Replace every other occurence on the same line - sed

To get pair coordinates, I need to bundle them with a comma.
Example input (space divided)
161768.669 467646.462 161716.059 467515.711 161683.343 467434.35 161696.535 467427.091 161713.457 467416.235
161719.332 467412.183 161732.58 467403.047 161763.757 467381.118 161791.136 467367.094 161808.447 467361.612 161834.383 467353.808
161842.354 467351.71 161870.358 467344.337
161886.521 467338.28 161905.239 467330.484 161908.865 467328.974 161972.72 467301.688 161984.372 467296.404 162000.0 467288.568 162056.152 467260.414
162065.771 467254.859 162102.992 467233.363 162138.102 467213.086
162146.165 467209.29 162165.68 467200.102 162165.684 467200.1 162270.285 467154.666 162318.855 467133.104 162358.101 467117.844 162165.004 466820.119 162176.208 466809.875
This attempt only replaces the first occurrence on every line
sed 's/ /,/1' file
161768.669,467646.462 161716.059 467515.711 161683.343 467434.35 161696.535 467427.091 161713.457 467416.235
161719.332,467412.183 161732.58 467403.047 161763.757 467381.118 161791.136 467367.094 161808.447 467361.612 161834.383 467353.808
161842.354,467351.71 161870.358 467344.337
161886.521,467338.28 161905.239 467330.484 161908.865 467328.974 161972.72 467301.688 161984.372 467296.404 162000.0 467288.568 162056.152 467260.414
162065.771,467254.859 162102.992 467233.363 162138.102 467213.086
162146.165,467209.29 162165.68 467200.102 162165.684 467200.1 162270.285 467154.666 162318.855 467133.104 162358.101 467117.844 162165.004 466820.119 162176.208 466809.875
Desired output
161768.669,467646.462 161716.059,467515.711 161683.343,467434.35 161696.535,467427.091 161713.457,467416.235
161719.332,467412.183 161732.58,467403.047 161763.757,467381.118 161791.136,467367.094 161808.447,467361.612 161834.383,467353.808
161842.354,467351.71 161870.358,467344.337
161886.521,467338.28 161905.239,467330.484 161908.865,467328.974 161972.72,467301.688 161984.372,467296.404 162000.0,467288.568 162056.152,467260.414
162065.771,467254.859 162102.992,467233.363 162138.102,467213.086
162146.165,467209.29 162165.68,467200.102 162165.684,467200.1 162270.285,467154.666 162318.855,467133.104 162358.101,467117.844 162165.004,466820.119 162176.208,466809.875
Can this be done with sed or otherwise?

You want a number followed by a space and a number to be replaced with a number followed by a comma and a number, for all occurences (the flag g).
sed 's/\([0-9]\+\.[0-9]\+\) \([0-9]\+\.[0-9]\+\)/\1,\2/g'
You might want to change how exactly a "number" is defined but this should work as long as you have digits before and after the decimal point, and you always have a decimal point.
If you don't care at all about whether these are number at all, you could do it even simpler.

I would go for awk, because allows you to loop through fields quite easily:
awk '{for (i=1; i<=NF; i+=2) printf "%s,%s%s", $i, $(i+1), (i==NF-1?"\n":" ")' file
Explanation
The idea is to loop through fields in blocks of two. Hence:
for (i=1; i<=NF; i+=2) normal loop with a seed of 2. (Is it correct to call it "seed"? Just guessing)
printf "%s,%s%s", $i, $(i+1), (i==NF-1?"\n":" ") prints two fields separated by comma. It adds a space or new line, depending on the loop position.
This would also work:
awk '{for (i=1; i<=NF; i+=2) printf "%s,%s ", $i, $(i+1); print ""}'
Test
$ awk '{for (i=1; i<=NF; i+=2) printf "%s,%s ", $i, $(i+1); print ""}' file
161768.669,467646.462 161716.059,467515.711 161683.343,467434.35 161696.535,467427.091 161713.457,467416.235
161719.332,467412.183 161732.58,467403.047 161763.757,467381.118 161791.136,467367.094 161808.447,467361.612 161834.383,467353.808
161842.354,467351.71 161870.358,467344.337
161886.521,467338.28 161905.239,467330.484 161908.865,467328.974 161972.72,467301.688 161984.372,467296.404 162000.0,467288.568 162056.152,467260.414
162065.771,467254.859 162102.992,467233.363 162138.102,467213.086
162146.165,467209.29 162165.68,467200.102 162165.684,467200.1 162270.285,467154.666 162318.855,467133.104 162358.101,467117.844 162165.004,466820.119 162176.208,466809.875

