Conditional editing of line by one-liner - perl

My question is more of an optimization one, rather then a "howto".
I have a lef file, with thousands of lines in the form of:
RECT 429.336 273.821 426.246 274.721 ;
I wanted to move left by 4 um all rects above a certain point using a one-liner:
perl -lane '$F[2] > 1200 ? print $F[0]," ", ($F[1] - 4)," ", $F[2]," ", ($F[3] -4)," ", $F[4], " ;" : print $_' trial.lef
Thing is, this is UGLY.
Is there a nicer way of editing the file?
I'm not picky and will be happy to have answers with other languages (awk, sed, etc.) as long as they are nicer than what I wrote.
Additional input:
LAYER M12 ;
RECT 0 411.214 1 412.214 ; <-- shouldn't change, because 411.214 < 1200
END
END kuku_pin
PIN gaga_pin
DIRECTION OUTPUT ;
USE SIGNAL ;
PORT
LAYER M11 ;
RECT 43.1045 1203.138 43.1805 1207.29 ; <-- should change to "RECT 39.1045 1203.138 39.1805 1207.29"
END

There really is not much room for improvement, but you can replace -n with -p to skip the extra print. Further, you can edit the array elements and use join for a bit prettier code:
perl -lape'if ($F[2] > 1200) { $F[1] -= 4; $F[3] -= 4; $_ = join " ", #F }'
-a autosplit mode, splits the line $_ on space and puts the values in the predefined #F array. This switch is used with -n or -p.
-p loops around the <> operator input, file or standard input
-= decreases the LHS by amount in RHS
join joins the line back together after math has been done
-l can be skipped in this case, since we never touch the line endings, but keeping it makes the code more flexible if we decide to edit the last field.
When the condition is not met, original line is printed unchanged. Otherwise, it is replaced with the joined values in #F.
If you decide to keep the leading whitespace before RECT you can surround your if-statement with
if (($pre) = /^(\s*RECT)/)
To store the beginning of the line, making the one-liner:
perl -lape'if (($pre) = /^(\s*RECT)/) { if ($F[2] > 1200) { $F[1] -= 4; $F[3] -= 4; $F[0] = $pre; $_ = join " ", #F }}'

Related

Append to line that is preceded AND followed by empty line

I need to append an asterisk to a line, but only if said line is preceded and followed by empty lines (FYI, said empty lines will NOT have any white space in them).
Suppose I have the following file:
foo
foo
foo
foo
foo
I want the output to look like this:
foo
foo
foo
foo*
foo
I tried modifying the following awk command (found here):
awk 'NR==1 {l=$0; next}
/^$/ {gsub(/test/,"xxx", l)}
{print l; l=$0}
END {print l}' file
to suit my uses, but got all tied up in knots.
Sed or Perl solutions are, of course, welcome also!
UPDATE:
It turned out that the question I asked was not quite correct. What I really needed was code that would append text to non-empty lines that do not start with whitespace AND are followed, two lines down, by non-empty lines that also do not start with whitespace.
For this revised problem, suppose I have the following file:
foo
third line foo
fifth line foo
this line starts with a space foo
this line starts with a space foo
ninth line foo
eleventh line foo
this line starts with a space foo
last line foo
I want the output to look like this:
foobar
third line foobar
fifth line foo
this line starts with a space foo
this line starts with a space foo
ninth line foobar
eleventh line foo
this line starts with a space foo
last line foo
For that, this sed one-liner does the trick:
sed '1N;N;/^[^[:space:]]/s/^\([^[:space:]].*\o\)\(\n\n[^[:space:]].*\)$/\1bar\2/;P;D' infile
Thanks to Benjamin W.'s clear and informative answer below, I was able to cobble this one-liner together!
A sed solution:
$ sed '1N;N;s/^\(\n.*\)\(\n\)$/\1*\2/;P;D' infile
foo
foo
foo
foo*
foo
N;P;D is the idiomatic way to look at two lines at the same time by appending the next one to the pattern space, then printing and deleting the first line.
1N;N;P;D extends that to always having three lines in the pattern space, which is what we want here.
The substitution matches if the first and last line are empty (^\n and \n$) and appends one * to the line between the empty lines.
Notice that this matches and appends a * also for the second line of three empty lines, which might not be what you want. To make sure this doesn't happen, the first capture group has to have at least one non-whitespace character:
sed '1N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
Question from comment
Can we not append the * if the line two above begins with abc?
