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.
Related
I have a file 0.txt containing the following value fields contents in parentheses:
(bread,milk,),
(rice,brand B,),
(pan,eggs,Brandc,),
I'm looking in OS and elsewhere for how to prepend the letter x to the beginning of each value between commas so that my output file becomes (using bash unix):
(xbread,xmilk,),
(xrice,xbrand B,),
(xpan,xeggs,xBrand C,),
the only thing I've really tried but not enough is:
awk '{gsub(/,/,",x");print}' 0.txt
for all purposes the prefix should not be applied to the last commas at the end of each line.
With awk
awk 'BEGIN{FS=OFS=","}{$1="(x"substr($1,2);for(i=2;i<=NF-2;i++){$i="x"$i}}1'
Explanation:
# Before you start, set the input and output delimiter
BEGIN{
FS=OFS=","
}
# The first field is special, the x has to be inserted
# after the opening (
$1="(x"substr($1,2)
# Prepend 'x' from field 2 until the previous to last field
for(i=2;i<=NF-2;i++){
$i="x"$i
}
# 1 is always true. awk will print in that case
1
The trick is to anchor the regexp so that it matches the whole comma-terminated substring you want to work with, not just the comma (and avoids other “special” characters in the syntax).
awk '{ gsub(/[^,()]+,/, "x&") } 1' 0.txt
sed -r 's/([^,()]+,)/x\1/g' 0.txt
I have a huge file (this is just a sample) and I would like to select all lines with "Ph_gUFAC1083" and all after until reach one that doesn't have the code (in this example Ph_gUFAC1139)
>uce_353_Ph_gUFAC1083 |uce_353
TTTAGCCATAGAAATGCAGAAATAATTAGAAGTGCCATTGTGTACAGTGCCTTCTGGACT
GGGCTGAAGGTGAAGGAGAAAGTATCATACTATCCTTGTCAGCTGCAAGGGTAATTACTG
CTGGCTGAAATTACTCAACATTTGTTTATAAGCTCCCCAGAGCATGCTGTAAATAGATTG
TCTGTTATAGTCCAATCACATTAAAACGCTGCTCCTTGCAAACTGCTACCTCCTGTTTTC
TGTAAGCTAGACAGAGAAAGCCTGCTGCTCACTTACTGAGCACCAAGCACTGAAGAGCTA
TGTTTAATGTGATTGTTTTCATTAGCTCTTCTCTGTCTGATATTACATTTATAATTTGCT
GGGCTTGAAGACTGGCATGTTGCATTGCTTTCATTTACTGTAGTAAGAGTGAATAGCTCT
AT
>uce_101_Ph_gUFAC1083 |uce_101
TTGGGCTTTATTTCCACCTTAAAATCTTTACCTGGCCGTGATCTGTTGTTCCATTACTGG
AGGGCAAAAATGGGAGGAATTGTCTGGGCTAAATTGCAATTAGGCAGCCCTGAGAGAGGC
TGGCACCAGTTAACTTGGGATATTGGAGTGAAAAGGCCCGTAATCAGCCTTCGGTCATGT
AGAACAATGCATAAAATTAAATTGACATTAATGAATAATTGTGTAATGAAAATGGAAGAG
GAGAGTTAATTGCATGTTACAGTGAGTGTAATGCCTAGATAACCTTGCATTTAATGCTAT
TCTTAGCCCTGCTGCCAAGACTTCTACAGAGCCTCTCTCTGCAGGAAGTCATTAAAGCTG
TGAGTAGATAATGCAGGCTCAGTGAAACCTAAGTGGCAACAATATA
>uce_171_Ph_gUFAC1083 |uce_171
CATGGAAAACGAGGAAAAGCCATATCTTCCAGGCCATTAATATTACTACGGAGACGTCTT
CATATCGCCGTAATTACAGCAGATCTCAAAGTGGCACAACCAAGACCAGCACCAAAGCTA
AAATAACTCGCAGGAGCAGGCGAGCTGCTTTTGCAGCCCTCAGTCCCAGAAATGCTCGGT
AGCTTTTCTTAAAATAGACAGCCTGTAAATAAGGTCTGTGAACTCAATTGAAGGTGGCTG
