I have a simple .csv file with lines that holds 't' values. Here is the example:
2ABC;t;t;t;tortuga;fault;t;t;bored
I want to replace them to '1' using sed.
If I make sed "s/;t;/;1;/g" I get the next result:
2ABC;1;t;1;tortuga;fault;1;t;bored
As you can see, consecutive ';t;' have been replaced through one. Yes, I can replace all ';t;' by sed -e "s/;t;/;1;/g" -e "s/;t;/;1;/g" but this is boring.
How can I make the replacement by one sed command?
If there is something to replace, branch to replace again.
sed ': again; /;t;/{ s//;1;/; b again }'
Overall, parsing cvs with sed is crude. Consider awk.
awk -F';' -v OFS=';' '{ for(i=1;i<=NF;++i) if ($i=="t") $i=1 } 1'
Lookarounds is helpful in such cases:
$ s='t;2ABC;t;t;t;tortuga;fault;t;t;bored;t'
$ echo "$s" | perl -lpe 's/(?<![^;])t(?![^;])/1/g'
1;2ABC;1;1;1;tortuga;fault;1;1;bored;1
echo '2ABC;t;t;t;tortuga;fault;t;t;bored' |
— gawk-specific solution
gawk -be '(ORS = RT)^!(NF = NF)' FS='^t$' OFS=1 RS=';'
— cross-awk-solution
{m,g,n}awk 'gsub(FS, OFS, $!(NF = NF))^_' FS=';t;' OFS=';1;' RS=
2ABC;1;1;1;tortuga;fault;1;1;bored
Related
This is mostly by curiosity, I am trying to have the same behavior as:
echo -e "test1:test2:test3"| sed 's/:/\n/g' | grep 1
in a single sed command.
I already tried
echo -e "test1:test2:test3"| sed -e "s/:/\n/g" -n "/1/p"
But I get the following error:
sed: can't read /1/p: No such file or directory
Any idea on how to fix this and combine different types of commands into a single sed call?
Of course this is overly simplified compared to the real usecase, and I know I can get around by using multiple calls, again this is just out of curiosity.
EDIT: I am mostly interested in the sed tool, I already know how to do it using other tools, or even combinations of those.
EDIT2: Here is a more realistic script, closer to what I am trying to achieve:
arch=linux64
base=https://chromedriver.storage.googleapis.com
split="<Contents>"
curl $base \
| sed -e 's/<Contents>/<Contents>\n/g' \
| grep $arch \
| sed -e 's/^<Key>\(.*\)\/chromedriver.*/\1/' \
| sort -V > out
What I would like to simplify is the curl line, turning it into something like:
curl $base \
| sed 's/<Contents>/<Contents>\n/g' -n '/1/p' -e 's/^<Key>\(.*\)\/chromedriver.*/\1/' \
| sort -V > out
Here are some alternatives, awk and sed based:
sed -E "s/(.*:)?([^:]*1[^:]*).*/\2/" <<< "test1:test2:test3"
awk -v RS=":" '/1/' <<< "test1:test2:test3"
# or also
awk 'BEGIN{RS=":"} /1/' <<< "test1:test2:test3"
Or, using your logic, you would need to pipe a second sed command:
sed "s/:/\n/g" <<< "test1:test2:test3" | sed -n "/1/p"
See this online demo. The awk solution looks cleanest.
Details
In sed solution, (.*:)?([^:]*1[^:]*).* pattern matches an optional sequence of any 0+ chars and a :, then captures into Group 2 any 0 or more chars other than :, 1, again 0 or more chars other than :, and then just matches the rest of the line. The replacement just keeps Group 2 contents.
In awk solution, the record separator is set to : and then /1/ regex is used to only return the record having 1 in it.
This might work for you (GNU sed):
sed 's/:/\n/;/^[^\n]*1/P;D' file
Replace each : and if the first line in the pattern space contains 1 print it.
Repeat.
An alternative:
sed -Ez 's/:/\n/g;s/^[^1]*$//mg;s/\n+/\n/;s/^\n//' file
This slurps the whole file into memory and replaces all colons by newlines. All lines that do not contain 1 are removed and surplus newlines deleted.
