gnuplot for cycle and spaces in filename - bash

I have small script in bash, which is generating graphs via gnuplot.
Everything works fine until names of input files contain space(s).
Here's what i've got:
INPUTFILES=("data1.txt" "data2 with spaces.txt" "data3.txt")
...
#MAXROWS is set earlier, not relevant.
for LINE in $( seq 0 $(( MAXROWS - 1 )) );do
gnuplot << EOF
reset
set terminal png
set output "out/graf_${LINE}.png"
filenames="${INPUTFILES[#]}"
set multiplot
plot for [file in filenames] file every ::0::${LINE} using 1:2 with line title "graf_${LINE}"
unset multiplot
EOF
done
This code works, but only without spaces in names of input files.
In the example gnuplot evaluate this:
1 iteration: file=data1.txt - CORRECT
2 iteration: file=data2 - INCORRECT
3 iteration: file=with - INCORRECT
4 iteration: file=spaces.txt - INCORRECT

The quick answer is that you can't do exactly what you want to do. Gnuplot splits the string in an iteration on spaces and there's no way around that (AFIK). Depending on what you want, there may be a "Work-around". You can write a (recursive) function in gnuplot to replace a character string with another --
#S,C & R stand for STRING, CHARS and REPLACEMENT to help this be a little more legible.
replace(S,C,R)=(strstrt(S,C)) ? \
replace( S[:strstrt(S,C)-1].R.S[strstrt(S,C)+strlen(C):] ,C,R) : S
Bonus points to anyone who can figure out how to do this without recursion...
Then your (bash) loop looks something like:
INPUTFILES_BEFORE=("data1.txt" "data2 with spaces.txt" "data3.txt")
INPUTFILES=()
#C style loop to avoid changing IFS -- Sorry SO doesn't like the #...
#This loop pre-processes files and changes spaces to '#_#'
for (( i=0; i < ${#INPUTFILES_BEFORE[#]}; i++)); do
FILE=${INPUTFILES_BEFORE[${i}]}
INPUTFILES+=( "`echo ${FILE} | sed -e 's/ /#_#/g'`" ) #replace ' ' with '#_#'
done
which preprocesses your input files to add '#_#' to the filenames which have spaces in them... Finally, the "complete" script:
...
INPUTFILES_BEFORE=("data1.txt" "data2 with spaces.txt" "data3.txt")
INPUTFILES=()
for (( i=0; i < ${#INPUTFILES_BEFORE[#]}; i++)); do
FILE=${INPUTFILES_BEFORE[${i}]}
INPUTFILES+=( "`echo ${FILE} | sed -e 's/ /#_#/g'`" ) #replace ' ' with '#_#'
done
for LINE in $( seq 0 $(( MAXROWS - 1 )) );do
gnuplot <<EOF
filenames="${INPUTFILES[#]}"
replace(S,C,R)=(strstrt(S,C)) ? \
replace( S[:strstrt(S,C)-1].R.S[strstrt(S,C)+strlen(C):] , C ,R) : S
#replace '#_#' with ' ' in filenames.
plot for [file in filenames] replace(file,'#_#',' ') every ::0::${LINE} using 1:2 with line title "graf_${LINE}"
EOF
done
However, I think the take-away here is that you shouldn't use spaces in filenames ;)

Escape the spaces:
"data2\ with\ spaces.txt"
EDIT
It seems that even with escape sequences, as you have mentioned, the bash for will always parse the input on the spaces.
Can you convert your script to work in a while loop fashion:
http://ubuntuforums.org/showthread.php?t=83424
This also may be a solution, but it's new to me and I'm still playing with it to understand exactly what it's doing:
http://www.cyberciti.biz/tips/handling-filenames-with-spaces-in-bash.html

