#!/bin/bash
{ cat sample.txt; echo; } | while read -r -a A_Name; do
if [ ! -z "${A_Name[0]}" ]; then
echo " ${A_Name[0]%.isx} "
fi
done
I am trying to display contents of a text file (which includes .isx files) using while loop but when i try to eliminate extension with %, it doesnt work.
Output
.isx is appearing for first two values:
./test.sh
abc.isx
def.isx
ghi
Input
sample.txt file:
abc.isx
def.isx
ghi.isx
Please, assist. Thank you.
why so complicated?
#!/bin/bash
cat sample.txt | while read line; do
echo "${line%.isx}"
done
or with sed
sed "s/\.isx//" sample.txt >output.txt
or with sed and inplace replacement
sed -i "s/\.isx//" sample.txt
Related
I am at my wits end as to why this loop is failing to concatenate the files the way I need it. Basically, lets say we have following files:
AB124661.lane3.R1.fastq.gz
AB124661.lane4.R1.fastq.gz
AB124661.lane3.R2.fastq.gz
AB124661.lane4.R2.fastq.gz
What we want is:
cat AB124661.lane3.R1.fastq.gz AB124661.lane4.R1.fastq.gz > AB124661.R1.fastq.gz
cat AB124661.lane3.R2.fastq.gz AB124661.lane4.R2.fastq.gz > AB124661.R2.fastq.gz
What I tried (and didn't work):
Create and save file names (AB124661) to a ID file:
ls -1 R1.gz | awk -F '.' '{print $1}' | sort | uniq > ID
This creates an ID file that stores the samples/files name.
Run the following loop:
for i in `cat ./ID`; do cat $i\.lane3.R1.fastq.gz $i\.lane4.R1.fastq.gz \> out/$i\.R1.fastq.gz; done
for i in `cat ./ID`; do cat $i\.lane3.R2.fastq.gz $i\.lane4.R2.fastq.gz \> out/$i\.R2.fastq.gz; done
The loop fails and concatenates into empty files.
Things I tried:
Yes, the ID file is definitely in the folder
When I run with echo it shows the cat command correct
Any help will be very much appreciated,
Best,
AC
why are you escaping the \> ? That's going to result in a cat: '>': No such file or directory instead of a redirection.
Don't read lines with for
while IFS= read -r id; do
cat "${id}.lane3.R1.fastq.gz" "${id}.lane4.R1.fastq.gz" > "out/${id}.R1.fastq.gz"
cat "${id}.lane3.R2.fastq.gz" "${id}.lane4.R2.fastq.gz" > "out/${id}.R2.fastq.gz"
done < ./ID
Let say you have id stored in file ./ID per line
while read -r line; do
cat "$line".lane3.R1.fastq.gz "$line".lane4.R1.fastq.gz > "$line".R1.fastq.gz
cat "$line".lane3.R2.fastq.gz "$line".lane4.R2.fastq.gz > "$line".R2.fastq.gz
done < ./ID
A pure shell solution could be like that:
for file in *.fastq.gz; do
id=${file%%.*}
[ -e "$id".R1.fastq.gz ] || cat "$id".*.R1.fastq.gz > "$id".R1.fastq.gz
[ -e "$id".R2.fastq.gz ] || cat "$id".*.R2.fastq.gz > "$id".R2.fastq.gz
done
Alternatively:
printf '%s\n' *.fastq.gz | cut -d. -f1 | sort -u |
while IFS= read -r id; do
cat "$id".*.R1.fastq.gz > "$id".R1.fastq.gz
cat "$id".*.R2.fastq.gz > "$id".R2.fastq.gz
done
This solution assumes filenames of interest don't contain newline characters.
How do I add a string after each line in a file using bash? Can it be done using the sed command, if so how?
If your sed allows in place editing via the -i parameter:
sed -e 's/$/string after each line/' -i filename
If not, you have to make a temporary file:
typeset TMP_FILE=$( mktemp )
touch "${TMP_FILE}"
cp -p filename "${TMP_FILE}"
sed -e 's/$/string after each line/' "${TMP_FILE}" > filename
I prefer echo. using pure bash:
cat file | while read line; do echo ${line}$string; done
I prefer using awk.
If there is only one column, use $0, else replace it with the last column.
One way,
awk '{print $0, "string to append after each line"}' file > new_file
or this,
awk '$0=$0"string to append after each line"' file > new_file
If you have it, the lam (laminate) utility can do it, for example:
$ lam filename -s "string after each line"
Pure POSIX shell and sponge:
suffix=foobar
while read l ; do printf '%s\n' "$l" "${suffix}" ; done < file |
sponge file
xargs and printf:
suffix=foobar
xargs -L 1 printf "%s${suffix}\n" < file | sponge file
Using join:
suffix=foobar
join file file -e "${suffix}" -o 1.1,2.99999 | sponge file
Shell tools using paste, yes, head
& wc:
suffix=foobar
paste file <(yes "${suffix}" | head -$(wc -l < file) ) | sponge file
Note that paste inserts a Tab char before $suffix.
Of course sponge can be replaced with a temp file, afterwards mv'd over the original filename, as with some other answers...
This is just to add on using the echo command to add a string at the end of each line in a file:
cat input-file | while read line; do echo ${line}"string to add" >> output-file; done
Adding >> directs the changes we've made to the output file.
Sed is a little ugly, you could do it elegantly like so:
hendry#i7 tmp$ cat foo
bar
candy
car
hendry#i7 tmp$ for i in `cat foo`; do echo ${i}bar; done
barbar
candybar
carbar
I need to output an array to a file in the following format.