Example input file:
foo
foo
abc
foo
foo
foo
foo
There are three foo between empty lines, but the first one should not get the * appended because the line two above starts with abc. This can be done as follows:
$ sed '1{N;N};N;/^abc/!s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
foo
foo
abc
foo
foo*
foo*
foo
This keeps four lines at a time in the pattern space and only makes the substitution if the pattern space does not start with abc:
1 { # On the first line
N # Append next line to pattern space
N # ... again, so there are three lines in pattern space
}
N # Append fourth line
/^abc/! # If the pattern space does not start with abc...
s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/ # Append '*' to 3rd line in pattern space
P # Print first line of pattern space
D # Delete first line of pattern space, start next cycle
Two remarks:
BSD sed requires an extra semicolon: 1{N;N;} instead of 1{N;N}.
If the first and third line of the file are empty, the second line does not get an asterisk appended because we only start checking once there are four lines in the pattern space. This could be solved by adding an extra substitution into the 1{} block:
1{N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/}
(remember the extra ; for BSD sed), but trying to cover all edge cases makes sed even less readable, especially in one-liners:
sed '1{N;N;s/^\(\n[^[:space:]].*\)\(\n\)$/\1*\2/};N;/^abc/!s/^\(.*\n\n[^[:space:]].*\)\(\n\)$/\1*\2/;P;D' infile
One way to think about these problems is as a state machine.
start: state = 0
0: /* looking for a blank line */
if (blank line) state = 1
1: /* leading blank line(s)
if (not blank line) {
nonblank = line
state = 2
}
2: /* saw non-blank line */
if (blank line) {
output noblank*
state = 0
} else {
state = 1
}
And we can translate this pretty directly to an awk program:
BEGIN {
state = 0; # start in state 0
}
state == 0 { # looking for a (leading) blank line
print;
if (length($0) == 0) { # found one
state = 1;
next;
}
}
state == 1 { # have a leading blank line
if (length($0) > 0) { # found a non-blank line
saved = $0; # save it
state = 2;
next;
} else {
print; # multiple leading blank lines (ok)
}
}
state == 2 { # saw the non-blank line
if (length($0) == 0) { # followed by a blank line
print saved "*"; # BINGO!
state = 1; # to the saw a blank-line state
} else { # nope, consecutive non-blank lines
print saved; # as-is
state = 0; # to the looking for a blank line state
}
print;
next;
}
END { # cleanup, might have something saved to show
if (state == 2) print saved;
}
This is not the shortest way, nor likely the fastest, but it's probably the most straightforward and easy to understand.
EDIT
Here is a comparison of Ed's way and mine (see the comments under his answer for context). I replicated the OP's input a million-fold and then timed the runnings:
# ls -l
total 22472
-rw-r--r--. 1 root root 111 Mar 13 18:16 ed.awk
-rw-r--r--. 1 root root 23000000 Mar 13 18:14 huge.in
-rw-r--r--. 1 root root 357 Mar 13 18:16 john.awk
# time awk -f john.awk < huge.in > /dev/null
2.934u 0.001s 0:02.95 99.3% 0+0k 112+0io 1pf+0w
# time awk -f ed.awk huge.in huge.in > /dev/null
14.217u 0.426s 0:14.65 99.8% 0+0k 272+0io 2pf+0w
His version took about 5 times as long, did twice as much I/O, and (not shown in this output) took 1400 times as much memory.
EDIT from Ed Morton:
For those of us unfamiliar with the output of whatever time command John used above, here's the 3rd-invocation results from the normal UNIX time program on cygwin/bash using GNU awk 4.1.3:
$ wc -l huge.in
1000000 huge.in
$ time awk -f john.awk huge.in > /dev/null
real 0m1.264s
user 0m1.232s
sys 0m0.030s
$ time awk -f ed.awk huge.in huge.in > /dev/null
real 0m1.638s
user 0m1.575s
sys 0m0.030s
so if you'd rather write 37 lines than 3 lines to save a third of a second on processing a million line file then John's answer is the right one for you.
