Add x^2 to every "nonzero" coefficient with sed/awk - bash

Question

I have to write, as easy as possible, a script or command which has to use awk or/and sed.
Input file:
23 12 0 33
3 4 19
1st line n=3
2nd line n=2
In each line of file we have string of numbers. Each number is coefficient and we have to add x^n where n is the highest power (sum of spaces between numbers in each line (no space after last number in each line)) and if we have "0" in our string we have to skip it.
So for that input we will have output like:
23x^3+12x^2+33
3x^2+4x+19
Please help me to write a short script solving that problem. Thank you so much for your time and all the help :)
My idea:
linescount=$(cat numbers|wc -l)
linecounter=1
While[linecounter<=linescount];
do
i=0
for i in spaces=$(cat numbers|sed 1p | sed " " )
do
sed -i 's/ /x^spaces/g'
i=($(i=i-1))
done
linecounter=($(linecounter=linecounter-1))
done

Answer 1

Following awk may help you on same too.
awk '{for(i=1;i<=NF;i++){if($i!="" && $i){val=(val?val "+" $i:$i)(NF-i==0?"":(NF-i==1?"x":"x^"NF-i))} else {pointer++}};if(val){print val};val=""} pointer==NF{print;} {pointer=""}' Input_file
Adding a non-one liner form of solution too here.
awk '
{
for(i=1;i<=NF;i++){
if($i!="" && $i){
val=(val?val "+" $i:$i)(NF-i==0?"":(NF-i==1?"x":"x^"NF-i))}
else {
pointer++}};
if(val) {
print val};
val=""
}
pointer==NF {
print}
{
pointer=""
}
' Input_file
EDIT: Adding explanation too here for better understanding of OP and all people's learning here.
awk '
{
for(i=1;i<=NF;i++){ ##Starting a for loop from variable 1 to till the value of NF here.
if($i!="" && $i){ ##checking if variable i value is NOT NULL then do following.
val=(val?val "+" $i:$i)(NF-i==0?"":(NF-i==1?"x":"x^"NF-i))} ##creating variable val here and putting conditions here if val is NULL then
##simply take value of that field else concatenate the value of val with its
##last value. Second condition is to check if last field of line is there then
##keep it like that else it is second last then print "x" along with it else keep
##that "x^" field_number-1 with it.
else { ##If a field is NULL in current line then come here.
pointer++}}; ##Increment the value of variable named pointer here with 1 each time it comes here.
if(val) { ##checking if variable named val is NOT NULL here then do following.
print val}; ##Print the value of variable val here.
val="" ##Nullifying the variable val here.
}
pointer==NF { ##checking condition if pointer value is same as NF then do following.
print} ##Print the current line then, seems whole line is having zeros in it.
{
pointer="" ##Nullifying the value of pointer here.
}
' Input_file ##Mentioning Input_file name here.

Answer 2

Offering a Perl solution since it has some higher level constucts than bash that make the code a little simpler:
use strict;
use warnings;
use feature qw(say);
my #terms;
while (my $line = readline(*DATA)) {
chomp($line);
my $degree = () = $line =~ / /g;
my #coefficients = split / /, $line;
my #terms;
while ($degree >= 0) {
my $coefficient = shift #coefficients;
next if $coefficient == 0;
push #terms, $degree > 1
? "${coefficient}x^$degree"
: $degree > 0
? "${coefficient}x"
: $coefficient;
}
continue {
$degree--;
}
say join '+', #terms;
}
__DATA__
23 12 0 33
3 4 19
Example output:
hunter#eros  ~  perl test.pl
23x^3+12x^2+33
3x^2+4x+19
Any information you want on any of the builtin functions used above: readline, chomp, push, shift, split, say, and join can be found in perldoc with perldoc -f <function-name>

Answer 3

$ cat a.awk
function print_term(i) {
# Don't print zero terms:
if (!$i) return;
# Print a "+" unless this is the first term:
if (!first) { printf " + " }
# If it's the last term, just print the number:
if (i == NF) printf "%d", $i
# Leave the coefficient blank if it's 1:
coef = ($i == 1 ? "" : $i)
# If it's the penultimate term, just print an 'x' (not x^1):
if (i == NF-1) printf "%sx", coef
# Print a higher-order term:
if (i < NF-1) printf "%sx^%s", coef, NF - i
first = 0
}
{
first = 1
# print all the terms:
for (i=1; i<=NF; ++i) print_term(i)
# If we never printed any terms, print a "0":
print first ? 0 : ""
}
Example input and output:
$ cat file
23 12 0 33
3 4 19
0 0 0
0 1 0 1
17
$ awk -f a.awk file
23x^3 + 12x^2 + 33
3x^2 + 4x + 19
0
x^2 + 1
17

Answer 4

$ cat ip.txt
23 12 0 33
3 4 19
5 3 0
34 01 02
$ # mapping each element except last to add x^n
$ # -a option will auto-split input on whitespaces, content in #F array
$ # $#F will give index of last element (indexing starts at 0)
$ # $i>0 condition check to prevent x^0 for last element
$ perl -lane '$i=$#F; print join "+", map {$i>0 ? $_."x^".$i-- : $_} #F' ip.txt
23x^3+12x^2+0x^1+33
3x^2+4x^1+19
5x^2+3x^1+0
34x^2+01x^1+02
$ # with post processing
$ perl -lape '$i=$#F; $_ = join "+", map {$i>0 ? $_."x^".$i-- : $_} #F;
s/\+0(x\^\d+)?\b|x\K\^1\b//g' ip.txt
23x^3+12x^2+33
3x^2+4x+19
5x^2+3x
34x^2+01x+02