File: a.txt
b.txt
I tried doing the following :
declare -a files=("a.txt" "b.txt")
empty=""
printf "File:" >> files.txt
for i in "${files[#]}"
do
printf "%-7s %-30s \n" "$empty" "$i" >> files.txt
done
But, I get the output as
File: a.txt
b.txt
Can anyone please help me to get the output in the required format.
#!/bin/bash
files=( 'a.txt' 'b.txt' 'c.txt' 'd.txt' )
set -- "${files[#]}"
printf 'File: %s\n' "$1"
shift
printf ' %s\n' "$#"
Output:
File: a.txt
b.txt
c.txt
d.txt
This uses the fact that printf will reuse its formatting string for all its other command line parameters.
We set the positional parameters to the list and then output the first element with the File: string prepended. We then shift $1 off the list of positional parameters and print the rest with a spacer string inserted.
Using sed
#!/bin/bash
declare -a files=("a.txt" "b.txt")
for i in "${files[#]}"
do
echo "$i" >> files.txt
done
sed -i '1 s/^/File: /' files.txt
sed -i '1 ! s/^/ /' files.txt
If you are using Mac, you have to modify sed commands in this way
sed -i '' '1 s/^/File: /' files.txt
sed -i '' '1 ! s/^/ /' files.txt
The output will be:
File: a.txt
b.txt
First of all we put into txt file all the file names (for loop). After that, via first sed command we add File: to the first line and via second sed command we add to all lines, except the first, six spaces equal to length of string File:
You could always started with a variable containing File: for the first iteration, and overwrite it with the correct number of spaces each time. The repeated assignment won't induce too much overhead.
prefix="File:"
for i in "${files[#]}"
do
printf "%-7s %-30s \n" "$prefix" "$i"
prefix=
done > Files.txt
I have been working in bash, and need to create a string argument. bash is a newish for me, to the point that I dont know how to build a string in bash from a list.
// foo.txt is a list of abs file names.
/foo/bar/a.txt
/foo/bar/b.txt
/delta/test/b.txt
should turn into: a.txt,b.txt,b.txt
OR: /foo/bar/a.txt,/foo/bar/b.txt,/delta/test/b.txt
code
s = ""
for file in $(cat foo.txt);
do
#what goes here? s += $file ?
done
myShellScript --script $s
I figure there was an easy way to do this.
with for loop:
for file in $(cat foo.txt);do echo -n "$file",;done|sed 's/,$/\n/g'
with tr:
cat foo.txt|tr '\n' ','|sed 's/,$/\n/g'
only sed:
sed ':a;N;$!ba;s/\n/,/g' foo.txt
This seems to work:
#!/bin/bash
input="foo.txt"
while IFS= read -r var
do
basename $var >> tmp
done < "$input"
paste -d, -s tmp > result.txt
output: a.txt,b.txt,b.txt
basename gets you the file names you need and paste will put them in the order you seem to need.
The input field separator can be used with set to create split/join functionality:
# split the lines of foo.txt into positional parameters
IFS=$'\n'
set $(< foo.txt)
# join with commas
IFS=,
echo "$*"
For just the file names, add some sed:
IFS=$'\n'; set $(sed 's|.*/||' foo.txt); IFS=,; echo "$*"
I want convert a column of data in a txt file to a row of a csv file using unix commands.
example:
ApplChk1,
ApplChk2,
v_baseLoanAmountTI,
v_plannedClosingDateField,
downPaymentTI,
this is a column which present in a txt file
I want output as follows in a csv file
ApplChk1,ApplChk2,v_baseLoanAmountTI,v_plannedClosingDateField,downPaymentTI,
Please let me know how to do it.
Thanks in advance
If that's a single column, which you want to convert to row, then there are many possibilities:
tr -d '\n' < filename ; echo # option 1 OR
xargs echo -n < filename ; echo # option 2 (This option however, will shrink spaces & eat quotes) OR
while read x; do echo -n "$x" ; done < filename; echo # option 3
Please let us know, how the input would look like, for multi-line case.
A funny pure bash solution (bash ≥ 4.1):
mapfile -t < file.txt; printf '%s' "${MAPFILE[#]}" $'\n'
Done!
for i in `< file.txt` ; do echo -n $i; done; echo ""
gives the output
ApplChk1,ApplChk2,v_baseLoanAmountTI,v_plannedClosingDateField,downPaymentTI,
To send output to a file:
{ for i in `< file.txt` ; do echo -n $i ; done; echo; } > out.csv
When I run it, this is what happens:
[jenny#jennys:tmp]$ more file.txt
ApplChk1,
ApplChk2,
v_baseLoanAmountTI,
v_plannedClosingDateField,
downPaymentTI,
[jenny#jenny:tmp]$ { for i in `< file.txt` ; do echo -n $i ; done; echo; } > out.csv
[jenny#jenny:tmp]$ more out.csv
ApplChk1,ApplChk2,v_baseLoanAmountTI,v_plannedClosingDateField,downPaymentTI,
perl -pe 's/\n//g' your_file
the above will output to stdout.
if you want to do it in place:
perl -pi -e 's/\n//g' your_file
You could use the Linux command sed to replace line \n breaks by commas , or space :
sed -z 's/\n/,/g' test.txt > test.csv
You could also add the -i option if you want to change file in-place :
sed -i -z 's/\n/,/g' test.txt