EDIT#3
It's the standard "time" built-in from tcsh/csh. And even if you didn't recognize it, the output should be intuitively obvious. And yes, boys and girls, my solution can also be written as a short incomprehensible mess:
s == 0 { print; if (length($0) == 0) { s = 1; next; } }
s == 1 { if (length($0) > 0) { p = $0; s = 2; next; } else { print; } }
s == 2 { if (length($0) == 0) { print p "*"; s = 1; } else { print p; s = 0; } print; next; }
END { if (s == 2) print p; }
Here's a perl filter version, for the sake of illustration — hopefully it's clear to see how it works. It would be possible to write a version that has a lower input-output delay (2 lines instead of 3) but I don't think that's important.
my #lines;
while (<>) {
# Keep three lines in the buffer, print them as they fall out
push #lines, $_;
print shift #lines if #lines > 3;
# If a non-empty line occurs between two empty lines...
if (#lines == 3 && $lines[0] =~ /^$/ && $lines[2] =~ /^$/ && $lines[1] !~ /^$/) {
# place an asterisk at the end
$lines[1] =~ s/$/*/;
}
}
# Flush the buffer at EOF
print #lines;
A perl one-liner
perl -0777 -lne's/(?<=\n\n)(.*?)(\n\n)/$1\*$2/g; print' ol.txt
The -0777 "slurps" in the whole file, assigned to $_, on which the (global) substitution is run and which is then printed.
The lookbehind (?<=text) is needed for repeating patterns, [empty][line][empty][line][empty]. It is a "zero-width assertion" that only checks that the pattern is there without consuming it. That way the pattern stays available for next matches.
Such consecutive repeating patterns trip up the /(\n\n)(.*?)(\n\n)/$1$2\*$3/, posted initially, since the trailing \n\n are not considered for the start of the very next pattern, having been just matched.
Update: My solution also fails after two consecutive matches as described above and needs the same lookback: s/(?<=\n\n)(\w+)\n\n/\1\2*\n\n/mg;
The easiest way is to use multi-line match:
local $/; ## slurp mode
$file = <DATA>;
$file =~ s/\n\n(\w+)\n\n/\n\n\1*\n\n/mg;
printf $file;
__DATA__
foo
foo
foo
foo
foo
It's simplest and clearest to do this in 2 passes:
$ cat tst.awk
NR==FNR { nf[NR]=NF; nr=NR; next }
FNR>1 && FNR<nr && NF && !nf[FNR-1] && !nf[FNR+1] { $0 = $0 "*" }
{ print }
$ awk -f tst.awk file file
foo
foo
foo
foo*
foo
The above takes one pass to record the number of fields on each line (NF is zero for an empty line) and then the second pass just checks your requirements - the current line is not the first or last in the file, it is not empty and the lines before and after are empty.
alternative awk solution (single pass)
$ awk 'NR>2 && !pp && !NF {p=p"*"}
NR>1{print p}
{pp=length(p);p=$0}
END{print p}' foo
foo
foo
foo
foo*
foo
Explanation: defer printing to next line for decision making, so need to keep previous line in p and state of the second previous line in pp (length zero assumed to be empty). Do the bookkeeping assignments and at the end print the last line.

Skipping particular positions in a string using substitution operator in perl

Yesterday, I got stuck in a perl script. Let me simplify it, suppose there is a string (say ABCDEABCDEABCDEPABCDEABCDEPABCDEABCD), first I've to break it at every position where "E" comes, and secondly, break it specifically where the user wants to be at. But, the condition is, program should not cut at those sites where E is followed by P. For example there are 6 Es in this sequence, so one should get 7 fragments, but as 2 Es are followed by P one will get 5 only fragments in the output.
I need help regarding the second case. Suppose user doesn't wants to cut this sequence at, say 5th and 10th positions of E in the sequence, then what should be the corresponding script to let program skip these two sites only? My script for first case is:
my $otext = 'ABCDEABCDEABCDEPABCDEABCDEPABCDEABCD';
$otext=~ s/([E])/$1=/g; #Main cut rule.
$otext=~ s/=P/P/g;
#output = split( /\=/, $otext);
print "#output";
Please do help!
To split on "E" except where it's followed by "P", you should use Negative look-ahead assertions.
From perldoc perlre "Look-Around Assertions" section:
(?!pattern)
A zero-width negative look-ahead assertion.
