UPDATED:
Using sed, how can I insert (NOT SUBSTITUTE) a new line on only the first match of keyword for each file.
Currently I have the following but this inserts for every line containing Matched Keyword and I want it to only insert the New Inserted Line for only the first match found in the file:
sed -ie '/Matched Keyword/ i\New Inserted Line' *.*
For example:
Myfile.txt:
Line 1
Line 2
Line 3
This line contains the Matched Keyword and other stuff
Line 4
This line contains the Matched Keyword and other stuff
Line 6
changed to:
Line 1
Line 2
Line 3
New Inserted Line
This line contains the Matched Keyword and other stuff
Line 4
This line contains the Matched Keyword and other stuff
Line 6
You can sort of do this in GNU sed:
sed '0,/Matched Keyword/s//New Inserted Line\n&/'
But it's not portable. Since portability is good, here it is in awk:
awk '/Matched Keyword/ && !x {print "Text line to insert"; x=1} 1' inputFile
Or, if you want to pass a variable to print:
awk -v "var=$var" '/Matched Keyword/ && !x {print var; x=1} 1' inputFile
These both insert the text line before the first occurrence of the keyword, on a line by itself, per your example.
Remember that with both sed and awk, the matched keyword is a regular expression, not just a keyword.
UPDATE:
Since this question is also tagged bash, here's a simple solution that is pure bash and doesn't required sed:
#!/bin/bash
n=0
while read line; do
if [[ "$line" =~ 'Matched Keyword' && $n = 0 ]]; then
echo "New Inserted Line"
n=1
fi
echo "$line"
done
As it stands, this as a pipe. You can easily wrap it in something that acts on files instead.
If you want one with sed*:
sed '0,/Matched Keyword/s//Matched Keyword\nNew Inserted Line/' myfile.txt
*only works with GNU sed
This might work for you:
sed -i -e '/Matched Keyword/{i\New Inserted Line' -e ':a;n;ba}' file
You're nearly there! Just create a loop to read from the Matched Keyword to the end of the file.
After inserting a line, the remainder of the file can be printed out by:
Introducing a loop place holder :a (here a is an arbitrary name).
Print the current line and fetch the next into the pattern space with the ncommand.
Redirect control back using the ba command which is essentially a goto to the a place holder. The end-of-file condition is naturally taken care of by the n command which terminates any further sed commands if it tries to read passed the end-of-file.
With a little help from bash, a true one liner can be achieved:
sed $'/Matched Keyword/{iNew Inserted Line\n:a;n;ba}' file
Alternative:
sed 'x;/./{x;b};x;/Matched Keyword/h;//iNew Inserted Line' file
This uses the Matched Keyword as a flag in the hold space and once it has been set any processing is curtailed by bailing out immediately.
If you want to append a line after first match only, use AWK instead of SED as below
awk '{print} /Matched Keyword/ && !n {print "New Inserted Line"; n++}' myfile.txt
Output:
Line 1
Line 2
Line 3
This line contains the Matched Keyword and other stuff
New Inserted Line
Line 4
This line contains the Matched Keyword and other stuff
Line 6
I have a .csv file that contains double quoted multi-line fields. I need to convert the multi-line cell to a single line. It doesn't show in the sample data but I do not know which fields might be multi-line so any solution will need to check every field. I do know how many columns I'll have. The first line will also need to be skipped. I don't how much data so performance isn't a consideration.
I need something that I can run from a bash script on Linux. Preferably using tools such as awk or sed and not actual programming languages.
The data will be processed further with Logstash but it doesn't handle double quoted multi-line fields hence the need to do some pre-processing.
I tried something like this and it kind of works on one row but fails on multiple rows.
sed -e :0 -e '/,.*,.*,.*,.*,/b' -e N -e '1n;N;N;N;s/\n/ /g' -e b0 file.csv
CSV example
First name,Last name,Address,ZIP
John,Doe,"Country
City
Street",12345
The output I want is
First name,Last name,Address,ZIP
John,Doe,Country City Street,12345
Jane,Doe,Country City Street,67890
etc.
etc.
First my apologies for getting here 7 months late...