TTTCTGAATTAGTCAGCCCTCACAAGGCTCTCGGCCTACATGCTAGTACATAAATTGTCC
ACTTTACCACCAGACAAGAAAGATTAGAGTAATAAACACGGGGCATTAGCTCAGCTAGAG
AAACACACCAGCCGTTACGCACACGCGGGATTGCCAAGAACTGTTAACCCCACTCTCCAG
AAACGCACACAAAAAAACAAGTTAAAGCCATGACATCATGGGAA
>uce_4300_Ph_gUFAC1139 |uce_4300
ATTAAAAATACAATCCTCATGTTTGCATTTTGCAGTCGTCAACAAGAAATTGAAGAGAAA
CTCATAGAGGAAGAAACTGCTCGAAGGGTGGAAGAACTTGTAGCTAAACGCGTGGAAGAA
GAGCTGGAGAAAAGAAAGGATGAGATTGAGCGAGAGGTTCTCCGCAGGGTGGAGGAGGCT
AAGCGCATCATGGAAAAACAGTTGCTCGAAGAACTCGAGCGACAGCGACAAGCTGAACTT
GCAGCACAAAAAGCCAGAGAGGTAACGCTCGGTCGTTTGGAAAGTAGAGACAGTCCATGG
CAAAACTTTCAGTGTCGGTTTGTGCCTCCTGTTCGGTTCAGAAAGAGATGGAATACAGCA
AATCTAATTCCCTTCTCATATAAACTTGCATTGCTGCGAAACTTAATTTCTAGCCTATTC
AGAGGAGCTCACTGATATTTAAACAGTTACTCTCCTAAAACCTGAACAAGGATACTTGAT
TCTTAATGGAACTGACCTACATATTTCAGAATTGTTTGAAACTTTTGCCATGGCTGCAGG
ATTATTCAGCAGTCCTTTCATTTT
>uce_1039_Ph_gUFAC1139 |uce_1039
ATTAGTGGAATACAAATATGCAAAAACCAAACAGTTTGGTGCTATAATGTGAAAAGAAAT
TTACACCAATCTTATTTTTAATTTGTATGGGAACATTTTTACCACAAATTCCATATTTTA
ATAATACTATCCCAACTCTATTTTTTAGACTCATTTTGTCACTGTTTTGTAACAGAAACA
CTGTAAATATTATAGATGTGGTAAACTATTATACTTGTTTTCTTATAAATGAAATGATCT
GTGCCAACACTGACAAAATGAATTAATGTGTTACTAAGGCAACAGTCACATTATATGCTT
TCTCTTTCACAGTATGCGGTAGAGCATATGGTTTACTCTTAATGGAACACTAGCTTCTCA
TTAACATACCAGTAGCAATGTCAGAACTTACAAACCAGCATAACAGAGAAATGGAAAAAC
TTATAAATTAGACCCTTTCAGTATTATTGAGTAGAAAATGACTGATGTTCCAAGGTACAA
TATTTAGCTAATACAGTGCCCTTTTCTGCATCTTTCTTCTCAAAGGAAAAAAAAATCCTC
AAAAAAAACCAGAGCAAGAAACCTAACTTTTTCTTGT
I already tried several alternatives without success, the closest I reached was
sed -n '/Ph_gUFAC1083/, />/p' file.txt
that gave me that:
>uce_2347_Ph_gUFAC1083 |uce_2347
GCTTTTCTATGCAGATTTTTTCTAATTCTCTCCCTCCCCTTGCTTCTGTCAGTGTGAAGC
CCACACTAAGCATTAACAGTATTAAAAAGAGTGTTATCTATTAGTTCAATTAGACATCAG
ACATTTACTTTCCAATGTATTTGAAGACTGATTTGATTTGGGTCCAATCATTTAAAAATA
AGAGAGCAGAACTGTGTACAGAGCTGTGTACAGATATCTGTAGCTCTGAAGTCTTAATTG
CAAATTCAGATAAGGATTAGAAGGGGCTGTATCTCTGTAGACCAAAGGTATTTGCTAATA
CCTGAGATATAAAAGTGGTTAAATTCAATATTTACTAATTTAGGATTTCCACTTTGGATT
TTGATTAAGCTTTTTGGTTGAAAACCCCACATTATTAAGCTGTGATGAGGGAAAAAGCAA
CTCTTTCATAAGCCTCACTTTAACGCTTTATTTCAAATAATTTATTTTGGACCTTCTAAA
G
>uce_353_Ph_gUFAC1083 |uce_353
>uce_101_Ph_gUFAC1083 |uce_101
TTGGGCTTTATTTCCACCTTAAAATCTTTACCTGGCCGTGATCTGTTGTTCCATTACTGG
AGGGCAAAAATGGGAGGAATTGTCTGGGCTAAATTGCAATTAGGCAGCCCTGAGAGAGGC
TGGCACCAGTTAACTTGGGATATTGGAGTGAAAAGGCCCGTAATCAGCCTTCGGTCATGT
AGAACAATGCATAAAATTAAATTGACATTAATGAATAATTGTGTAATGAAAATGGAAGAG
GAGAGTTAATTGCATGTTACAGTGAGTGTAATGCCTAGATAACCTTGCATTTAATGCTAT
TCTTAGCCCTGCTGCCAAGACTTCTACAGAGCCTCTCTCTGCAGGAAGTCATTAAAGCTG
TGAGTAGATAATGCAGGCTCAGTGAAACCTAAGTGGCAACAATATA
>uce_171_Ph_gUFAC1083 |uce_171
Do you know how to do it using grep, sed or awk?