An alternative to the really ugly sed is: grep -o '\w*2\w*'
$ printf "test1:test2:test3\nbob3:bob2:fred2\n" | grep -o '\w*2\w*'
test2
bob2
fred2
grep -o: only matching
Or: grep -o '[^:]*2[^:]*'
echo -e "test1:test2:test3" | sed -En 's/:/\n/g;/^[^\n]*2[^\n]*(\n|$)/P;//!D'
sed -n doesn't print unless told to
sed -E allows using parens to match (\n|$) which is newline or the end of the pattern space
P prints the pattern buffer up to the first newline.
D trims the pattern buffer up to the first newline
[^\n] is a character class that matches anything except a newline
// is sed shorthand for repeating a match
//! is then matching everything that didn't match previously
So, after you split into newlines, you want to make sure the 2 character is between the start of the pattern buffer ^ and the first newline.
And, if there is not the character you are looking for, you want to D delete up to the first newline.
At that point, it works for one line of input, with one string containing the character you're looking for.
To expand to several matches within a line, you have to ta, conditionally branch back to label :a:
$ printf "test1:test2:test3\nbob3:bob2:fred2\n" | \
sed -En ':a s/:/\n/g;/^[^\n]*2[^\n]*(\n|$)/P;D;ta'
test2
bob2
fred2
This is simply NOT a job for sed. With GNU awk for multi-char RS:
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' '/1/'
test1
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' 'NR%2'
test1
test3
test5
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS='[:\n]' '!(NR%2)'
test2
test4
test6
$ echo "foo1:bar1:foo2:bar2:foo3:bar3" | awk -v RS='[:\n]' '/foo/ || /2/'
foo1
foo2
bar2
foo3
With any awk you'd just have to strip the \n from the final record before operating on it:
$ echo "test1:test2:test3:test4:test5:test6"| awk -v RS=':' '{sub(/\n$/,"")} /1/'
test1
I have the followiing input file and I need to remove all the characters from the strings that appear after the last '/'. I'll also show my expected output below.
input:
/start/one/two/stopone.js
/start/one/two/three/stoptwo.js
/start/one/stopxyz.js
expected output:
/start/one/two/
/start/one/two/three/
/start/one/
I have tried to use sed but with no luck so far.
You could simply use good old grep:
grep -o '.*/' file.txt
This simple expression takes advantage of the fact that grep is matching greedy. Meaning it will consume as much characters as possible, including /, until the last / in path.
Original Answer:
You can use dirname:
while read line ; do
echo dirname "$line"
done < file.txt
or sed:
sed 's~\(.*/\).*~\1~' file.txt
perl -lne 'print $1 if(/(.*)\//)' your_file
Try this GNU sed command,
$ sed -r 's~^(.*\/).*$~\1~g' file
/start/one/two/
/start/one/two/three/
/start/one/
Through awk,
awk -F/ '{sub(/.*/,"",$NF); print}' OFS="/" file
How to replace '.' with '_' within the part of the line before the '=' char in the input below
Need single sed command to do all three
echo "few.num.dots=/home/user/.hidden/folder.dot" | sed 's/\./_/g'
required output => few_num_dots=/home/user/.hidden/folder.dot
echo "var=nodot" | sed 's/\./_/g'
required output => var=nodot
echo "var.one=onedot.notthis" | sed 's/\./_/g'
required output => var_one=onedot.notthis
You can use a conditional branching using the t command. It does a loop until the substitution command fails, and the command replaces any . character followed by an equal sign:
echo "few.num.dots=/home/user/.hidden/folder.dot" |
sed ':a; s/\.\([^=]*=\)/_\1/; ta'
It yields:
few_num_dots=/home/user/.hidden/folder.dot
perl?
echo "few.num.dots=/home/user/.hidden/folder.dot" |
perl -pe 's/^[^=]+/ ($x=$&) =~ tr{.}{_}; $x /e'
few_num_dots=/home/user/.hidden/folder.dot
awk?
awk -F= -v OFS='=' '{gsub(/\./,"_",$1)} 1'
You can do it this way as well,
echo "few.num.dots=/home/user/.hidden/folder.dot" |
sed -e '1,/./s/\./_/' -e '1,/./s/\./_/'
few_num_dots=/home/user/.hidden/folder.dot
First -e replaces the first occurrence of the pattern ., next -e replaces the next one...