For example /foo(?!bar)/ matches any occurrence of "foo" that isn't followed by "bar".
my $otext = 'ABCDEABCDEABCDEPABCDEABCDEPABCDEABCD';
# E E EP E EP E
my #output=split(/E(?!P)/, $otext);
use Data::Dumper; print Data::Dumper->Dump([\#output]);"
$VAR1 = [
'ABCD',
'ABCD',
'ABCDEPABCD',
'ABCDEPABCD',
'ABCD'
];
Now, in order to NOT cut at occurences #2 and #4, you can do 2 things:
Concoct a really fancy regex that automatically fails to match on given occurence. I will leave that to someone else to attempt in an answer for completeness sake.
Simply stitch together the correct fragments.
I'm too brain-dead to come up with a good idiomatic way of doing it, but the simple and dirty way is either:
my %no_cuts = map { ($_=>1) } (2,4); # Do not cut in positions 2,4
my #output_final;
for(my $i=0; $i < #output; $i++) {
if ($no_cuts{$i}) {
$output_final[-1] .= $output[$i];
} else {
push #output_final, $output[$i];
}
}
print Data::Dumper->Dump([\#output_final];
$VAR1 = [
'ABCD',
'ABCDABCDEPABCD',
'ABCDEPABCDABCD'
];
Or, simpler:
my %no_cuts = map { ($_=>1) } (2,4); # Do not cut in positions 2,4
for(my $i=0; $i < #output; $i++) {
$output[$i-1] .= $output[$i];
$output[$i]=undef; # Make the slot empty
}
my #output_final = grep {$_} #output; # Skip empty slots
print Data::Dumper->Dump([\#output_final];
$VAR1 = [
'ABCD',
'ABCDABCDEPABCD',
'ABCDEPABCDABCD'
];
Here's a dirty trick that exploits two facts:
normal text strings never contain null bytes (if you don't know what a null byte is, you should as a programmer: http://en.wikipedia.org/wiki/Null_character, and nb. it is not the same thing as the number 0 or the character 0).
perl strings can contain null bytes if you put them there, but be careful, as this may screw up some perl internal functions.
The "be careful" is just a point to be aware of. Anyway, the idea is to substitute a null byte at the point where you don't want breaks:
my $s = "ABCDEABCDEABCDEPABCDEABCDEPABCDEABCD";
my #nobreak = (4,9);
foreach (#nobreak) {
substr($s, $_, 1) = "\0";
}
"\0" is an escape sequence representing a null byte like "\t" is a tab. Again: it is not the character 0. I used 4 and 9 because there were E's in those positions. If you print the string now it looks like:
ABCDABCDABCDEPABCDEABCDEPABCDEABCD
Because null bytes don't display, but they are there, and we are going to swap them back out later. First the split:
my #a = split(/E(?!P)/, $s);
Then swap the zero bytes back:
$_ =~ s/\0/E/g foreach (#a);
If you print #a now, you get:
ABCDEABCDEABCDEPABCD
ABCDEPABCD
ABCD
Which is exactly what you want. Note that split removes the delimiter (in this case, the E); if you intended to keep those you can tack them back on again afterward. If the delimiter is from a more dynamic regex it is slightly more complicated, see here:
http://perlmeme.org/howtos/perlfunc/split_function.html
"Example 9. Keeping the delimiter"
If there is some possibility that the #nobreak positions are not E's, then you must also keep track of those when you swap them out to make sure you replace with the correct character again.

How can I write this sed/bash command in awk or perl (or python, or ...)?

I need to replace instances of Progress (n,m) and Progress label="some text title" (n,m) in a scripting language with new values (N,M) where
N= integer ((n/m) * normal)
M= integer ( normal )
The progress statement can be anywhere on the script line (and worse, though not with current scripts, split across lines).
The value normal is a specified number between 1 and 255, and n and m are floating point numbers
So far, my sed implementation is below. It only works on Progress (n,m) formats and not Progress label="Title" (n,m) formats, but its just plain nuts:
#!/bin/bash
normal=$1;
file=$2
for n in $(sed -rn '/Progress/s/Progress[ \t]+\(([0-9\. \t]+),([0-9\. \t]+)\).+/\1/p' "$file" )
do
m=$(sed -rn "/Progress/s/Progress[ \t]+\(${n},([0-9\. \t]+).+/\1/p" "$file")
N=$(echo "($normal * $n)/$m" | bc)
M=$normal
sed -ri "/Progress/s/Progress[ \t]+\($n,$m\)/Progress ($N,$M)/" "$file"
done
Simply put: This works, but, is there a better way?
