Within bash shell, I need to add numbers in the format : 0.13281E-04 and -0.79254E-04.
So, in the script, I have:
tt1=`echo $var_t1| sed -e 's/[eE]+*/\\*10\\^/'`
tt2=`echo $var_t2| sed -e 's/[eE]+*/\\*10\\^/'`
var_t=`echo "($tt1 + $tt2)/2.0" | bc -l`
which gives result as: -.000032986500000
Is there any straight forward way getting var_t as -3.29865E-05?
You can use the option -v of printf like so
printf -v var_t "%g\n" -.00003298650000
echo $var_t
Related
Why doesn't the follow bash script work? I would like it to output
two lines like this:
XXXXXXX
YYYYYYY
It works if I change the sed line to use a filename instead of the variable, but I want to use the variable.
#!/bin/bash
input=$(echo -e '=======\n-------\n')
for sym in = -; do
if [ "$sym" == '-' ]; then
replace=Y
else
replace=X
fi
printf "%s\n" "s/./$replace/g"
done | sed -f- <<<"$input"
The main problem is that you're giving sed two sources to read standard input from: the for loop that is fed through the pipe, and the variable coming through the here-string. As it turns out, the here-string gets precedence and sed complains that there are extra characters after a command (= is a command).
Instead of a here-string, you could use process substitution:
for sym in = -; do
if [ "$sym" == '-' ]; then
replace=Y
else
replace=X
fi
printf "%s\n" "s/./$replace/g"
done | sed -f- <(printf '%s\n' '=======' '-------')
You'll notice that the output isn't what you want, though, namely
YYYYYYY
YYYYYYY
This is because the sed script you end up with looks like this:
s/./X/g
s/./Y/g
No matter what you do first, the last command replaces everything with Y.
It has been long time since my last bash script.
I m just trying to extract the content of a file from the the start variable to the stop one.
My source file is night4.info and it contains a list of .jpg files.The structure of this file is similar to:
./2014-11-02/18h/00mn/2014-11-02T18-00-00.048000-depth.jpg
./2014-11-02/18h/00mn/2014-11-02T18-00-00.182000-depth.jpg
./2014-11-02/18h/00mn/2014-11-02T18-00-00.316000-depth.jpg
This is the code so far :
#! /bin/bash
start=$(grep -n $1 night4.info | cut -d : -f 1)
stop=$(grep -n $2 night4.info | cut -d : -f 1)
echo "1" >> list.info
sed -n -e "$start,$stop p" night4.info >> list.info
And this is how I m running my script:
./script1.sh 2014-11-02T18-00-00.048000 2014-11-03T06-59-59.981000
There is no error message and the code doesn't give the right output.
You could use a Perl one-liner with the range operator:
perl -ne "print if /\Q$1\E/../\Q$2\E/" night4.info >> list.info
There's a file with some lines containing some text and either date or time stamp:
...
string1-20141001
string2-1414368000000
string3-1414454400000
...
I want to quickly convert time stamps to dates, like this:
$ date -d #1414368000 +"%Y%m%d"
20141027
and I want to do this dynamically with sed or some similar command line tool. For testing I unsuccessfully use this:
$ echo "something-1414454400000" | sed "s/-\(..........\)...$/-$(date -d #\\1 +'%Y%m%d')/"
date: invalid date '#\\1'
something-
but echoing seems to be working:
$ echo "something-1414454400000" | sed "s/-\(..........\)...$/-$(echo \\1)/"
something-1414454400
so what could be done?
It's interesting what's happening here. Some pointers:
Always single-quote your regex for sed, if possible, when using BASH (etc), especially if using special characters like$. This is why date is being run (with -d #\\1) before sed even gets involved.
Your "working" echo example isn't, actually (I believe): echo \\1 produces \1 (and as above, will do so before sed even gets invoked). This then happens to valid sed replacement syntax, so will substitute your group on the LHS, which is why the output looks about right.
Note that by using -r, you can use easier / more advanced regex syntax.