A ruby one-liner:
ruby -ane 'puts $F.each_slice(2).map {|*pair| pair.join ","}.join " "' file

# posix version (non GNU sed or use --posix)
sed 's/\([^ ]\{1,\}\) \([^ ]\{1,\}\)/\1,\2/g' YourFile
GNU sed version
sed 's/\([^ ]+\) \([^ ]+\)/\1,\2/g' YourFile
Replace any space between a peer of non blank group by a comma. it doesn not replace space between 2 peer because sed start the next substitution AFTER the last one so next non blank group is no including the following space

Related

Replacing all occurrence after nth occurrence in a line in perl

I need to replace all occurrences of a string after nth occurrence in every line of a Unix file.
My file data:
:account_id:12345:6789:Melbourne:Aus
:account_id:98765:43210:Adelaide:Aus
My output data:
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
tried using sed: sed 's/://3g' test.txt
Unfortunately, the g option with the occurrence is not working as expected. instead, it is replacing all the occurrences.
Another approach using awk
awk -v c=':' -v n=2 'BEGIN{
FS=OFS=""
}
{
j=0;
for(i=0; ++i<=NF;)
if($i==c && j++>=n)$i=""
}1' file
$ cat file
:account_id:12345:6789:Melbourne:Aus
:account_id:98765:43210:Adelaide:Aus
$ awk -v c=':' -v n=2 'BEGIN{FS=OFS=""}{j=0;for(i=0; ++i<=NF;)if($i==c && j++>=n)$i=""}1' file
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
With GNU awk, using gensub please try following. This is completely based on your shown samples, where OP wants to remove : from 3rd occurrence onwards. Using gensub to segregate parts of matched values and removing all colons from 2nd part(from 3rd colon onwards) in it as per OP's requirement.
awk -v regex="^([^:]*:)([^:]*:)(.*)" '
{
firstPart=restPart=""
firstPart=gensub(regex, "\\1 \\2", "1", $0)
restPart=gensub(regex,"\\3","1",$0)
gsub(/:/,"",restPart)
print firstPart restPart
}
' Input_file
I have inferred based on the limited data you've given us, so it's possible this won't work. But I wouldn't use regex for this job. What you have there is colon delimited fields.
So I'd approach it using split to extract the data, and then some form of string formatting to reassemble exactly what you like:
#!/usr/bin/perl
use strict;
use warnings;
while (<DATA>) {
chomp;
my ( undef, $first, #rest ) = split /:/;
print ":$first:", join ( "", #rest ),"\n";
}
__DATA__
:account_id:12345:6789:Melbourne:Aus
:account_id:98765:43210:Adelaide:Aus
This gives you the desired result, whilst IMO being considerably clearer for the next reader than a complicated regex.
You can use the perl solution like
perl -pe 's~^(?:[^:]*:){2}(*SKIP)(?!)|:~~g if /^:account_id:/' test.txt
See the online demo and the regex demo.
The ^(?:[^:]*:){2}(*SKIP)(?!)|: regex means:
^(?:[^:]*:){2}(*SKIP)(?!) - match
^ - start of string (here, a line)
(?:[^:]*:){2} - two occurrences of any zero or more chars other than a : and then a : char
(*SKIP)(?!) - skip the match and go on to search for the next match from the failure position
| - or
: - match a : char.
And only run the replacement if the current line starts with :account_id: (see if /^:account_id:/').
Or an awk solution like
awk 'BEGIN{OFS=FS=":"} /^:account_id:/ {result="";for (i=1; i<=NF; ++i) { result = result (i > 2 ? $i : $i OFS)}; print result}' test.txt
See this online demo. Details:
BEGIN{OFS=FS=":"} - sets the input/output field separator to :
/^:account_id:/ - line must start with :account_id:
result="" - sets result variable to an empty string
for (i=1; i<=NF; ++i) { result = result (i > 2 ? $i : $i OFS)}; print result} - iterates over the fields and if the field number is greater than 2, just append the current field value to result, else, append the value + output field separator; then print the result.
I would use GNU AWK following way if n fixed and equal 2 following way, let file.txt content be
:account_id:12345:6789:Melbourne:Aus
:account_id:98765:43210:Adelaide:Aus
then
awk 'BEGIN{FS=":";OFS=""}{$2=FS $2 FS;print}' file.txt
output
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
Explanation: use : as field separator and nothing as output field separator, this itself does remove all : so I add : which have to be preserved: 1st (before second column) and 2nd (after second column). Beware that I tested it solely for this data, so if you would want to use it you should firstly test it with more possible inputs.
(tested in gawk 4.2.1)
This might work for you (GNU sed):
sed 's/:/\n/3;h;s/://g;H;g;s/\n.*\n//' file
Replace the third occurrence of : by a newline.
Make a copy of the line.
Delete all occurrences of :'s.
Append the amended line to the copy.
Join the two lines by removing everything from third occurrence of the copy to the third occurrence of the amended line.
N.B. The use of the newline is the best delimiter to use in the case of sed, as the line presented to seds commands are initially devoid of newlines. However the important property of the delimiter is that it is unique and therefore can be any such character as long as it is not found anywhere in the data set.
An alternative solution uses a loop to remove all :'s after the first two:
sed -E ':a;s/^(([^:]*:){2}[^:]*):/\1/;ta' file
With GNU awk for the 3rd arg to match() and gensub():
$ awk 'match($0,/(:[^:]+:)(.*)/,a){ $0=a[1] gensub(/:/,"","g",a[2]) } 1' file
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus
and with any awk in any shell on every Unix box:
$ awk 'match($0,/:[^:]+:/){ tgt=substr($0,1+RLENGTH); gsub(/:/,"",tgt); $0=substr($0,1,RLENGTH) tgt } 1' file
:account_id:123456789MelbourneAus
:account_id:9876543210AdelaideAus