My toolbox has sed and bash scripting in it, and not so much perl, awk and the like which I think this problem is more suited to.
Edit Sample input.
Progress label="qt-xx-95" (0, 50) thermal label "qt-xx-95" ramp(slew=.75,sp=95,closed) Progress (20, 50) Pause 5 Progress (25, 50) Pause 5 Progress (30, 50) Pause 5 Progress (35, 50) Pause 5 Progress (40, 50) Pause 5 Progress (45, 50) Pause 5 Progress (50, 50)
Progress label="qt-95-70" (0, 40) thermal label "qt-95-70" hold(sp=70) Progress (10, 40) Pause 5 Progress (15, 40) Pause 5 Progress (20, 40) Pause 5 Progress (25, 40) Pause 5
awk has good splitting capabilities, so it might be a good choice for this problem.
Here is a solution that works for the supplied input, let's call it update_m_n_n.awk. Run it like this in bash: awk -f update_m_n_n.awk -v normal=$NORMAL input_file.
#!/usr/bin/awk
BEGIN {
ORS = RS = "Progress"
FS = "[)(]"
if(normal == 0) normal = 10
}
NR == 1 { print }
length > 1 {
split($2, A, /, */)
N = int( normal * A[1] / A[2] )
M = int( normal )
sub($2, N ", " M)
print $0
}
Explanation
ORS = RS = "Progress": Split sections at Progress and include Progress in the output.
FS = "[)(]": Separate fields at parenthesis.
NR == 1 { print }: Insert ORS before the first section.
split($2, A, /, */): Assuming there is only on parenthesized item between occurrences of Progress, this splits m and n into the A array.
sub($2, N ", " M): Substitute the new values the into current record.
This is somewhat brittle but it seems to do the trick? It could be changed to a one-line with perl -pe but I think this is clearer:
use 5.16.0;
my $normal = $ARGV[0];
while(<STDIN>){
s/Progress +(label=\".+?\")? *( *([0-9. ]+) *, *([0-9. ]+) *)/sprintf("Progress $1 (%d,%d)", int(($2/$3)*$normal),int($normal))/eg;
print $_;
}
The basic idea is to optionally capture the label clause in $1, and to capture n and m into $2 and $3. We use perl's ability to replace the matched string with an evaluated piece of code by providing the "e" modifier. It's going to fail dramatically if the label clause has any escaped quotes or contains the string that matches something that looks like a Progress toekn, so its not ideal. I agree that you need an honest to goodness parser here, though you could modify this regex to correct some of the obvious deficiencies like the weak number matching for n and m.
My initial thought was to try sed with recursive substitutions (t command), however I suspected that would get stuck.
This perl code might work for statements that are not split across lines. For splits across lines, perhaps it makes sense to write a separate pre-processor to join disparate lines.
The code splits "Progress" statements into separate line-segments, applies any replacement rules then rejoins the segments into one line and prints. Non-matching lines are simply printed. The matching code uses back-references and becomes somewhat unreadable. I have assumed your "normal" parameter can take floating values as the spec didn't seem clear.
#!/usr/bin/perl -w
use strict;
die("Wrong arguments") if (#ARGV != 2);
my ($normal, $file) = #ARGV;
open(FILE, '<', $file) or die("Cannot open $file");
while (<FILE>) {
chomp();
my $line = $_;
# Match on lines containing "Progress"
if (/Progress/) {
$line =~ s/(Progress)/\n$1/go; # Insert newlines on which to split
my #segs = split(/\n/, $line); # Split line into segments containing possibly one "Progress" clause
# Apply text-modification rules
#segs = map {
if (/(Progress[\s\(]+)([0-9\.]+)([\s,]+)([0-9\.]+)(.*)/) {
my $newN = int($2/$4 * $normal);
my $newM = int($normal);
$1 . $newN . $3 . $newM . $5;
} elsif (/(Progress\s+label="[^"]+"[\s\(]+)([0-9\.]+)([\s,]+)([0-9\.]+)(.*)/) {
my $newN = int($2/$4 * $normal);
my $newM = int($normal);
$1 . $newN . $3 . $newM . $5;
} else {
$_; # Segment doesn't contain "Progress"
}
} #segs;
$line = join("", #segs); # Reconstruct the single line
}
print($line,"\n"); # Print all lines
}

How to quickly find and replace many items on a list without replacing previously replaced items in BASH?