Hard to say exactly what to do without a bit more context, but to fix the immediate problems, try something like:
echo "something-1414454400000" | sed -re 's/-([0-9]{10,}).+/-$(date -d #\1 +"%Y%m%d")/'
which produces: $(date -d #1414454400) (which you can then pipe to sh)
Or for a more complete solution, you can change the regex to produce a shell command directly, and pipe it:
echo "something-1414454400000" | sed -re 's/(.*-)([0-9]{10,10}).+/echo \1$(date -d #\2 \"+%Y%M%d\")/' | sh
..producing something-20140028
You can do this in BASH:
while read -r p; do
if [[ "$p" =~ ^(.+-)([0-9]{10}).{3}$ ]]; then
echo -n "${BASH_REMATCH[1]}"
date -d "#${BASH_REMATCH[2]}" +"%Y%m%d"
else
echo "$p"
fi
done < file
OUTPUT:
string1-20141001
string2-20141026
string3-20141027
awk -F- 'BEGIN { OFS=FS }
$2 ~ /^[0-9]{13}$/ {
"date -d#" $2/1000 " +%Y%m%d " | getline t; $2=t }1'
Just try this command. I have checked it. It is working on your inputs.
cat file | sed -E "s,(.*)-(.*),\1-`date -d #1414368000 +'%Y%m%d'`,g"
I have written the following filter as a function in my ~/.bash_profile:
hilite() {
export REGEX_SED=$(echo $1 | sed "s/[|()]/\\\&/g")
while read line
do
echo $line | egrep "$1" | sed "s/$REGEX_SED/\x1b[7m&\x1b[0m/g"
done
exit 0
}
to find lines of anything piped into it matching a regular expression, and highlight matches using ANSI escape codes on a VT100-compatible terminal.
For example, the following finds and highlights the strings bin, U or 1 which are whole words in the last 10 lines of /etc/passwd:
tail /etc/passwd | hilite "\b(bin|[U1])\b"
However, the script runs very slowly as each line forks an echo, egrep and sed.
In this case, it would be more efficient to do egrep on the entire input, and then run sed on its output.
How can I modify my function to do this? I would prefer to not create any temporary files if possible.
P.S. Is there another way to find and highlight lines in a similar way?
sed can do a bit of grepping itself: if you give it the -n flag (or #n instruction in a script) it won't echo any output unless asked. So
while read line
do
echo $line | egrep "$1" | sed "s/$REGEX_SED/\x1b[7m&\x1b[0m/g"
done
could be simplified to
sed -n "s/$REGEX_SED/\x1b[7m&\x1b[0m/gp"
EDIT:
Here's the whole function:
hilite() {
REGEX_SED=$(echo $1 | sed "s/[|()]/\\\&/g");
sed -n "s/$REGEX_SED/\x1b[7m&\x1b[0m/gp"
}
That's all there is to it - no while loop, reading, grepping, etc.
If your egrep supports --color, just put this in .bash_profile:
hilite() { command egrep --color=auto "$#"; }
(Personally, I would name the function egrep; hence the usage of command).
I think you can replace the whole while loop with simply
sed -n "s/$REGEX_SED/\x1b[7m&\x1b[0m/gp"
because sed can read from stdin line-by-line so you don't need read
I'm not sure if running egrep and piping to sed is faster than using sed alone, but you can always compare using time.
Edit: added -n and p to sed to print only highlighted lines.
Well, you could simply do this:
egrep "$1" $line | sed "s/$REGEX_SED/\x1b[7m&\x1b[0m/g"
But I'm not sure that it'll be that much faster ; )
Just for the record, this is a method using a temporary file:
hilite() {
export REGEX_SED=$(echo $1 | sed "s/[|()]/\\\&/g")
export FILE=$2
if [ -z "$FILE" ]
then
export FILE=~/tmp
echo -n > $FILE
while read line
do
echo $line >> $FILE
done
fi
egrep "$1" $FILE | sed "s/$REGEX_SED/\x1b[7m&\x1b[0m/g"
return $?
