A question to extend previous one here. (I prefer asking new question rather editing first one. I may be wrong)
EDIT : ok, I was wrong, I should edit my first question. My bad (SO question is an art, difficult to master)
I have csv file, with semi-column as field delimiter. Here is an extract of csv file :
...;field;(:);10000(n,d);(:);field;....
...;field;123.12(b);123(a);123.00(:);....
Here is the desired output :
...;field;(:);(n,d) 10000;(:);field;....
...;field;(b) 123.12;(a) 123;(:) 123.00;....
I search a solution to swap 2 patterns in each field.
pattern 1 : any digit, with optional decimal mark (.) and optional decimal digit
e.g : 1 / 1111.00 / 444444444.3 / 32 / 32.6666666 / 1.0 / ....
pattern 2 : any string that begin with left parenthesis, follow by one or more character, ending with right parenthesis
e.g : (n,a,p) / (:) / (llll) / (d) / (123) / (1;2;3) ...
Solutions provided in first question are right for simple file that contain only one column. If I try the solution within csv file, I face multiple failures.
So I try awk similar solution, which is (I think) more "column-oriented".
I have try
awk -F";" '{print gensub(/([[:digit:].]*)(\(.*\))/, "\\2 \\1", "g")}' file
I though by fixing field delimiter (;), "my regex swap" will succes in every field. It was a mistake.
Here is an exemple of failure
;(:);7320000(n,d);(:)
desired output --> ;(:);(n,d) 7320000;(:)
My questions (finally) : why awk fail when it success with one-column file. what is the best tool to face this challenge ?
sed with very long regex ?
awk with very long regex ?
for loop ?
other tools ?
PS : I know I am not clear. I have 2 problems (English language, technical limitations). Sorry.
Your "question" is far too long, cluttered, and containing too many separate questions to wade through but here's how to get the output you want from the input you provided with any sed:
$ sed 's/\([0-9][0-9.]*\)\(([^)]*)\)/\2 \1/g' file
...;field;(:);(n,d) 10000;(:);field;....
...;field;(b) 123.12;(a) 123;(:) 123.00;....
Well, when parsing simple delimetered files without any quoted values, usually awk comes to the rescue:
awk -vFS=';' -vOFS=';' '{
for (i = 1; i < NF; i++) {
split($i, t, "(")
if (length(t[1]) != 0 && length(t[2]) != 0) {
$i="("t[2]" "t[1]
}
}
print
}' <<EOF
...;field;(:);10000(n,d);(:);field;....
...;field;123.12(b);123(a);123.00(:);....
EOF
However this will fail if fields are quoted, ie. the separator ; comes inside the values...
First we set input and output seapartor as ;
We iterate through all the fields in the line for (i = 1; i < NF; i++)
We split the line over ( character
If the first field splitted over ( is nonzero length and the second field has also nonzero length
We swap the firelds for this fields and add a space (we also remember about the removed ( on the beginning).
And then the line get's printed.
A solution using sed and xargs, but you need to know the number of fields in advance:
{
sed 's/;/\n/g' |
sed 's/\([^(]\{1,\}\)\((.*)\)/\2 \1/' |
xargs -d '\n' -n7 -- printf "%s;%s;%s;%s;%s;%s;%s\n"
} <<EOF
...;field;(:);10000(n,d);(:);field;....
...;field;123.12(b);123(a);123.00(:);....
EOF
For each ; i do a newline
For each line i substitute the string with at least on character before ( and a string inside ).
I then merge 7 lines using ; as separator with xargs and printf.
This might work for you (GNU sed):
sed -r 's/([0-9]+(\.[0-9]+)?)(\([^)]*\))/\3 \1/g' file
Look for group of numbers (possibly with a decimal point) followed by a pair of parens and rearrange them in the desired fashion, globally through out each line.
Related
I have this sed command which add's 3 zero's to an id (this occurs only if the id is 13 characters long):
sed 's/^\(.\{14\}\)\([0-9]\{13\}[^0-9]\)/\1000\2/' file
My input looks like this:
A:AAAA:AA: :A:**0123456789ABC **:AAA:AAA : :AA: : :
And my output is this one:
A:AAAA:AA: :A:**0000123456789ABC **:AAA:AAA : :AA: : :
I want to get rid off the 3 whitespaces after the id number. I can't delete the entire column because I have different data on other records so I want to delete the spaces just in the records/lines I expanded previously. So maybe I just need to add something to the existing command.
As you can see there are other whitespaces on the record, but I just want to delete the ones next to de ID(bold one).