I want to perform about many find and replace operations on some text. I have a UTF-8 CSV file containing what to find (in the first column) and what to replace it with (in the second column), arranged from longest to shortest.
E.g.:
orange,fruit2
carrot,vegetable1
apple,fruit3
pear,fruit4
ink,item1
table,item2
Original file:
"I like to eat apples and carrots"
Resulting output file:
"I like to eat fruit3s and vegetable1s."
However, I want to ensure that if one part of text has already been replaced, that it doesn't mess with text that was already replaced. In other words, I don't want it to appear like this (it matched "table" from within vegetable1):
"I like to eat fruit3s and vegeitem21s."
Currently, I am using this method which is quite slow, because I have to do the whole find and replace twice:
(1) Convert the CSV to three files, e.g.:
a.csv b.csv c.csv
orange 0001 fruit2
carrot 0002 vegetable1
apple 0003 fruit3
pear 0004 fruit4
ink 0005 item1
table 0006 item 2
(2) Then, replace all items from a.csv in file.txt with the matching column in b.csv, using ZZZ around the words to make sure there is no mistake later in matching the numbers:
a=1
b=`wc -l < ./a.csv`
while [ $a -le $b ]
do
for i in `sed -n "$a"p ./b.csv`; do
for j in `sed -n "$a"p ./a.csv`; do
sed -i "s/$i/ZZZ$j\ZZZ/g" ./file.txt
echo "Instances of '"$i"' replaced with '"ZZZ$j\ZZZ"' ("$a"/"$b")."
a=`expr $a + 1`
done
done
done
(3) Then running this same script again, but to replace ZZZ0001ZZZ with fruit2 from c.csv.
Running the first replacement takes about 2 hours, but as I must run this code twice to avoid editing the already replaced items, it takes twice as long. Is there a more efficient way to run a find and replace that does not perform replacements on text already replaced?
Here's a perl solution which is doing the replacement in "one phase".
#!/usr/bin/perl
use strict;
my %map = (
orange => "fruit2",
carrot => "vegetable1",
apple => "fruit3",
pear => "fruit4",
ink => "item1",
table => "item2",
);
my $repl_rx = '(' . join("|", map { quotemeta } keys %map) . ')';
my $str = "I like to eat apples and carrots";
$str =~ s{$repl_rx}{$map{$1}}g;
print $str, "\n";
Tcl has a command to do exactly this: string map
tclsh <<'END'
set map {
"orange" "fruit2"
"carrot" "vegetable1"
"apple" "fruit3"
"pear" "fruit4"
"ink" "item1"
"table" "item2"
}
set str "I like to eat apples and carrots"
puts [string map $map $str]
END
I like to eat fruit3s and vegetable1s
This is how to implement it in bash (requires bash v4 for the associative array)
declare -A map=(
[orange]=fruit2
[carrot]=vegetable1
[apple]=fruit3
[pear]=fruit4
[ink]=item1
[table]=item2
)
str="I like to eat apples and carrots"
echo "$str"
i=0
while (( i < ${#str} )); do
matched=false
for key in "${!map[#]}"; do
if [[ ${str:$i:${#key}} = $key ]]; then
str=${str:0:$i}${map[$key]}${str:$((i+${#key}))}
((i+=${#map[$key]}))
matched=true
break
fi
done
$matched || ((i++))
done
echo "$str"
I like to eat apples and carrots
I like to eat fruit3s and vegetable1s
This will not be speedy.
Clearly, you may get different results if you order the map differently. In fact, I believe the order of "${!map[#]}" is unspecified, so you might want to specify the order of the keys explicitly:
keys=(orange carrot apple pear ink table)
# ...
for key in "${keys[#]}"; do
One way to do it would be to do a two-phase replace:
phase 1:
s/orange/##1##/
s/carrot/##2##/
...
phase 2:
s/##1##/fruit2/
s/##2##/vegetable1/
...
The ##1## markers should be chosen so that they don't appear in the original text or the replacements of course.