Thx
$ awk '/^>/{if(match($0,"Ph_gUFAC1083")){s=1} else s=0}s' file
I made a simple criteria for your request,
If the the start of the line is >, we're going to judge if "Ph_gUFAC1083" existed, if yes, set s=1, set s=0 otherwise.
For the line that doesn't start with >, the value of s would be retained.
The final s in the awk command decide if the line to be printed (s=1) or not (s=0).
If what you want is every line with Ph_gUFAC1139 plus block of lines after that line until the next line starting with >, then the following awk snippet might do:
$ awk 'BEGIN {RS=ORS=">"} /Ph_gUFAC1139/' file.txt
This uses the > character as a record separator, then simply displays records that contain the text you're interested in.
If you wanted to be able to provide the search string using a variable, you'd do it something like this:
$ val="Ph_gUFAC1139"
$ awk -v s="$val" 'BEGIN {RS=ORS=">"} $0 ~ s' file.txt
UPDATE
A comment mentions that the solution above shows trailing record separators rather than leading ones. You can adapt your output to match your input by reversing this order manually:
awk 'BEGIN { RS=ORS=">" } /Ph_gUFAC1139/ { printf "%s%s",ORS,$0 }' file.txt
Note that in the initial examples, a "match" of the regex would invoke awk's default "action", which is to print the line. The default action is invoked if no action is specified within the script. The code (immediately) above includes an action .. which prints the record, preceded by the separator.
This might work for you (GNU sed):
sed '/^>/h;G;/Ph_gUFAC1083/P;d' file
Store each line beginning with > in the hold space (HS) and then append the HS to every line. If any line contains the string Ph_gUFAC1083 print the first line in the pattern space (PS) and discard the everything else.
N.B. the regexp for the match may be amended to /\n.*Ph_gUFAC1083/ if the string match may occur in any line.
This program is used to find the block which starts with Ph_gUFAC1083 and ends with any statement other than Ph_gUFAC1139
cat inp.txt |
awk '
BEGIN{begin=0}
{
# Ignore blank lines
if( $0 ~ /^$/ )
{
print $0
next
}
# mark the line that contains Ph_gUFAC1083 and print it
if( $0 ~ /Ph_gUFAC1083/ )
{
begin=1
print $0
}
else
{
# if the line contains Ph_gUFAC1083 and Ph_gUFAC1139 was found before it, print it
if( begin == 1 && ( $0 ~ /Ph_gUFAC1139/ ) )
{
print $0
}
else
{
# found a line which doesnt contain Ph_gUFAC1139 , mark the end of the block.
begin = 0
}
}
}'
How to remove part of the middle of a line/string by matching two known patterns, one in front of text to be removed and one behind the text to be removed?
I have a Linux text file with thousands of one line, comma delimited records. unfortunately, all records are not the same format. Each line may have as many as four comma delimited fields of which only the first and last are constant, the two middle fields may, or may not, be present.
Examples of existing line (record) formats. Messy data but the first field is always present, as is the last field, starts with word ADDED.
