
float/h.e.com
EDIT:
If you want to run it as one-liner:
awk '{for(n=split($0,a,"[./]",s); n>=1; n--); printf "%s%s", a[n], (n==1?ORS:s[(n-1)])}' foo.txt

Regex with awk or gawk

I'm a beginner user of awk/gawk.
If I run below, the shell gives me nothing. Please help!
echo "A=1,B=2,3,C=,D=5,6,E=7,8,9"|awk 'BEGIN{
n = split($0, arr, /,(?=\\w+=)/)
for (x=1; x<n; x++) printf "arr[%d]=%s\n", x, arr[x]
}'
.....................................................
I am trying to parse:
A=1,B=2,3,C=,D=5,6,E=7,8,9
Expected Output:
A=1
B=2,3
C=
D=5,6
E=7,8,9
I bet there's something wrong with my awk.
gawk doesn't support look-ahead.
if you want gawk to parse it as you expected, try this:
awk '{n=split(gensub(/,([A-Z])/, " \\1","g" ),arr," ");for(x=1;x<=n;x++)print arr[x]}'
test with your example:
kent$ echo "A=1,B=2,3,C=,D=5,6,E=7,8,9"|awk '{n=split(gensub(/,([A-Z])/, " \\1","g" ),arr," ");for(x=1;x<=n;x++)print arr[x]}'
A=1
B=2,3
C=
D=5,6
E=7,8,9
This might be easier with sed:
$ echo "A=1,B=2,3,C=,D=5,6,E=7,8,9" | sed 's/,\(\w\+=\)/\n\1/g'
A=1
B=2,3
C=
D=5,6
E=7,8,9
If you are using gnu awk, you could do:
awk '{printf $0 "\n" substr( RT, 2 )}' RS=,[A-Z]
As nhahtdh, theres is no lookahead in awk... But you can use a different separator for the assignments. Why not "A=1;B=2,3,4;C=5..."?
If your input must have that format, try flex...
You could also use comma as the record separator:
echo "A=1,B=2,3,C=,D=5,6,E=7,8,9" |
awk -v RS=, '{sep=","} /=/ {sep="\n"} NR==1 {sep=""} {printf "%s%s", sep, $0}'
outputs
A=1
B=2,3
C=
D=5,6
E=7,8,9
You have two problems. First, you don't want a BEGIN clause; you just want this to run on every input line. Second, you are trying to use regular expression features that AWK does not support.
Instead of trying to use a fancy pattern that splits the string, loop and call match() to parse out the features you want.
echo "A=1,B=2,3,C=,D=5,6,E=7,8,9"|awk '
{
line = $0
for (i = 0;;)
{
i = match(line, /([A-Z]+)=([0-9,]*)(,|$)/, arr)
if (0 == i)
break
key = arr[1]
value = arr[2]
l = length(key "=" value ",") + 1
line = substr(line, l)
printf "DEBUG: key '%s' value '%s'\n", key, value
}
}'
This prints:
DEBUG: key A value 1
DEBUG: key B value 2,3
DEBUG: key C value
DEBUG: key D value 5,6
DEBUG: key E value 7,8,9
Other way using awk
awk '{print gensub(/,([A-Z]+=)/, "\n\\1","g")}' temp.txt
Output
A=1
B=2,3
C=
D=5,6
E=7,8,9