Using awk
$ echo "few.num.dots=/home/user/.hidden/folder.dot" |awk '/=/{gsub(/\./,"_",$1)}1' FS="=" OFS="="
few_num_dots=/home/user/.hidden/folder.dot
$ echo "var.one=onedot.notthis" |awk '/=/{gsub(/\./,"_",$1)}1' FS="=" OFS="="
var_one=onedot.notthis
This might work for you (GNU sed):
sed 's/=/\n&/;h;y/./_/;G;s/\n.*\n.*\n//' file
Insert a marker to divide the line, copy the line, translate the characters, append the original line and using the marker reconstitute the line.
I have a line:
<random junk>TYPE=snp;<more random junk>
and I need to return everything between the end of TYPE= and the ; (in this case snp but it could be any of a number of text strings.
I tried various sed / awk solutions but I can't seem to get it working. I have the feeling this is a simple problem so, sorry about that.
This seems to work:
sed 's/.*TYPE=\(.*\);.*/\1/'
EDIT:
Ah, so there can be semicolons in the random junk. Try this:
sed 's/.*TYPE=\([^;]*\);.*/\1/'
requires GNU grep:
grep -Po '(?<=TYPE=)[^;]+'
meaning: preceded by "TYPE=", find some non-semicolon characters
One way using GNU sed:
sed -r 's/.*TYPE=([^;]+).*/\1/' file.txt
Since you also tagged this awk:
$ text='<random junk>TYPE=snp;<more random junk>'
$ echo "$text" | awk -FTYPE= '{sub(/;.*/,"",$2); print $2}'
snp
$ text='foo=bar;baz=fnu;TYPE=snp;XAI=0;XAM=0'
$ echo "$text" | awk -FTYPE= '{sub(/;.*/,"",$2); print $2}'
snp
(Only using the variable to keep the lines from wrapping.)
Or, to parse this as set of variable=value pairs rather than just a string of text:
$ echo "$text" | awk -vRS=";" -F= '$1=="TYPE" {print $2}'
snp
You can also do this in pure bash, if you want:
$ t="red=blue;TYPE=snp;XAI=0.0037843;XAM=0.0170293;XAS=0.013245;XRI=0;XRM=0"
$ t=${t#*TYPE=}
$ t=${t%%;*}
$ echo $t
snp
I can chain multiple sed substitutions and a awk operation to achieve this, but is there a single sed substitution that can do it?
Also is there any other tool that is more suitable for this parsing task?
You could try:
sed -r 's!rec:id=(.*?)&name=(.*?)&age=(.*?)!\1 \2 \3!' input_file
If you don't know the rec:id etc in advance but you know there's three, you could try:
sed -r 's![^=]+=(.*?)&[^=]+=(.*?)&[^=]+=(.*?)!\1 \2 \3!' input_file
If you don't know how many &name=value pairs you're after in advance but want to output all the values, you could try something like:
grep -P -o '(?<==)([^&]*)(?=&|$)' | xargs
where the -P means 'perl regex', the regex says "find the string followed by an & (or end of string) and preceded by and equals sign", the -o means to print just the matches (ie the 1, zz, and 21) each on their own line, and the | xargs moves these from their own line to one line and space separated (ie 1\nzz\n21 to 1 zz 21).
This might work for you:
echo "rec:id=1&name=zz&age=21" | sed 's/[^=]*=\([^&]*\)/\1 /g'
1 zz 21
However this leaves an extra space at the end, to solve this use:
echo "rec:id=1&name=zz&age=21"|sed 's/[^=]*=\([^&]*\)/\1 /g:;s/ $//'
1 zz 21
How about parsing the values directly into variables?
inbound="rec:id=1&name=zz&age=21"
eval $(echo $inbound | cut -c5- | tr \& "\n")
echo "Name:$name, ID:$id, Age:$age"
Or even better, though slightly more arcane:
inbound="rec:id=1&name=zz&age=21"
IFS=\& eval $(cut -c5- <<< $inbound)
echo "Name:$name, ID:$id, Age:$age"