FNAME LNAME, SOME COMMENT, JOINED DATE, ADDED TO DB DATE
FNAME LNAME, ADDED TO DB DATE
FNAME LNAME, SOME COMMENT, ADDED TO DB DATE
FNAME LNAME, JOINED DATE, ADDED TO DB DATE
Objective is to keep field one including the comma, throw away everything following the first comma, keeping the word "ADDED" and everything that follows to the end of line and insert a space between the first comma and the word ADDED.
For each line in parse the file from start of line to the first comma (keep this).
Parse rest of line up to the space before the word “Added” and throw it away.
Keep everything from the space before the word “ADDED” to end of line and concatenate the first part and last part to form one record per line with two fields separated by a comma and a space.
(if record is already in desired format, change nothing)
Final file to look like:
FNAME LNAME, ADDED TO DB DATE
or
Fred Flintstone, ADDED on January 1st 2015 By Barney Rubble
Thanks!
If you don't care about blank lines:
awk '{print $1,$NF}' FS=, OFS=, input
(Blank lines will be output as a single comma)
If you want to just skip blank lines, use:
awk 'NF>1{print $1,$NF}' FS=, OFS=, input
If you want to keep them:
awk '{printf( "%s%s\n", $1, NF>1 ? ","$NF : "")}' FS=, OFS=, input
Note that this will not ensure a single space after the comma, but will retain the spacing as in the final column of the original file. (that is, if there are 3 spaces after the final column in the original, you'll get 3 in the output). It's not clear to me from the description, but that seems like desirable behavior.
A Perl solution
perl -ne 'print join ", ", (split /,\s*/)[0,-1]' myfile
or
perl -pe 's/,.*(?=,)//' myfile
Both of those solutions work fine for me with the data you have given, but you may like to try
perl -pe 's/,.*(?=,\s*ADDED)//' myfile
You can use backreference:
sed 's/\(^[^,]*,\).* ADDED/\1 ADDED/' file
one more approach with awk could help here.
awk -F, '{val=$1;sub(/FNAME.*\,/,",");print val $0}' Input_file
Where I am making field separator as (,) then saving first field to variable named val, now substituting FNAME till comma with (,) in current line, now printing the value of variable val and new edited current line.
Using perl
#!/usr/bin/perl
use strict;
use warnings;
open my $fh, "<", "file.txt" or die "$!: couldn't open file\n";
while(<$fh>) {
my #arr = split(/,/);
my $text = $arr[0] . ", " . $arr[$#arr];
print "$text\n";
}
XYZNA0000778800Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
XYZNA0000778900Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
I have above file format from which I want to find a matching record. For example, match a number(7789) on line starting with XYZ and once matched look for a matching number (7345) in lines below starting with 1 until it reaches to line starting with 9. retrieve the entire line record. How can I accomplish this using shell script, awk, sed or any combination.
Expected Output:
XYZNA0000778900Z
17345000012300324000000004000000000000000
With sed one can do:
$ sed -n '/^XYZ.*7789/,/^9$/{/^1.*7345/p}' file
17345000012300324000000004000000000000000
Breakdown:
sed -n ' ' # -n disabled automatic printing
/^XYZ.*7789/, # Match line starting with XYZ, and
# containing 7789
/^1.*7345/p # Print line starting with 1 and
# containing 7345, which is coming
# after the previous match
/^9$/ { } # Match line that is 9
range { stuff } will execute stuff when it's inside range, in this case the range is starting at /^XYZ.*7789/ and ending with /^9$/.
.* will match anything but newlines zero or more times.
If you want to print the whole block matching the conditions, one can use:
$ sed -n '/^XYZ.*7789/{:s;N;/\n9$/!bs;/\n1.*7345/p}' file
XYZNA0000778900Z
16123000012300321000000008000000000000000
16124000012300322000000007000000000000000
17234000012300323000000005000000000000000
17345000012300324000000004000000000000000
17456000012300325000000003000000000000000
9
This works by reading lines between ^XYZ.*7779 and ^9$ into the pattern
space. And then printing the whole thing if ^1.*7345 can be matches:
sed -n ' ' # -n disables printing
/^XYZ.*7789/{ } # Match line starting
# with XYZ that also contains 7789
:s; # Define label s
N; # Append next line to pattern space
/\n9$/!bs; # Goto s unless \n9$ matches
/\n1.*7345/p # Print whole pattern space
# if \n1.*7345 matches
I'd use awk:
awk -v rid=7789 -v fid=7345 -v RS='\n9\n' -F '\n' 'index($1, rid) { for(i = 2; i < $NF; ++i) { if(index($i, fid)) { print $i; next } } }' filename
This works as follows:
-v RS='\n9\n' is the meat of the whole thing. Awk separates its input into records (by default lines). This sets the record separator to \n9\n, which means that records are separated by lines with a single 9 on them. These records are further separated into fields, and
-F '\n' tells awk that fields in a record are separated by newlines, so that each line in a record becomes a field.