I only found ways to delete entire columns, but I haven't been able to find a way to delete specific characters.
Just add three spaces after the closing \):
sed 's/^\(.\{14\}\)\([0-9]\{13\}[^0-9]\) /\1000\2/'
To make it work for your example, you also need to extend [0-9] to [0-9A-C].
You can use
sed 's/^\(.\{14\}\)\([[:alnum:]]\{13\}\)[[:space:]]*:/\1000\2:/' file
See the online demo:
#!/bin/bash
s='A:AAAA:AA: :A:0123456789ABC :AAA:AAA : :AA: : :'
sed 's/^\(.\{14\}\)\([[:alnum:]]\{13\}\)[[:space:]]*:/\1000\2:/' <<< "$s"
Output:
A:AAAA:AA: :A:0000123456789ABC:AAA:AAA : :AA: : :
Notes:
[[:alnum:]]\{13\} - matches 13 alphanumeric chars, not just digits
[[:space:]]*: matches zero or more whitespaces and a : (hence, the : must be added into the replacement pattern).
Since you are working with delimited fields, when one of the fields before the one you are working with inevitably changes in size just counting from the start a fixed length will break this.
Consider using awk instead and work solely with the 6th field. First strip out spaces, then check the length. If 13, add the leading 3 zeroes. Lastly print out the line.
$ awk -F ':' 'BEGIN { OFS=":"}{ gsub(" ", "", $6) };{if(length($6) == 13)$6="000"$6;print $0}' file.txt
A:AAAA:AA: :A:0000123456789ABC:AAA:AAA : :AA: : :
$
I have a file named list.txt containing a (supplier,product) pair and I must show the number of products from every supplier and their names using Linux terminal
Sample input:
stationery:paper
grocery:apples
grocery:pears
dairy:milk
stationery:pen
dairy:cheese
stationery:rubber
And the result should be something like:
stationery: 3
stationery: paper pen rubber
grocery: 2
grocery: apples pears
dairy: 2
dairy: milk cheese
Save the input to file, and remove the empty lines. Then use GNU datamash:
datamash -s -t ':' groupby 1 count 2 unique 2 < file
Output:
dairy:2:cheese,milk
grocery:2:apples,pears
stationery:3:paper,pen,rubber
The following pipeline shoud do the job
< your_input_file sort -t: -k1,1r | sort -t: -k1,1r | sed -E -n ':a;$p;N;s/([^:]*): *(.*)\n\1:/\1: \2 /;ta;P;D' | awk -F' ' '{ print $1, NF-1; print $0 }'
where
sort sorts the lines according to what's before the colon, in order to ease the successive processing
the cryptic sed joins the lines with common supplier
awk counts the items for supplier and prints everything appropriately.
Doing it with awk only, as suggested by KamilCuk in a comment, would be a much easier job; doing it with sed only would be (for me) a nightmare. Using both is maybe silly, but I enjoyed doing it.
If you need a detailed explanation, please comment, and I'll find time to provide one.
Here's the sed script written one command per line:
:a
$p
N
s/([^:]*): *(.*)\n\1:/\1: \2 /
ta
P
D
and here's how it works:
:a is just a label where we can jump back through a test or branch command;
$p is the print command applied only to the address $ (the last line); note that all other commands are applied to every line, since no address is specified;
N read one more line and appends it to the current pattern space, putting a \newline in between; this creates a multiline in the pattern space
s/([^:]*): *(.*)\n\1:/\1: \2 / captures what's before the first colon on the line, ([^:]*), as well as what follows it, (.*), getting rid of eccessive spaces, *;
ta tests if the previous s command was successful, and, if this is the case, transfers the control to the line labelled by a (i.e. go to step 1);
P prints the leading part of the multiline up to and including the embedded \newline;
D deletes the leading part of the multiline up to and including the embedded \newline.
This should be close to the only awk code I was referring to:
< os awk -F: '{ count[$1] += 1; items[$1] = items[$1] " " $2 } END { for (supp in items) print supp": " count[supp], "\n"supp":" items[supp]}'
The awk script is more readable if written on several lines:
awk -F: '{ # for each line
# we use the word before the : as the key of an associative array
count[$1] += 1 # increment the count for the given supplier
items[$1] = items[$1] " " $2 # concatenate the current item to the previous ones
}
END { # after processing the whole file
for (supp in items) # iterate on the suppliers and print the result
print supp": " count[supp], "\n"supp":" items[supp]
}
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
Below is my sample data in the csv .
20160711,"M","N1","F","S","A","good data with.....some special character and space
space ..