}
which also takes a file/pathname as the second argument, for case like
cat /etc/passwd | hilite "\b(bin|[U1])\b"
my question seems to be general, but i can't find any answers.
In sed command, how can you replace the substitution pattern by a value returned by a simple bash function.
For instance, I created the following function :
function parseDates(){
#Some process here with $1 (the pattern found)
return "dateParsed;
}
and the folowing sed command :
myCatFile=`sed -e "s/[0-3][0-9]\/[0-1][0-9]\/[0-9][0-9]/& parseDates &\}/p" myfile`
I found that the caracter '&' represents the current pattern found, i'd like it to be passed to my bash function and the whole pattern to be substituted by the pattern found +dateParsed.
Does anybody have an idea ?
Thanks
you can use the "e" option in sed command like this:
cat t.sh
myecho() {
echo ">>hello,$1<<"
}
export -f myecho
sed -e "s/.*/myecho &/e" <<END
ni
END
you can see the result without "e":
cat t.sh
myecho() {
echo ">>hello,$1<<"
}
export -f myecho
sed -e "s/.*/myecho &/" <<END
ni
END
Agree with Glenn Jackman.
If you want to use bash function in sed, something like this :
sed -rn 's/^([[:digit:].]+)/`date -d #&`/p' file |
while read -r line; do
eval echo "$line"
done
My file here begins with a unix timestamp (e.g. 1362407133.936).
Bash function inside sed (maybe for other purposes):
multi_stdin(){ #Makes function accepet variable or stdin (via pipe)
[[ -n "$1" ]] && echo "$*" || cat -
}
sans_accent(){
multi_stdin "$#" | sed '
y/àáâãäåèéêëìíîïòóôõöùúûü/aaaaaaeeeeiiiiooooouuuu/
y/ÀÁÂÃÄÅÈÉÊËÌÍÎÏÒÓÔÕÖÙÚÛÜ/AAAAAAEEEEIIIIOOOOOUUUU/
y/çÇñÑߢÐð£Øø§µÝý¥¹²³ªº/cCnNBcDdLOoSuYyY123ao/
'
}
eval $(echo "Rogério Madureira" | sed -n 's#.*#echo & | sans_accent#p')
or
eval $(echo "Rogério Madureira" | sed -n 's#.*#sans_accent &#p')
Rogerio
And if you need to keep the output into a variable:
VAR=$( eval $(echo "Rogério Madureira" | sed -n 's#.*#echo & | desacentua#p') )
echo "$VAR"
do it step by step. (also you could use an alternate delimiter , such as "|" instead of "/"
function parseDates(){
#Some process here with $1 (the pattern found)
return "dateParsed;
}
value=$(parseDates)
sed -n "s|[0-3][0-9]/[0-1][0-9]/[0-9][0-9]|& $value &|p" myfile
Note the use of double quotes instead of single quotes, so that $value can be interpolated
I'd like to know if there's a way to do this too. However, for this particular problem you don't need it. If you surround the different components of the date with ()s, you can back reference them with \1 \2 etc and reformat however you want.
For instance, let's reverse 03/04/1973:
echo 03/04/1973 | sed -e 's/\([0-9][0-9]\)\/\([0-9][0-9]\)\/\([0-9][0-9][0-9][0-9]\)/\3\/\2\/\1/g'
sed -e 's#[0-3][0-9]/[0-1][0-9]/[0-9][0-9]#& $(parseDates &)#' myfile |
while read -r line; do
eval echo "$line"
done
You can glue together a sed-command by ending a single-quoted section, and reopening it again.
sed -n 's|[0-3][0-9]/[0-1][0-9]/[0-9][0-9]|& '$(parseDates)' &|p' datefile
However, in contrast to other examples, a function in bash can't return strings, only put them out:
function parseDates(){
# Some process here with $1 (the pattern found)
echo dateParsed
}