Here's a proof-of-concept implementation in perl:
#!/usr/bin/perl -w
#
my $repls = $ARGV[0];
die ("first parameter must be the replacement list file") unless defined ($repls);
my $tmpFmt = "###%d###";
open(my $replsFile, "<", $repls) || die("$!: $repls");
shift;
my #replsList;
my $i = 0;
while (<$replsFile>) {
chomp;
my ($from, $to) = /\"([^\"]*)\",\"([^\"]*)\"/;
if (defined($from) && defined($to)) {
push(#replsList, [$from, sprintf($tmpFmt, ++$i), $to]);
}
}
while (<>) {
foreach my $r (#replsList) {
s/$r->[0]/$r->[1]/g;
}
foreach my $r (#replsList) {
s/$r->[1]/$r->[2]/g;
}
print;
}
I would guess that most of your slowness is coming from creating so many sed commands, which each need to individually process the entire file. Some minor adjustments to your current process would speed this up a lot by running 1 sed per file per step.
a=1
b=`wc -l < ./a.csv`
while [ $a -le $b ]
do
cmd=""
for i in `sed -n "$a"p ./a.csv`; do
for j in `sed -n "$a"p ./b.csv`; do
cmd="$cmd ; s/$i/ZZZ${j}ZZZ/g"
echo "Instances of '"$i"' replaced with '"ZZZ${j}ZZZ"' ("$a"/"$b")."
a=`expr $a + 1`
done
done
sed -i "$cmd" ./file.txt
done
Doing it twice is probably not your problem. If you managed to just do it once using your basic strategy, it would still take you an hour, right? You probably need to use a different technology or tool. Switching to Perl, as above, might make your code a lot faster (give it a try)
But continuing down the path of other posters, the next step might be pipelining. Write a little program that replaces two columns, then run that program twice, simultaneously. The first run swaps out strings in column1 with strings in column2, the next swaps out strings in column2 with strings in column3.
Your command line would be like this
cat input_file.txt | perl replace.pl replace_file.txt 1 2 | perl replace.pl replace_file.txt 2 3 > completely_replaced.txt
And replace.pl would be like this (similar to other solutions)
#!/usr/bin/perl -w
my $replace_file = $ARGV[0];
my $before_replace_colnum = $ARGV[1] - 1;
my $after_replace_colnum = $ARGV[2] - 1;
open(REPLACEFILE, $replace_file) || die("couldn't open $replace_file: $!");
my #replace_pairs;
# read in the list of things to replace
while(<REPLACEFILE>) {
chomp();
my #cols = split /\t/, $_;
my $to_replace = $cols[$before_replace_colnum];
my $replace_with = $cols[$after_replace_colnum];
push #replace_pairs, [$to_replace, $replace_with];
}
# read input from stdin, do swapping
while(<STDIN>) {
# loop over all replacement strings
foreach my $replace_pair (#replace_pairs) {
my($to_replace,$replace_with) = #{$replace_pair};
$_ =~ s/${to_replace}/${replace_with}/g;
}
print STDOUT $_;
}
A bash+sed approach:
count=0
bigfrom=""
bigto=""
while IFS=, read from to; do
read countmd5sum x < <(md5sum <<< $count)
count=$(( $count + 1 ))
bigfrom="$bigfrom;s/$from/$countmd5sum/g"
bigto="$bigto;s/$countmd5sum/$to/g"
done < replace-list.csv
sed "${bigfrom:1}$bigto" input_file.txt
I have chosen md5sum, to get some unique token. But some other mechanism can also be used to generate such token; like reading from /dev/urandom or shuf -n1 -i 10000000-20000000
A awk+sed approach:
awk -F, '{a[NR-1]="s/####"NR"####/"$2"/";print "s/"$1"/####"NR"####/"}; END{for (i=0;i<NR;i++)print a[i];}' replace-list.csv > /tmp/sed_script.sed
sed -f /tmp/sed_script.sed input.txt
A cat+sed+sed approach:
cat -n replace-list.csv | sed -rn 'H;g;s|(.*)\n *([0-9]+) *[^,]*,(.*)|\1\ns/####\2####/\3/|;x;s|.*\n *([0-9]+)[ \t]*([^,]+).*|s/\2/####\1####/|p;${g;s/^\n//;p}' > /tmp/sed_script.sed
sed -f /tmp/sed_script.sed input.txt
Mechanism:
Here, it first generates the sed script, using the csv as input file.
Then uses another sed instance to operate on input.txt
Notes:
The intermediate file generated - sed_script.sed can be re-used again, unless the input csv file changes.