-v rid=7789 -v fid=7345 sets two awk variables rid and fid (meant by me as record identifier and field identifier, respectively. The names are arbitrary.) to your search strings. You could encode these in the awk script directly, but this way makes it easier and safer to replace the values with those of a shell variables (which I expect you'll want to do).
Then the code:
index($1, rid) { # In records whose first field contains rid
for(i = 2; i < $NF; ++i) { # Walk through the fields from the second
if(index($i, fid)) { # When you find one that contains fid
print $i # Print it,
next # and continue with the next record.
} # Remove the "next" line if you want all matching
} # fields.
}
Note that multi-character record separators are not strictly required by POSIX awk, and I'm not certain if BSD awk accepts it. Both GNU awk and mawk do, though.
EDIT: Misread question the first time around.
an extendable awk script can be
$ awk '/^9$/{s=0} s&&/7345/; /^XYZ/&&/7789/{s=1} ' file
set flag s when line starts with XYZ and contains 7789; reset when line is just 9, and print when flag is set and contains pattern 7345.
This might work for you (GNU sed):
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^XYZ[^\n]*7789/!b;/7345/p' file
Use the option -n for the grep-like nature of sed. Gather up records beginning with XYZ and ending in 9. Reject any records which do not have 7789 in the header. Print any remaining records that contain 7345.
If the 7345 will always follow the header,this could be shortened to:
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^XYZ[^\n]*7789.*7345/p' file
If all records are well-formed (begin XYZ and end in 9) then use:
sed -n '/^XYZ/h;//!H;/^9/!b;x;/^[^\n]*7789.*7345/p' file
The input is following
Title: Aoo Boo
Author: First Last
I am trying to output
Aoo Boo, First Last, "
by using awk like this
awk 'BEGIN { FS="[:[:space:]]+" }
/Title/ { sub(/^Title: /,""); t = $0; } # save title
/Author/{ sub(/^Author: /,""); printf "%s,%s,\"\n", t, $0}
' t.txt
But the output is like ,"irst Last. Basically it prints everything from the beginning of the sentence.
But if I change $0 to $2, the output is as expected which is Boo,Last,"
Why is it incorrect? What is the right way to do?
You need to get rid of the Windows line endings in your text file if you want to use Unix utilities.
If you're lucky, you'll find you have the dos2unix program installed, and you'll only need to do this:
dos2unix t.txt
If not, you could do it with tr:
tr -d '\r' < t.txt > new_t.txt
For reference, what is going on is that Windows files have \r\n at the end of every line (actually, a CR control code followed by a NL control code). On Linux, the lines ends with the \n, so the \r is part of the data; when you print it out, the terminal interprets as a "carriage return", which moves the cursor to the beginning of the current line, rather than advancing to the next line. Since the value of t ends with a \r, the following text overwrites the value of t.
It works with $2 because you've reassigned FS to include [:space:]; that definition of field separators is more generous than the awk default, since it includes \r and \f, neither of which are default field separators. Consequently, $2 does not contain the \r, but $0 does.
This assumes there are no colons in titles or names...
awk -F': *' '
$1=="Title" {
sub(/[^[:print:]]/,"");
t=$2;
}
$1=="Author" {
sub(/[^[:print:]]/,"");
printf("%s, %s\n", t, $2);
}
' inputfile.txt
This works by finding the title and storing it in a variable, then finding the author and using that as a trigger to print everything according to your format. You can alter the format as you see fit.
It may break if there are extra colons on the line, as the colon is being used to split fields. It may also break if your input doesn't match your example.
Perhaps the most important thing in this example is the sub(...) functions, which strip off non-printable characters like the carriage return that rici noticed you have. The regular expression [^[:print:]] matches "printable" characters, which the carriage return is not. This script will substitute them into oblivion if they're there, but should do no harm if they are not.