I came across a problem similar to yours today, with multiple fields with multi-line types. I was glad to find your question but at least for my case I have the complexity that, as more than one field is conflicting, quotes might open, close and open again on the same line... anyway, reading a lot and combining answers from different posts I came up with something like this:
First I count the quotes in a line, to do that, I take out everything but quotes and then use wc:
quotes=`echo $line | tr -cd '"' | wc -c` # Counts the quotes
If you think of a single multi-line field, knowing if the quotes are 1 or 2 is enough. In a more generic scenario like mine I have to know if the number of quotes is odd or even to know if the line completes the record or expects more information.
To check for even or odd you can use the mod operand (%), in general:
even % 2 = 0
odd % 2 = 1
For the first line:
Odd means that the line expects more information on the next line.
Even means the line is complete.
For the subsequent lines, I have to know the status of the previous one. for instance in your sample text:
First name,Last name,Address,ZIP
John,Doe,"Country
City
Street",12345
You can say line 1 (John,Doe,"Country) has 1 quote (odd) what means the status of the record is incomplete or open.
When you go to line 2, there is no quote (even). Nevertheless this does not mean the record is complete, you have to consider the previous status... so for the lines following the first one it will be:
Odd means that record status toggles (incomplete to complete).
Even means that record status remains as the previous line.
What I did was looping line by line while carrying the status of the last line to the next one:
incomplete=0
cat file.csv | while read line; do
quotes=`echo $line | tr -cd '"' | wc -c` # Counts the quotes
incomplete=$((($quotes+$incomplete)%2)) # Check if Odd or Even to decide status
if [ $incomplete -eq 1 ]; then
echo -n "$line " >> new.csv # If line is incomplete join with next
else
echo "$line" >> new.csv # If line completes the record finish
fi
done
Once this was executed, a file in your format generates a new.csv like this:
First name,Last name,Address,ZIP
John,Doe,"Country City Street",12345
I like one-liners as much as everyone, I wrote that script just for the sake of clarity, you can - arguably - write it in one line like:
i=0;cat file.csv|while read l;do i=$((($(echo $l|tr -cd '"'|wc -c)+$i)%2));[[ $i = 1 ]] && echo -n "$l " || echo "$l";done >new.csv
I would appreciate it if you could go back to your example and see if this works for your case (which you most likely already solved). Hopefully this can still help someone else down the road...
Recovering the multi-line fields
Every need is different, in my case I wanted the records in one line to further process the csv to add some bash-extracted data, but I would like to keep the csv as it was. To accomplish that, instead of joining the lines with a space I used a code - likely unique - that I could then search and replace:
i=0;cat file.csv|while read l;do i=$((($(echo $l|tr -cd '"'|wc -c)+$i)%2));[[ $i = 1 ]] && echo -n "$l ~newline~ " || echo "$l";done >new.csv
the code is ~newline~, this is totally arbitrary of course.
Then, after doing my processing, I took the csv text file and replaced the coded newlines with real newlines:
sed -i 's/ ~newline~ /\n/g' new.csv
References:
Ternary operator: https://stackoverflow.com/a/3953666/6316852
Count char occurrences: https://stackoverflow.com/a/41119233/6316852
Other peculiar cases: https://www.linuxquestions.org/questions/programming-9/complex-bash-string-substitution-of-csv-file-with-multiline-data-937179/
TL;DR
Run this:
i=0;cat file.csv|while read l;do i=$((($(echo $l|tr -cd '"'|wc -c)+$i)%2));[[ $i = 1 ]] && echo -n "$l " || echo "$l";done >new.csv
... and collect results in new.csv
I hope it helps!
If Perl is your option, please try the following:
perl -e '
while (<>) {
$str .= $_;
}
while ($str =~ /("(("")|[^"])*")|((^|(?<=,))[^,]*((?=,)|$))/g) {
if (($el = $&) =~ /^".*"$/s) {
$el =~ s/^"//s; $el =~ s/"$//s;
$el =~ s/""/"/g;
$el =~ s/\s+(?!$)/ /g;
}
push(#ary, $el);
}
foreach (#ary) {
print /\n$/ ? "$_" : "$_,";
}' sample.csv
sample.csv:
First name,Last name,Address,ZIP
John,Doe,"Country
City
Street",12345
John,Doe,"Country
City
Street",67890
Result:
First name,Last name,Address,ZIP
John,Doe,Country City Street,12345
John,Doe,Country City Street,67890
This might work for you (GNU sed):
sed ':a;s/[^,]\+/&/4;tb;N;ba;:b;s/\n\+/ /g;s/"//g' file
Test each line to see that it contains the correct number of fields (in the example that was 4). If there are not enough fields, append the next line and repeat the test. Otherwise, replace the newline(s) by spaces and finally remove the "'s.