....","M","072","00126"
20160711,"M","N1","F","S","A","R","M","072","00126"
20160711,"M","N1","F","S","A","R","M","072","00126"
In above in a field I have good data along with junk data and line splited to new line .
I want to remove this special character (due to this special char and space,the line was moved to the next line) as well as merge this split line to a single line.
currently I am using something like below which is taking lots of time :
tr -cd '\11\12\15\40-\176' | gawk -v RS='"' 'NR % 2 == 0 { gsub(/\n/, "") } { printf("%s%s", $0, RT) }' MY_FILE.csv > MY_FILE.csv.tmp
attached a screenshot of original data in the file .
You could use
tr -c '[:print:]\r\n' ' ' <bad.csv >better.csv
to get rid of the non-printable chars…
sed '/[^"]$/ { N ; s/\n// }' better.csv | sed '/[^"]$/ { N ; s/\n// }' >even_better.csv
would cover most cases (i.e. would fail to trap an extra line break just after a random quote)
– Samson Scharfrichter
One problem that you will likely have with a traditional unix tool like awk is that while it supports field separators, it does not support quote+comma-style CSV formatting like the one in your screenshot or sample data. Awk can separate fields in a record using a field separator, but it has no concept of quote armour around your fields, so embedded commas are also considered field separators.
If you're comfortable with that because none of your plaintext data includes commas, and none of your "non-printable" data includes commas by accident, then you can just consider the quotes to be part of the field. They're printable characters, after all.
If you want to join your multi-line records into a single line and strip any non-printable characters, the following awk one-liner might do:
awk -F, 'NF<10{$0=last $0;last=$0} NF<10{next} {last="";sub(/[^[:print:]]/,"")} 1' inputfile
Note that this works except in cases where the line break is between the last comma and the content of the last field because from awk's perspective an empty field is valid and there's no need to join. If this logic doesn't match your data, you get another fun programming task as a result. :)
Let's break out the awk script and see what it does.
awk -F, ' # Set comma as the field separator...
NF<10 { # For any lines that have fewer than 10 fields...
$0=last $0 # Insert the last "saved" line here,
last=$0 # and save the newly joined line for the next round.
}
NF<10 { # If we still have fewer than 10 lines,
next # repeat.
}
{
sub(/[^[:print:]]/,"") # finally, substitute an empty string
} # for all non-printables,
1' inputfile # And print the current line.
I have some awk code that is running really slow. The format of my file is tab delimited 5 column ASCII. I am operating on column 5 to get a count of appropriate characters to alter the value in column 4.
Example input line:
10 5134832 N 28 Aaaaa*AAAAaAAAaAAAAaAAAA^]a^]a^Fa^]a
If I find any "^" in $5 I want to not count it, or the following character.
Then I want to find out how many characters are ">" or "<" or "*" and remove them from the count. I'm guessing using a gsub, and 3 splits is less than ideal, especially since column 5 can occasionally be a very very long string.
awk '{l=$4; if($5~/>/ || $5~/</ || $5~/*/ ) {gsub(/\^./,"");l-=split($5,a,"<")-1;l-=split($5,a,">")-1;l-=split($5,a,"*")-1}
If the code runs successfully on the line above, l will be 27.
I am omitting the surrounding parts of the command to try and focus on the part I have a question about.
So, what is the best step to make this run faster?
Well as I see, your gsub pattern will not work, as the / was not closed. Anyway, if I get it correctly and you want the character count of $5 without some characters, I'd go with:
count=length(gensub("[><A-Z^]","","g",$5))
You should list your skippable characters between [ and ], and do not start with ^!
Do you need to use awk, or will this work instead?
cut -f 5 < $file | grep -v '^[A-Z]' | tr -d '<>*\n' | wc -c
Translation:
Extract the 5th field from the tab-delimited $file.
Remove all fields starting with a capital letter.
Remove the characters <, >, *, and newlines.
Count the remaining characters.
Here's a guess:
awk '
BEGIN {FS = OFS = "\t"}
{
str = $5
gsub(/\^.|[><*]/, "", str)
l = length(str)
}
'
This might work for you:
echo "10 5134832 N 28 Aaaaa*AAAAaAAAaAAAAaAAAA^]a^]a^Fa^]a" |
awk '/[><*^]/{t=$5;gsub(/[><*]|[\^]./,"",t);$4=length(t)}1'
10 5134832 N 27 Aaaaa*AAAAaAAAaAAAAaAAAA^]a^]a^Fa^]a
if you want to show the amended fifth field:
awk '/[><*^]/{gsub(/[><*]|[\^]./,"",$5);$4=length($5)}1'