How to apply one command into another sed command?

I have one command which is used to extract lines between two string patterns 'string1' and 'string2'. This is stored in variable called 'var1'.
var1=$(awk '/string1/{flag=1; next} /string2/{flag=0} flag' text.txt)
This command works well and the output is a set of lines.
Do you hear the people sing?
Singing a song of angry men?
It is the music of a people
Who will not be slaves again
I want the output of the above command to be inserted after a string pattern 'string3' in another file called stat.txt. I used sed as follows
sed '/string3/a'$var1'' stat.txt
I am having trouble getting the new output. Here, the $var1 seems to be working partially i.e. only one line -
string3
Do you hear the people sing?
Any other suggestions to solve this?
I would be tempted to use sed to extract the lines, and awk to insert them into the other text:
lines=$(sed -n '/string1/,/string2/ p' text.txt)
awk -v new="$lines" '{print} /string3/ {print new}' stat.txt
or perhaps both tasks in a single awk call
awk '
NR == FNR && /string1/ {flag = 1}
NR == FNR && /string2/ {flag = 0}
NR == FNR && flag {lines = lines $0 ORS}
NR == FNR {next}
{print}
/string3/ {printf "%s", lines} # it already ends with a newline
' text.txt stat.txt
It's a data format problem...
Appending a multi-line block of text with the sed append command requires that every line in the block to be appended ends with a \ -- except for the last line of that block. So if we take the two lines of code that didn't work in the question, and reformat the text as required by the append command, the original code should work as expected:
var1=$(awk '/string1/{flag=1; next} /string2/{flag=0} flag' text.txt)
var1="$(sed '$!s/$/\\/' <<< "$var1")"
sed '/string3/a'$var1'' stat.txt
Note that the 2nd line above contains a bashism. A more portable version would be:
var1="$(echo "$var1" | sed '$!s/$/\\/')"
Either variant would convert $var1 to:
Do you hear the people sing?\
Singing a song of angry men?\
It is the music of a people\
Who will not be slaves again