N.B. This may be fraught with problems such as ,'s between "'s and quoted "'s.
Try cat -v file.csv. When the file was made with Excel, you might have some luck: When the newlines in a field are a simple \n and the newline at the end is a \r\n (which will look like ^M), parsing is simple.
# delete all newlines and replace the ^M with a new newline.
tr -d "\n" < file.csv| tr "\r" "\n"
# Above two steps with one command
tr "\n\r" " \n" < file.csv
When you want a space between the joined line, you need an additional step.
tr "\n\r" " \n" < file.csv | sed '2,$ s/^ //'
EDIT: #sjaak commented this didn't work is his case.
When your broken lines also have ^M you still can be a lucky (wo-)man.
When your broken field is always the first field in double quotes and you have GNU sed 4.2.2, you can join 2 lines when the first line has exactly one double quote.
sed -rz ':a;s/(\n|^)([^"]*)"([^"]*)\n/\1\2"\3 /;ta' file.csv
Explanation:
-z don't use \n as line endings
:a label for repeating the step after successful replacement
(\n|^) Search after a newline or the very first line
([^"]*) Substring without a "
ta Go back to label a and repeat
awk pattern matching is working.
answer in one line :
awk '/,"/{ORS=" "};/",/{ORS="\n"}{print $0}' YourFile
if you'd like to drop quotes, you could use:
awk '/,"/{ORS=" "};/",/{ORS="\n"}{print $0}' YourFile | sed 's/"//gw NewFile'
but I prefer to keep it.
to explain the code:
/Pattern/ : find pattern in current line.
ORS : indicates the output line record.
$0 : indicates the whole of the current line.
's/OldPattern/NewPattern/': substitude first OldPattern with NewPattern
/g : does the previous action for all OldPattern
/w : write the result to Newfile
I am working with a fasta file and need to add line-specific text to each of the headers. So for example if my file is:
>TER1
AGCATGCTAGCTAGTCGACTCGATCGCATGCTC
>TER2
AGCATGCTAGCTAGACGACTCGATCGCATGCTC
>URC1
AGCATGCTAGCTAGTCGACTCGATCGCATGCTC
>URC2
AGCATGCTACCTAGTCGACTCGATCGCATGCTC
>UCR3
AGCATGCTAGCTAGTCGACTCGATGGCATGCTC
I want a while loop that will read through each line; for those with a > at the start, I want to append |population: plus the first three characters after the >. So line one would be:
>TER1|population:TER
etc.
I can't figure out how to make this work. Here my best attempt so far.
filename="testfasta.fa"
while read -r line
do
if [[ "$line" == ">"* ]]; then
id=$(cut -c2-4<<<"$line")
printf $line"|population:"$id"\n" >>outfile
else
printf $line"\n">>outfile
fi
done <"$filename"
This produces a file with the original headers and following line each on a single line.
Can someone tell me where I'm going wrong? My if and else loop aren't working at all!
Thanks!
You could use a while loop if you really want,
but sed would be simpler:
sed -e 's/^>\(...\).*/&|population:\1/' "$filename"
That is, for lines starting with > (pattern: ^>),
capture the next 3 characters (with \(...\)),
and match the rest of the line (.*),
replace with the line as it was (&),
and the fixed string |population:,
and finally the captured 3 characters (\1).