####<number>#### is chosen as some pattern, which is not present in the input file. Change this pattern if required.
cat -n | is not UUOC :)
This might work for you (GNU sed):
sed -r 'h;s/./&\\n/g;H;x;s/([^,]*),.*,(.*)/s|\1|\2|g/;$s/$/;s|\\n||g/' csv_file | sed -rf - original_file
Convert the csv file into a sed script. The trick here is to replace the substitution string with one which will not be re-substituted. In this case each character in the substitution string is replaced by itself and a \n. Finally once all substitutions have taken place the \n's are removed leaving the finished string.
There are a lot of cool answers here already. I'm posting this because I'm taking a slightly different approach by making some large assumptions about the data to replace ( based on the sample data ):
Words to replace don't contain spaces
Words are replaced based on the longest, exactly matching prefix
Each word to replace is exactly represented in the csv
This a single pass, awk only answer with very little regex.
It reads the "repl.csv" file into an associative array ( see BEGIN{} ), then attempts to match on prefixes of each word when the length of the word is bound by key length limits, trying to avoid looking in the associative array whenever possible:
#!/bin/awk -f
BEGIN {
while( getline repline < "repl.csv" ) {
split( repline, replarr, "," )
replassocarr[ replarr[1] ] = replarr[2]
# set some bounds on the replace word sizes
if( minKeyLen == 0 || length( replarr[1] ) < minKeyLen )
minKeyLen = length( replarr[1] )
if( maxKeyLen == 0 || length( replarr[1] ) > maxKeyLen )
maxKeyLen = length( replarr[1] )
}
close( "repl.csv" )
}
{
i = 1
while( i <= NF ) { print_word( $i, i == NF ); i++ }
}
function print_word( w, end ) {
wl = length( w )
for( j = wl; j >= 0 && prefix_len_bound( wl, j ); j-- ) {
key = substr( w, 1, j )
wl = length( key )
if( wl >= minKeyLen && key in replassocarr ) {
printf( "%s%s%s", replassocarr[ key ],
substr( w, j+1 ), !end ? " " : "\n" )
return
}
}
printf( "%s%s", w, !end ? " " : "\n" )
}
function prefix_len_bound( len, jlen ) {
return len >= minKeyLen && (len <= maxKeyLen || jlen > maxKeylen)
}
Based on input like:
I like to eat apples and carrots
orange you glad to see me
Some people eat pears while others drink ink
It yields output like:
I like to eat fruit3s and vegetable1s
fruit2 you glad to see me
Some people eat fruit4s while others drink item1
Of course any "savings" of not looking the replassocarr go away when the words to be replaced goes to length=1 or if the average word length is much greater than the words to replace.

How do I repeat a character n times in a string?

I am learning Perl, so please bear with me for this noob question.
How do I repeat a character n times in a string?
I want to do something like below:
$numOfChar = 10;
s/^\s*(.*)/' ' x $numOfChar$1/;
By default, substitutions take a string as the part to substitute. To execute code in the substitution process you have to use the e flag.
$numOfChar = 10;
s/^(.*)/' ' x $numOfChar . $1/e;
This will add $numOfChar space to the start of your text. To do it for every line in the text either use the -p flag (for quick, one-line processing):
cat foo.txt | perl -p -e "$n = 10; s/^(.*)/' ' x $n . $1/e/" > bar.txt
or if it's a part of a larger script use the -g and -m flags (-g for global, i.e. repeated substitution and -m to make ^ match at the start of each line):
$n = 10;
$text =~ s/^(.*)/' ' x $n . $1/mge;
Your regular expression can be written as:
$numOfChar = 10;
s/^(.*)/(' ' x $numOfChar).$1/e;
but - you can do it with:
s/^/' ' x $numOfChar/e;
Or without using regexps at all:
$_ = ( ' ' x $numOfChar ) . $_;
You're right. Perl's x operator repeats a string a number of times.
print "test\n" x 10; # prints 10 lines of "test"
EDIT: To do this inside a regular expression, it would probably be best (a.k.a. most maintainer friendly) to just assign the value to another variable.
my $spaces = " " x 10;
s/^\s*(.*)/$spaces$1/;
There are ways to do it without an extra variable, but it's just my $0.02 that it'll be easier to maintain if you do it this way.
EDIT: I fixed my regex. Sorry I didn't read it right the first time.