I am trying to extract the numbers (0.000500) from a bunch of files alá eta_x2-0.000500.
I would think that below line would do that, but i get only 0 and not 0.000500.
How can I get the maximum match?
find eta* | sed 's/.*\([0-9.]\+\)/\1/g'
The .* is greedy, so it is going to match as many characters as possible. In this case the .* will match eta_x2-0.00050 with just the final 0 matched in your group.
Normally the answer here would be to just use a non-greedy match using .*?, but I don't think sed supports that.
You should be able to get this to work by requiring that there is one non-digit character before you start matching, this way the .* will have to stop before consuming the digits:
sed 's/.*[^0-9.]\([0-9.]\+\)/\1/g'
Of course if you know the digits you want will be immediately after a -, you can replace [^0-9.] with - and it will work the same way.
add - in sed before the last numbers, and g is useless.
find eta* | sed 's/.*-\([0-9.]\+\)/\1/'
Don't need sed
find eta* | grep -oP '(?<=-)[\s.]+'
find eta* | cut -d'-' -f 2
This should work -
find eta* | sed 's/[^-]*-\([0-9]*\.[0-9]*\)/\1/'
Related
Input:
p45-322-16.jpg
Desired output:
p45
I'm trying to make a bash script with grep or awk or sed or something that could run on a bash shell.
Currently I'm stuck with this:
echo "p45-322-16.jpg" | sed 's/\(.*\)-.*/\1/'
Output:
p45-322
You can use cut command: echo "p45-322-16.jpg" | cut -d"-" -f1
You need to limit what you will accept. Right now, you are accepting too much by using ., and the greedy-by-default nature of regexes is consuming too many characters.
Try either limiting the accepted characters to digits only, or specifically excluding the dash:
\([0-9]*\).*
\([^-]*\).*
echo "p45-322-16.jpg" | sed 's/\([^-]*\).*/\1/'
The .* part of your regex is greedy, so it reads as far as possible so that the regex still matches. This means it claims everything up to the last -.
Use [^-]* to match everything until a -.
I want to extract number from string. This is the string
#all/30
All I want is 30. How can I extract?
I try to use :
echo "#all/30" | sed 's/.*\/([^0-9])\..*//'
But nothing happen.
How should I write for the regular expression?
Sorry for bad english.
You may consider using grep to extract the numbers from a simple string like this.
echo "#all/30" | grep -o '[0-9]\+'
-o option shows only the matching part that matches the pattern.
You could try the below sed command,
$ echo "#all/30" | sed 's/[^0-9]*\([0-9]\+\)[^0-9]*/\1/'
30
[^0-9]* [^...] is a negated character class. It matches any character but not the one inside the negated character class. [^0-9]* matches zero or more non-digit characters.
\([0-9]\+\) Captures one or more digit characters.
[^0-9]* Matches zero or more non-digit characters.
Replacing the matched characters with the chars inside group 1 will give you the number 30
echo "all/30" | sed 's/[^0-9]*\/\([0-9][0-9]*\)/\1/'
Avoid writing '.*' as it consumes entire string. Default matches are always greedy.
echo "all/30" | sed 's/[^0-9]*//g'
# OR
echo "all/30" | sed 's#.*/##'
# OR
echo "all/30" | sed 's#.*\([0-9]*\)#\1#'
without more info about possible input string we can only assume that structure is #all/ followed by the number (only)
Another question for the sed experts.
I have a string representing an pathname that will have two numbers in it. An example is:
./pentaray_run2/Trace_220560.dat
I need to extract the second of these numbers - ie 220560
I have (with some help from the forums) been able to extract all the numbers together (ie 2220560) with:
sed "s/[^0-9]//g"
or extract only the first number with:
sed -r 's|^([^.]+).*$|\1|; s|^[^0-9]*([0-9]+).*$|\1|'
But what I'm after is the second number!! Any help much appreciated.
PS the number I'm after is always the second number in the string.
is this ok?
sed -r 's/.*_([0-9]*)\..*/\1/g'
with your example:
kent$ echo "./pentaray_run2/Trace_220560.dat"|sed -r 's/.*_([0-9]*)\..*/\1/g'
220560
You can extract the last numbers with this:
sed -e 's/.*[^0-9]\([0-9]\+\)[^0-9]*$/\1/'
It is easier to think this backwards:
From the end of the string, match zero or more non-digit characters
Match (and capture) one or more digit characters
Match at least one non-digit character
Match all the characters to the start of the string
Part 3 of the match is where the "magic" happens, but it also limits your matches to have at least a non-digit before the number (ie. you can't match a string with only one number that is at the start of the string, although there is a simple workaround of inserting a non-digit to the start of the string).