Related

Using sed in order to change a specific character in a specific line

I'm a beginner in bash and here is my problem. I have a file just like this one:
Azzzezzzezzzezzz...
Bzzzezzzezzzezzz...
Czzzezzzezzzezzz...
I try in a script to edit this file.ABC letters are unique in all this file and there is only one per line.
I want to replace the first e of each line by a number who can be :
1 in line beginning with an A,
2 in line beginning with a B,
3 in line beginning with a C,
and I'd like to loop this in order to have this type of result
Azzz1zzz5zzz1zzz...
Bzzz2zzz4zzz5zzz...
Czzz3zzz6zzz3zzz...
All the numbers here are random int variables between 0 and 9. I really need to start by replacing 1,2,3 in first exec of my loop, then 5,4,6 then 1,5,3 and so on.
I tried this
sed "0,/e/s/e/$1/;0,/e/s/e/$2/;0,/e/s/e/$3/" /tmp/myfile
But the result was this (because I didn't specify the line)
Azzz1zzz2zzz3zzz...
Bzzzezzzezzzezzz...
Czzzezzzezzzezzz...
I noticed that doing sed -i "/A/ s/$/ezzz/" /tmp/myfile will add ezzz at the end of A line so I tried this
sed -i "/A/ 0,/e/s/e/$1/;/B/ 0,/e/s/e/$2/;/C/ 0,/e/s/e/$3/" /tmp/myfile
but it failed
sed: -e expression #1, char 5: unknown command: `0'
Here I'm lost.
I have in a variable (let's call it number_of_e_per_line) the number of e in either A, B or C line.
Thank you for the time you take for me.
Just apply s command on the line that matches A.
sed '
/^A/{ s/e/$1/; }
/^B/{ s/e/$2/; }
# or shorter
/^C/s/e/$3/
'
s command by default replaces the first occurrence. You can do for example s/s/$1/2 to replace the second occurrence, s/e/$1/g (like "Global") replaces all occurrences.
0,/e/ specifies a range of lines - it filters lines from the first up until a line that matches /e/.
sed is not part of Bash. It is a separate (crude) programming language and is a very standard command. See https://www.grymoire.com/Unix/Sed.html .
Continuing from the comment. sed is a poor choice here unless all your files can only have 3 lines. The reason is sed processes each line and has no way to keep a separate count for the occurrences of 'e'.
Instead, wrapping sed in a script and keeping track of the replacements allows you to handle any file no matter the number of lines. You just loop and handle the lines one at a time, e.g.
#!/bin/bash
[ -z "$1" ] && { ## valiate one argument for filename provided
printf "error: filename argument required.\nusage: %s filename\n" "./$1" >&2
exit 1
}
[ -s "$1" ] || { ## validate file exists and non-empty
printf "error: file not found or empty '%s'.\n" "$1"
exit 1
}
declare -i n=1 ## occurrence counter initialized 1
## loop reading each line
while read -r line || [ -n "$line" ]; do
[[ $line =~ ^.*e.*$ ]] || continue ## line has 'e' or get next
sed "s/e/1/$n" <<< "$line" ## substitute the 'n' occurence of 'e'
((n++)) ## increment counter
done < "$1"
Your data file having "..." at the end of each line suggests your files is larger than the snippet posted. If you have lines beginning 'A' - 'Z', you don't want to have to write 26 separate /match/s/find/replace/ substitutions. And if you have somewhere between 3 and 26 (or more), you don't want to have to rewrite a different sed expression for every new file you are faced with.
That's why I say sed is a poor choice. You really have no way to make the task a generic task with sed. The downside to using a script is it will become a poor choice as the number of records you need to process increase (over 100000 or so just due to efficiency)
Example Use/Output
With the script in replace-e-incremental.sh and your data in file, you would do:
$ bash replace-e-incremental.sh file
Azzz1zzzezzzezzz...
Bzzzezzz1zzzezzz...
Czzzezzzezzz1zzz...
To Modify file In-Place
Since you make multiple calls to sed here, you need to redirect the output of the file to a temporary file and then replace the original by overwriting it with the temp file, e.g.
$ bash replace-e-incremental.sh file > mytempfile && mv -f mytempfile file
$ cat file
Azzz1zzzezzzezzz...
Bzzzezzz1zzzezzz...
Czzzezzzezzz1zzz...

