How can I remove useless ".0" strings in a txt file of numbers?
If I have this file:
43.65 24.0 889.5 5.0
32.14 32.0 900.0 6.0
38.27 43.0 899.4 5.0
I want to get:
43.65 24 889.5 5
32.14 32 900 6
38.27 43 899.4 5
I tried: sed 's|\.0 | |g' but that obviously does not work with new lines and EOF.
Any suggestion without getting into python, etc?
This might work for you (GNU sed):
sed 's/\.0\b//g' file
Or
sed 's/\.0\>//g' file
Remove any period followed by a zero followed by a word boundary.
You can use
sed -E 's,([0-9])\.0($| ),\1\2,g' file
Details:
-E - enables POSIX ERE syntax
([0-9])\.0($| ) - finds and captures into Group 1 a digit, then matches .0, and then matches and captures into Group 2 a space or end of string
\1\2 - replaces with Group 1 + Group 2 concatenated values
g - replaces all occurrences.
See the online demo:
s='43.65 24.0 889.5 5.0
32.14 32.0 900.0 6.0
38.27 43.0 899.4 5.0'
sed -E 's,([0-9])\.0($| ),\1\2,g' <<< "$s"
Output:
43.65 24 889.5 5
32.14 32 900 6
38.27 43 899.4 5
Related
I am trying to uncomment "/usr/lib/sa/sa1" or "/usr/lib/sa/sa2" entries from one of the configuration file.
Below is the regex:
"^[0-9].*/usr/lib/sa/sa(1|2)"
Eg:
#0 * * * * /usr/lib/sa/sa1 1200 3 &
#5 23 * * * /usr/lib/sa/sa2 -s 0:00 -e 23:01 -i 3600 -ubcwyaqvm &
Output should be:
0 * * * * /usr/lib/sa/sa1 1200 3 &
5 23 * * * /usr/lib/sa/sa2 -s 0:00 -e 23:01 -i 3600 -ubcwyaqvm &
Tried the below "sed" command but no luck:
sed -e '/^#.*\/usr\/lib\/sa\/sa\(1\|2\)/s/^#//g' adm
Can you please let me know where am I going wrong here.
Thanks in advance!
You can use
sed '/^#.*\/usr\/lib\/sa\/sa[12]/s/^#//'
Here, no grouping construct is used, no alternation operator is used either. It is possible as the 1 and 2 are single char alternatives, and bracket expressions are supported in all versions of sed.
Also, see this online demo.
I would like to extract all lines between INFO:root:id is and one line after the INFO:root:newId.
Can anyone advise how I can achieve this?
Currently I'm using
sed -n '/INFO:root:id is/,/INFO:root:newId/p' 1/python.log
and I'm trying to figure out how to print one line after the second pattern match.
INFO:root:id is
INFO:root:16836211
DEBUG:urllib3.connectionpool:Starting new HTTPS connection (1): abc.hh.com
DEBUG:urllib3.connectionpool:https://abc.hh.com:443 "POST /api/v2/import/row.json HTTP/1.1" 201 4310
INFO:root:newId
INFO:root:35047536
INFO:root:id is
INFO:root:46836211
DEBUG:urllib3.connectionpool:Starting new HTTPS connection (1): abc.hh.com
DEBUG:urllib3.connectionpool:https://abc.hh.com:443 "POST /api/v2/import/row.json HTTP/1.1" 201 4310
INFO:root:newId
INFO:root:55547536
If I am understanding question correctly
$ seq 10 | sed -n '/3/,/5/{/5/N;p;}'
3
4
5
6
/3/ is starting regex and /5/ is ending regex
/5/N get additional line for ending regex
tested on GNU sed, syntax might differ for other versions
With awk
$ seq 10 | awk '/3/{f=1} f; /5/{f=0; if((getline a)>0) print a}'
3
4
5
6
Unclear whether you want only the first set of lines after a match or all matches.