The magic is to counter-act the left-to-right greediness of the .* (part 4). Without part 3, part 4 would consume all it can, which includes the numbers, but with it, matching makes sure that it stops in order to allow at least a non-digit followed by a digit to be consumed by parts 1 and 2, allowing the number to be captured.
If grep is welcome :
$ echo './pentaray_run2/Trace_220560.dat' | grep -oP '\d+\D+\K\d+'
220560
And more portable with Perl with the same regex :
echo './pentaray_run2/Trace_220560.dat' | perl -lne 'print $& if /\d+\D+\K\d+/'
220560
I think the approach is cleaner & more robust than using sed
This might work for you (GNU sed):
sed -r 's/([^0-9]*([0-9]*)){2}.*/\2/' file
This extracts the second number:
sed -r 's/([^0-9]*([0-9]*)){1}.*/\2/' file
and this extracts the first.
I have this:
echo 12345 | grep -o '[[:digit:]]\{1,4\}'
Which gives this:
1234
5
I understand whats happening. How do I stop grep from trying to continue matching after 1 successful match?
How do I get only
1234
Do you want grep to stop matching or do you only care about the first match. You could use head if the later is true...
`grep stuff | head -n 1`
Grep is a line based util so the -m 1 flag tells grep to stop after it matches the first line which when combined with head is pretty good in practice.
You need to do the grouping: \(...\) followed by the exact number of occurrence: \{<n>\} to do the job:
maci:~ san$ echo 12345 | grep -o '\([[:digit:]]\)\{4\}'
1234
Hope it helps. Cheers!!
Use sed instead of grep:
echo 12345 | sed -n '/^\([0-9]\{1,4\}\).*/s//\1/p'
This matches up to 4 digits at the beginning of the line, followed by anything, keeps just the digits, and prints them. The -n prevents lines from being printed otherwise. If the digit string might also appear mid-line, then you need a slightly more complex command.
In fact, ideally you'll use a sed with PCRE regular expressions since you really need a non-greedy match. However, up to a reasonable approximation, you can use: (A semi-solution to a considerably more complex problem...now removed!)
Since you want the first string of up to 4 digits on the line, simply use sed to remove any non-digits and then print the digit string:
echo abc12345 | sed -n '/^[^0-9]*\([0-9]\{1,4\}\).*/s//\1/p'
This matches a string of non-digits followed by 1-4 digits followed by anything, keeps just the digits, and prints them.
If – as in your example – your numeric expression will appear at the beginning of the string you're starting with, you could just add a start-of-line anchor ^:
echo 12345 | grep -o '^\([[:digit:]]\)\{1,4\}'
Depending on which exact digits you want, an end-of-line anchor $ might help also.
grep manpage says on this topic (see chapter 'regular expressions'):
(…)
{n,}
The preceding item is matched n or more times.
{n,m}
The preceding item is matched at least n times, but not more than m times.
(…)
So the answer should be:
echo 12345 | grep -o '[[:digit:]]\{4\}'
I just tested it on cygwin terminal (2018) and it worked!
example file:
blahblah 123.a.site.com some-junk
yoyoyoyo 456.a.site.com more-junk
hihohiho 123.a.site.org junk-in-the-trunk
lalalala 456.a.site.org monkey-junk
I want to grep out all those domains in the middle of each line, they all have a common part a.site with which I can grep for, but I can't work out how to do it without returning the whole line?
Maybe sed or a regex is need here as a simple grep isn't enough?
You can do:
grep -o '[^ ]*a\.site[^ ]*' input
or
awk '{print $2}' input
or
sed -e 's/.*\([^ ]*a\.site[^ ]*\).*/\1/g' input
Try this to find anything in that position
$ sed -r "s/.* ([0-9]*)\.(.*)\.(.*)/\2/g"
[0-9]* - For match number zero or more time.
.* - Match anything zero or more time.
\. - Match the exact dot.
() - Which contain the value particular expression in parenthesis, it can be printed using \1,\2..\9. It contain only 1 to 9 buffer space. \0 means it contain all the expressed pattern in the expression.