Convert multi-line csv to single line using Linux tools

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

Adding a comma after $variable

I'm writing a for loop in bash to run a command and I need to add a comma after one of my variables. I can't seem to do this without an extra space added. When I move "," right next to $bams then it outputs *.sorted,
#!/bin/bash
bams=*.sorted
for i in $bams
do echo $bams ","
done;
Output should be this:
'file1.sorted','file2.sorted','file3.sorted'
The eventual end goal is to be able to insert a list of files into a --flag in the format above. Not sure how to do that either.
First, a literal answer (if your goal were to generate a string of the form 'foo','bar','baz', rather than to run a program with a command line equivalent to somecommand --flag='foo','bar','baz', which is quite different):
shopt -s nullglob # generate a null result if no matches exist
printf -v var "'%s'," *.sorted # put list of files, each w/ a comma, in var
echo "${var%,}" # echo contents of var, with last comma removed
Or, if you don't need the literal single quotes (and if you're passing your result to another program on its command line with the single quotes being syntactic rather than literal, you absolutely don't want them):
files=( *.sorted ) # put *.sorted in an array
IFS=, # set the comma character as the field separator
somecommand --flag "${files[*]}" # run your program with the comma-separated list
try this -
lst=$( echo *.sorted | sed 's/ /,/g' ) # stack filenames with commas
echo $lst
if you really need the single-ticks around each filename, then
lst="'$( echo *.sorted | sed "s/ /','/g" )'" # commas AND quotes
#!/bin/bash
bams=*.sorted
for i in $bams
do flag+="${flag:+,}'$i'"
done
echo $flag

sed from pattern till end of file inside a for loop

I'm writing a bash script that would allow me to take a certain amount of text from a file and add some other text before that for a list of files.
directory=$(pwd)
for f in *test.txt
do
filename=$(basename $f .txt)
printf "%%sum=4 \n"> input.temp
printf "file=$directory"/"$filename".txt" \n">> input.temp
printf "some commands \n">> input.temp
printf "\n" >> input.temp
printf "description \n">> input.temp
sed -n "/0 1/,$p" "$f" >> input.temp;
mv input.temp $filename.temp
done
I have a problem with the sed command inside the for loop. I looked around and people suggest adding double quotes which I did but to no avail. I think it might be the $p.
I hope this is clear enough. If it's not, I'll try to explain better.
sed -n "/0 1/,$p" "$f" >> input.temp; does not work
sed -n '/0 1/,$p' "$f" >> input.temp; does not work
sed -n "/0 1/,\$p" "$f" >> input.temp; does not work
FYI I'm not trying to find something else that works. I want to fix this exact input. I sound like an asshole I'm sure.
Sample input
%sum=8
file=otherpath/filename.txt
some other commands
another description
0 1
0.36920852 -0.56246512
0.77541848 0.05756533
2.05409026 0.62333039
2.92655258 0.56906375
2.52034254 -0.05096652
1.24167014 -0.61673008
-0.60708600 -0.99443872
0.10927459 0.09899803
3.90284624 1.00103940
3.18648588 -0.09239788
0.93151968 -1.09013674
2.50047427 1.30468389
2.19361322 2.54108378
3.18742399 0.34152442
3.38679424 1.11276220
1.56936488 3.27250306
1.81754180 4.19564055
1 2 1.5 6
2 3 1.5
3 4
4 5 1.5
5 6 1.5
6 11 1.0
7
8
9
10
11
12
13 16
14
15
16 17
17
My desired output is basically this file from "0 1" till the end preceded by the stuff I put inside the printf.
UPDATE: If you're interested, the two scripts tripleee and Ed Morton provided work perfectly well. The problem in my script was me leaving out the -i option from the sed line (for inplace).
sed -n "/0 1/,$p" "$f" >> input.temp
should be replaced by
sed -ni '/0 1/,$p' "$f"
I see you updated your question and provided some additional information in your comments so try this, uses GNU awk 4.* for -i inplace:
awk -i inplace -v directory="$(pwd)" '
FNR==1 {
print "%%sum=4 "
print "file=" directory "/" FILENAME
print "some commands "
print ""
print "description "
found = 0
}
/0 1/ { found = 1 }
found
' *text.txt
If you don't have GNU awk then the technically correct way to do it is using xargs but it's simpler using a shell loop for the file manipulation (moving) part:
for file in *test.txt
do
awk -v directory="$(pwd)" '
FNR==1 {
print "%%sum=4 "
print "file=" directory "/" FILENAME
print "some commands "
print ""
print "description "
found = 0
}
/0 1/ { found = 1 }
found
' "$file" > tmp && mv tmp "$file"
done
Like others have already commented, you basically just need to use single quotes instead of double, because $p in double quotes gets replaced with the value of the shell variable p by the shell, before sed executes (in practice, probably an empty string).
However, you might also want to investigate doing it all in sed. You might then instead stick with the double quotes (because there are other variables you do want to substitute) and instead escape the dollar sign in $p with a backslash to protect it from the shell.
directory=$(pwd) # just do this once before the loop; the value doesn't change
for f in *text.txt; do
# no braces
filename=$(basename "$f" .txt)
sed -n "1i\\
%sum=4\\
file=$directory/$filename.txt\\
some commands\\
\\
description
/0 1/,\$p" "$f" >inputout.temp2 # no pointless separate temp file
done
In practice, I imagine you would like for the output file to be different in each iteration (maybe "$filename.temp" instead?) but what you do about that is up to you, obviously. As it is now, the file will contain the output from the last iteration.