Sed - replace words

I have a problem with replacing string.
|Stm=2|Seq=2|Num=2|Svc=101|MsgSize(514)=514|MsgType=556|SymbolIndex=16631
I want to find occurrence of Svc till | appears and swap place with Stm till | appears.
My attempts went to replacing characters and this is not my goal.
awk -F'|' -v OFS='|'
'{a=b=0;
for(i=1;i<=NF;i++){a=$i~/^Stm=/?i:a;b=$i~/^Svc=/?i:b}
t=$a;$a=$b;$b=t}7' file
outputs:
|Svc=101|Seq=2|Num=2|Stm=2|MsgSize(514)=514|MsgType=556|SymbolIndex=16631
the code exchange the column of Stm.. and Svc.., no matter which one comes first.
If perl solution is okay, assumes only one column matches each for search terms
$ cat ip.txt
|Stm=2|Seq=2|Num=2|Svc=101|MsgSize(514)=514|MsgType=556|SymbolIndex=16631
$ perl -F'\|' -lane '
#i = grep { $F[$_] =~ /Svc|Stm/ } 0..$#F;
$t=$F[$i[0]]; $F[$i[0]]=$F[$i[1]]; $F[$i[1]]=$t;
print join "|", #F;
' ip.txt
|Svc=101|Seq=2|Num=2|Stm=2|MsgSize(514)=514|MsgType=556|SymbolIndex=16631
-F'\|' -lane split input line on |, see also Perl flags -pe, -pi, -p, -w, -d, -i, -t?
#i = grep { $F[$_] =~ /Svc|Stm/ } 0..$#F get index of columns matching Svc and Stm
$t=$F[$i[0]]; $F[$i[0]]=$F[$i[1]]; $F[$i[1]]=$t swap the two columns
Or use ($F[$i[0]], $F[$i[1]]) = ($F[$i[1]], $F[$i[0]]); courtesy How can I swap two Perl variables
print join "|", #F print the modified array
You need to use capture groups and backreferences in a string substition.
The below will swap the 2:
echo '|Stm=2|Seq=2|Num=2|Svc=101|MsgSize(514)=514|MsgType=556|SymbolIndex=16631' | sed 's/\(Stm.*|\)\(.*\)\(Svc.*|\)/\3\2\1/'
As pointed out in the comment from #Kent, this will not work if the strings were not in that order.

Remove newline depending on the format of the next line

I have a special file with this kind of format :
title1
_1 texthere
title2
_2 texthere
I would like all newlines starting with "_" to be placed as a second column to the line before
I tried to do that using sed with this command :
sed 's/_\n/ /g' filename
but it is not giving me what I want to do (doing nothing basically)
Can anyone point me to the right way of doing it ?
Thanks
Try following solution:
In sed the loop is done creating a label (:a), and while not match last line ($!) append next one (N) and return to label a:
:a
$! {
N
b a
}
After this we have the whole file into memory, so do a global substitution for each _ preceded by a newline:
s/\n_/ _/g
p
All together is:
sed -ne ':a ; $! { N ; ba }; s/\n_/ _/g ; p' infile
That yields:
title1 _1 texthere
title2 _2 texthere
If your whole file is like your sample (pairs of lines), then the simplest answer is
paste - - < file
Otherwise
awk '
NR > 1 && /^_/ {printf "%s", OFS}
NR > 1 && !/^_/ {print ""}
{printf "%s", $0}
END {print ""}
' file
This might work for you (GNU sed):
sed ':a;N;s/\n_/ /;ta;P;D' file
This avoids slurping the file into memory.
or:
sed -e ':a' -e 'N' -e 's/\n_/ /' -e 'ta' -e 'P' -e 'D' file
A Perl approach:
perl -00pe 's/\n_/ /g' file
Here, the -00 causes perl to read the file in paragraph mode where a "line" is defined by two consecutive newlines. In your example, it will read the entire file into memory and therefore, a simple global substitution of \n_ with a space will work.
That is not very efficient for very large files though. If your data is too large to fit in memory, use this:
perl -ne 'chomp;
s/^_// ? print "$l " : print "$l\n" if $. > 1;
$l=$_;
END{print "$l\n"}' file
Here, the file is read line by line (-n) and the trailing newline removed from all lines (chomp). At the end of each iteration, the current line is saved as $l ($l=$_). At each line, if the substitution is successful and a _ was removed from the beginning of the line (s/^_//), then the previous line is printed with a space in place of a newline print "$l ". If the substitution failed, the previous line is printed with a newline. The END{} block just prints the final line of the file.