If you want the first set between the matching patterns, it is easy if you use /INFO:root:id/ for your end match as well and then use head -n -1 to print everything but the last line.
$ sed -n '/INFO:root:id is/,/INFO:root:id/p' test.txt | head -n -1
INFO:root:id is
INFO:root:16836211
DEBUG:urllib3.connectionpool:Starting new HTTPS connection (1): abc.hh.com
DEBUG:urllib3.connectionpool:https://abc.hh.com:443 "POST /api/v2/import/row.json HTTP/1.1" 201 4310
INFO:root:newId
INFO:root:35047536
Just use flags to indicate when you've found the beginning and ending regexps and print accordingly:
$ seq 10 | awk 'e{print buf $0; buf=""; b=e=0} /3/{b=1} b{buf = buf $0 ORS; if (/5/) e=1}'
3
4
5
6
Note that this does not have the potential issue of printing lines when there's only the beginning or ending regexp present but not both. The other answers, including your currently accepted answer, have that problem:
$ seq 10 | sed -n '/3/,/27/{/27/N;p;}'
3
4
5
6
7
8
9
10
$ seq 10 | awk '/3/{f=1} f; /27/{f=0; if((getline a)>0) print a}'
3
4
5
6
7
8
9
10
$ seq 10 | awk 'e{print buf $0; buf=""; b=e=0} /3/{b=1} b{buf = buf $0 ORS; if (/27/) e=1}'
$
Note that the script I posted correctly didn't print anything because a block of text beginning with 3 and ending with 27 was not present in the input.
I've been trying to pull a field from a row in a file although each row may have plus or minus 2 or 3 fields per row. They aren't always equal in the number of fields per row.
Here is a snippet:
A orarpp 45286124 1 1 0 20 60 Nov 25 9-16:42:32 01:04:58 11176 117056 0 - oracleXXX (LOCAL=NO)
A orarpp 45351560 1 1 3 20 61 Nov 30 5-03:54:42 02:24:48 4804 110684 0 - ora_w002_XXX
A orarpp 45548236 1 1 22 20 71 Nov 26 8-19:36:28 00:56:18 10628 116508 0 - oracleXXX (LOCAL=NO)
A orarpp 45679190 1 1 0 20 60 Nov 28 6-23:42:20 00:37:59 10232 116112 0 - oracleXXX (LOCAL=NO)
A orarpp 45744808 1 1 0 20 60 10:52:19 23:08:12 00:04:58 11740 117620 0 - oracleXXX (LOCAL=NO)
A root 45810380 1 1 0 -- 39 Nov 25 9-19:54:34 00:00:00 448 448 0 - garbage
In the case of the first line, I'm interested in 9-16:42:32 and the similar fields for each row.
I've tried to pull it by using ':' as the field separator and then filter from there however, what I am trying to accomplish is to do something if the number before the dash (in the example it's 9) is greater than one.
cat file.txt | grep oracle | awk -F: '{print substr($1, length($1)-5)}'
This is because the number of fields on either side of the actual field I need can be different from line to line.
Definitely not the most efficient but I've been trying to do this with an awk one liner.
Hints or a direction would be appreciated to get me moving again. I am not opposed to doing in a better way than awk.
Thanks.
Maybe cut is the right tool for this job? For example, with your snippet:
$ cut -c 62-71 file.txt
9-16:42:32
5-03:54:42
8-19:36:28
6-23:42:20
23:08:12
9-19:54:34
The arguments tell cut to snip columns (-c) 62 through 71.
For additional processing, you can pipe it to awk.
You can also accomplish the whole thing in awk by accepting entire lines and then using substr to extract the columns you want. For example, this awk command produces the same output as the cut command above:
awk '{ print substr($0, 62, 10) }' file.txt
Whether you create a pipeline or do the processing entirely in awk is at least in part a matter of personal taste / style.