This will produce for your input:
>TER1|population:TER
AGCATGCTAGCTAGTCGACTCGATCGCATGCTC
>TER2|population:TER
AGCATGCTAGCTAGACGACTCGATCGCATGCTC
>URC1|population:URC
AGCATGCTAGCTAGTCGACTCGATCGCATGCTC
>URC2|population:URC
AGCATGCTACCTAGTCGACTCGATCGCATGCTC
>UCR3|population:UCR
AGCATGCTAGCTAGTCGACTCGATGGCATGCTC
Or you can use this awk, also producing the same output:
awk '{sub(/^>.*/, $0 "|population:" substr($0, 2, 3))}1' "$filename"
You can do this quickly in awk:
awk '$1~/^>/{$1=$1"|population:"substr($1,2,3)}{}1' infile.txt > outfile.txt
$ awk '$1~/^>/{$1=$1"|population:"substr($1,2,3)}{}1' testfile
>TER1|population:TER
AGCATGCTAGCTAGTCGACTCGATCGCATGCTC
>TER2|population:TER
AGCATGCTAGCTAGACGACTCGATCGCATGCTC
>URC1|population:URC
AGCATGCTAGCTAGTCGACTCGATCGCATGCTC
>URC2|population:URC
AGCATGCTACCTAGTCGACTCGATCGCATGCTC
>UCR3|population:UCR
AGCATGCTAGCTAGTCGACTCGATGGCATGCTC
Here awk will:
Test if the record starts with a > The $1 looks at the first field, but $0 for the entire record would work just as well in this case. The ~ will perform a regex test, and ^> means "Starts with >". Making the test: ($1~/^>/)
If so it will set the first field to the output you are looking for (using substr() to get the bits of the string you want. {$1=$1"|population:"substr($1,2,3)}
Finally it will print out the entire record (with the changes if applicable): {}1 which is shorthand for {print $0} or.. print the entire record.
Looking for a solution for a bash script using sed or awk to comment out a line, only if the previous line contains a matching string.
For example, a file containing:
...
if [ $V1 -gt 100 ]; then
some specific commands
else
some other specific commands
fi
...
I'd like to comment out the line containing else but ONLY if the previous line contains specific.
I've attempted piping multiple sed commands along with grep commands to no avail.
sed -E '/specific/{n;s/^([[:blank:]]*)else$/\1#else/}'
Output
...
if [ $V1 -gt 100 ]; then
some specific commands
#else
some other commands
fi
...
A retrospection
/specific/ look for the line containing the pattern specific
n add the next line to the pattern space. n auto prints the current pattern space.
Check if the next line is (one_or_more_spaces)else,if yes, substitute the line with a (one_or_more_spaces_found_previously)#else. Remember () is for pattern reuse and \1 is the previously matched pattern reused.
-E enable extended regex
-i is for inplace edit of the actual file
You can use this awk solution:
awk '/specific/{p=NR} NR==p+1{p=0; if (/^[[:blank:]]*else/) $0 = "#" $0} 1' file
if [ $V1 -gt 100 ]; then
some specific commands
#else
some other commands
fi
In this block /specific/p=NR we find specific and store current line # in p
Next block is executed for very next line due to p == NR+1 condition
We rest p=0 and if that line has else at start with optional whitespaces before we just comment it out.
I want to extract the first column of the last line of a text file. Instead of output the content of interest in another file and read it in again, can I just use some command to read it into a variable directly?
For exampole, if my file is like this:
...
123 456 789(this is the last line)
What I want is to read 123 into a variable in my shell script. How can I do that?
One approach is to extract the line you want, read its columns into an array, and emit the array element you want.
For the last line:
#!/bin/bash
# ^^^^- not /bin/sh, to enable arrays and process substitution
read -r -a columns < <(tail -n 1 "$filename") # put last line's columns into an array
echo "${columns[0]}" # emit the first column
Alternately, awk is an appropriate tool for the job:
line=2
column=1
var=$(awk -v line="$line" -v col="$column" 'NR == line { print $col }' <"$filename")
echo "Extracted the value: $var"
That said, if you're looking for a line close to the start of a file, it's often faster (in a runtime-performance sense) and easier to stick to shell builtins. For instance, to take the third column of the second line of a file:
{
read -r _ # throw away first line
read -r _ _ value _ # extract third value of second line
} <"$filename"
This works by using _s as placeholders for values you don't want to read.
I guess with "first column", you mean "first word", do you?
If it is guaranteed, that the last line doesn't start with a space, you can do
tail -n 1 YOUR_FILE | cut -d ' ' -f 1
You could also use sed:
$> var=$(sed -nr '$s/(^[^ ]*).*/\1/p' "file.txt")
The -nr tells sed to not output data by default (-n) and use extended regular expressions (-r to avoid needing to escape the paranthesis otherwise you have to write \( \))). The $ is an address that specifies the last line. The regular expression anchors the beginning of the line with the first ^, then matches everything that is not a space [^ ]* and puts that the result into a capture group ( ) and then gets rid of the rest of the line .* by replacing the line with the capture group \1, then print p to print the line.