divide each line in equal part

I would be happy if anyone can suggest me command (sed or AWK one line command) to divide each line of file in equal number of part. For example divide each line in 4 part.
Input:
ATGCATHLMNPHLNTPLML
Output:
ATGCA THLMN PHLNT PLML
This should work using GNU sed:
sed -r 's/(.{4})/\1 /g'
-r is needed to use extended regular expressions
.{4} captures every four characters
\1 refers to the captured group which is surrounded by the parenthesis ( ) and adds a space behind this group
g makes sure that the replacement is done as many times as possible on each line
A test; this is the input and output in my terminal:
$ echo "ATGCATHLMNPHLNTPLML" | sed -r 's/(.{4})/\1 /g'
ATGC ATHL MNPH LNTP LML
I suspect awk is not the best tool for this, but:
gawk --posix '{ l = sprintf( "%d", 1 + (length()-1)/4);
gsub( ".{"l"}", "& " ) } 1' input-file
If you have a posix compliant awk you can omit the --posix, but --posix is necessary for gnu awk and since that seems to be the most commonly used implementation I've given the solution in terms of gawk.
This might work for you (GNU sed):
sed 'h;s/./X/g;s/^\(.*\)\1\1\1/\1 \1 \1 \1/;G;s/\n/&&/;:a;/^\n/bb;/^ /s/ \(.*\n.*\)\n\(.\)/\1 \n\2/;ta;s/^.\(.*\n.*\)\n\(.\)/\1\2\n/;ta;:b;s/\n//g' file
Explanation:
h copy the pattern space (PS) to the hold space (HS)
s/./X/g replace every character in the HS with the same non-space character (in this case X)
s/^\(.*\)\1\1\1/\1 \1 \1 \1/ split the line into 4 parts (space separated)
G append a newline followed by the contents of the HS to the PS
s/\n/&&/ double the newline (to be later used as markers)
:a introduce a loop namespace
/^\n/bb if we reach a newline we are done and branch to the b namespace
/^ /s/ \(.*\n.*\)\n\(.\)/\1 \n\2/;ta; if the first character is a space add a space to the real line at this point and repeat
s/^.\(.*\n.*\)\n\(.\)/\1\2\n/;ta any other character just bump along and repeat
:b;s/\n//g all done just remove the markers and print out the result
This work for any length of line, however is the line is not exactly divisible by 4 the last portion will contain the remainder as well.
perl
perl might be a better choice here:
export cols=4
perl -ne 'chomp; $fw = 1 + int length()/$ENV{cols}; while(/(.{1,$fw})/gm) { print $1 . " " } print "\n"'
This re-calculates field-width for every line.
coreutils
A GNU coreutils alternative, field-width is chosen based on the first line of infile:
cols=4
len=$(( $(head -n1 infile | wc -c) - 1 ))
fw=$(echo "scale=0; 1 + $len / 4" | bc)
cut_arg=$(paste -d- <(seq 1 $fw 19) <(seq $fw $fw $len) | head -c-1 | tr '\n' ',')
Value of cut_arg is in the above case:
1-5,6-10,11-15,16-
Now cut the line into appropriate chunks:
cut --output-delimiter=' ' -c $cut_arg infile