Would this do?
awk -F: '/oracle/ {print substr($0,62,10)}' file.txt
9-16:42:32
8-19:36:28
6-23:42:20
23:08:12
This search for oracle and then print 10 characters starting from position 62
You can grab those identifiers with one of
grep -o '[[:digit:]]\+-[[:digit:]]\{2\}:[[:digit:]]\{2\}:[[:digit:]]\{2\}'
grep -oP '\d+-\d\d:\d\d:\d\d' # GNU grep
It sounds like you want to do something with the lines, not just find the ids. Please elaborate.
Using GNU awk:
gawk --re-interval '
/oracle/ && \
match($0, /([[:digit:]]+)-([[:digit:]]{2}:){2}[[:digit:]]{2}/, a) && \
a[1]>1 {
# do something with the matching line
print
}
' file
I have a large report file (about 20MB) that looks like this:
586 700006207 8,622.09 896
9,882.82 896
777 68607099 900.00 896
587 800006207 7,059.22 896
959.02 896
697.87 896
7 280667985 .00 899
On 1st and 2nd columns there are blanks if the values are the same as the line above. I need help with a grep/sed/powershell one-liner to fill out the empty spaces, so that it looks like this:
586 700006207 8,622.09 896
586 700006207 9,882.82 896
777 68607099 900.00 896
587 800006207 7,059.22 896
587 800006207 959.02 896
587 800006207 697.87 896
7 280667985 .00 899
Thanks.
This might work for you:
sed ':a;$!N;/^\(\( [0-9]\+ *[0-9]\+\).*\n\)\( \{15\}\)/{s//\1\2/;ta};P;D' file
From the data you have provided the line always begins with a space. If this is not the case then:
sed ':a;$!N;/^\(\([0-9]\+ *[0-9]\+\).*\n\)\( \{14\}\)/{s//\1\2/;ta};P;D' file
Assuming there are not blank lines and the inter-column delimiters are spaces, the following works (tested in Ubuntu/bash shell, using GNU sed)...
sed -r "/^ {15}/{G; s/^ {15}(.*)\n(.{15}).*/\2\1/};h" "report"
Assuming you need to keep the spacing and the first 2 columns are the first 15 chars:
awk '
NF==2 {print fill substr($0, 16); next}
{print; fill = substr($0, 1, 15)}
'
Q1: Sed specify the whole line and if the line is nothing but the string then delete
I have a file that contains several of the following numbers:
1 1
3 1
12 1
1 12
25 24
23 24
I want to delete numbers that are the same in each line. For that I have either been using:
sed '/1 1/d' < old.file > new.file
OR
sed -n '/1 1/!p' < old.file > new.file
Here is the main problem. If I search for pattern '1 1' that means I get rid of '1 12' as well. So for I want the pattern to specify the whole line and if it does, to delete it.
Q2: Automation of question 1
I am also trying to automate this problem. The range of numbers in the first column and the second column could be from 1 to 25.
So far this is what I got:
for ((i=1;i<26;i++)); do
sed "/'$i' '$i'/d" < oldfile > newfile; mv newfile oldfile;
done
This does nothing to the oldfile in the end. :(
This would be more readable with awk:
awk '$1 == $2 {next} {print}' oldfile > newfile
Update based on comment:
If the requirement is to remove lines where the two values are within 1 of each other:
awk '{d = $1-$2; if (-1 <= d && d <= 1) next; else print}' oldfile
Unfortunately, awk does not have abs() (at least nawk and gawk don't)
Just put the first number in a group (\([0-9]*\)) and then look for it with a backreference (\1). Since the line to delete should contain only the group, repeated, use the ^ to mark the beginning of line and the $ to mark the end of line. For example, for the following file:
$ cat input
1 1
3 1
12 1
1 12
12 12
12 13
13 13
25 24
23 24
...the result is:
$ sed '/^\([0-9]*\) \1$/d' input
3 1
12 1
1 12
12 13
25 24
23 24
You can also do it with grep:
grep -E -v "([0-9])*\s\1" testfile
Look for multiple digits in a row and remember them, followed by a single whitespace, followed by whatever digits you remembered.