Bash - extracting a string between two points

For example:
((
extract everything here, ignore the rest
))
I know how to ignore everything within, but I don't know how to do the opposite. Basically, it'll be a file and it needs to extract the data between the two points and then output it to another file. I've tried countless approaches, and all seem to tell me the indentation I'm stating doesn't exist in the file, when it does.
If somebody could point me in the right direction, I'd be grateful.
If your data are "line oriented", so the marker is alone (as in the example), you can try some of the following:
function getdata() {
cat - <<EOF
before
((
extract everything here, ignore the rest
someother text
))
after
EOF
}
echo "sed - with two seds"
getdata | sed -n '/((/,/))/p' | sed '1d;$d'
echo "Another sed solution"
getdata | sed -n '1,/((/d; /))/,$d;p'
echo "With GNU sed"
getdata | gsed -n '/((/{:a;n;/))/b;p;ba}'
echo "With perl"
getdata | perl -0777 -pe "s/.*\(\(\s*\\n(.*)?\)\).*/\$1/s"
Ps: yes, its looks like a dance of crazy toothpicks
Assuming you want to extract the string inside (( and )):
VAR="abc((def))ghi"
echo "$VAR"
VAR=${VAR##*((}
VAR=${VAR%%))*}
echo "$VAR"
## cuts away the longest string from the beginning; # cuts away the shortest string from the beginning; %% cuts away the longest string at the end; % cuts away the shortes string at the end
The file :
$ cat /tmp/l
((
extract everything here, ignore the rest
someother text
))
The script
$ awk '$1=="((" {p=1;next} $1=="))" {p=o;next} p' /tmp/l
extract everything here, ignore the rest
someother text
sed -n '/^((/,/^))/ { /^((/b; /^))/b; p }'
Brief explanation:
/^((/,/^))/: range addressing (inclusive)
{ /^((/b; /^))/b; p }: sequence of 3 commands
1. skip line with ^((
2. skip line with ^))
3. print
The line skipping is required to make the range selection exclusive.

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