Answer 5

One possibility is:
#!/usr/bin/env bash
line=1
linemax=$(grep -oEc '(( |^)[0-9]+)+' inputFile)
while [ $line -lt $linemax ]; do
degree=$(($(grep -oE ' +' - <<<$(grep -oE '(( |^)[0-9]+)+' inputFile | head -$line | tail -1) | cut -d : -f 1 | uniq -c)+1))
coeffs=($(grep -oE '(( |^)[0-9]+)+' inputFile | head -$line | tail -1))
i=0
while [ $i -lt $degree ]; do
if [ ${coeffs[$i]} -ne 0 ]; then
if [ $(($degree-$i-1)) -gt 1 ]; then
echo -n "${coeffs[$i]}x^$(($degree-$i-1))+"
elif [ $(($degree-$i-1)) -eq 1 ]; then
echo -n "${coeffs[$i]}x"
else
echo -n "${coeffs[$i]}"
fi
fi
((i++))
done
echo
((line++))
done
The most important lines are:
# Gets degree of the equation
degree=$(($(grep -oE ' +' - <<<$(grep -oE '(( |^)[0-9]+)+' inputFile | head -$line | tail -1) | cut -d : -f 1 | uniq -c)+1))
# Saves coefficients in an array
coeffs=($(grep -oE '(( |^)[0-9]+)+' inputFile | head -$line | tail -1))
Here, grep -oE '(( |^)[0-9]+)+' finds lines containing only numbers (see edit). grep -oE ' +' - ........... |cut -d : -f 1 |uniq counts the numbers of coefficients per line as explained in this question.
Edit: An improved regex for capturing lines with only numbers is
grep -E '(( |^)[0-9]+)+' inputfile | grep -v '[a-zA-Z]'

Answer 6

sed -r "s/(.*) (.*) (.*) (.*)/\1x^3+\2x^2+\3x+\4/; \
s/(.*) (.*) (.*)/\1x^2+\2x+\3/; \
s/\+0x(^.)?\+/+/g; \
s/^0x\^.[+]//g; \
s/\+0$//g;" koeffs.txt
Line 1: Handle 4 elements
Line 2: Handle 3
Line 3: Handle 0 in the middle
Line 5: Handle 0 at start
Line 5: Handle 0 at end

Answer 7

Here is a more bashy, less sedy answer which is better readable, than the sed one, I think:
#!/bin/bash
#
# 0 4 12 => 12x^3
# 2 4 12 => 12x
# 3 4 12 => 12
term () {
p=$1
leng=$2
fac=$3
pot=$((leng - 1 - p))
case $pot in
0) echo -n '+'${fac} ;;
1) echo -n '+'${fac}x ;;
*) echo -n '+'${fac}x^$pot ;;
esac
}
handleArray () {
# mapfile puts a counter into the array, starting with 0 for the 1st
# get rid of it!
shift
coeffs=($*)
# echo ${coeffs[#]}
cnt=0
len=${#coeffs[#]}
while (( cnt < len ))
do
if [[ ${coeffs[$cnt]} != 0 ]]
then
term $cnt $len ${coeffs[$cnt]}
fi
((cnt++))
done
echo # -e '\n' # extra line for dbg, together w. line 5 of the function.
}
mapfile -n 0 -c 1 -C handleArray < ./koeffs.txt coeffs | sed -r "s/^\++//;s/\++$//;"
The mapfile reads data and produces an array. See help mapfile for a brief syntax introduction.
We need some counting, to know, to which power to raise. Meanwhile we try to get rid of 0-terms.
In the end I use sed to remove leading and trailing plusses.

Answer 8

sh solution
while read line ; do
set -- $line
while test $1 ; do
i=$(($#-1))
case $1 in
0) ;;
*) case $i in
0) j="" ;;
1) j="x" ;;
*) j="x^$i" ;;
esac
result="$result$1$j+";;
esac
shift
done
echo "${result%+}"
result=""
done < infile

Answer 9

$ cat tst.awk
{
out = sep = ""
for (i=1; i<=NF; i++) {
if ($i != 0) {
pwr = NF - i
if ( pwr == 0 ) { sfx = "" }
else if ( pwr == 1 ) { sfx = "x" }
else { sfx = "x^" pwr }
out = out sep $i sfx
sep = "+"
}
}
print out
}
$ awk -f tst.awk file
23x^3+12x^2+33
3x^2+4x+19

Answer 10

First, my test set:
$ cat file
23 12 0 33
3 4 19
0 1 2
2 1 0
Then the awk script:
$ awk 'BEGIN{OFS="+"}{for(i=1;i<=NF;i++)$i=$i (NF-i?"x^" NF-i:"");gsub(/(^|\+)0(x\^[0-9]+)?/,"");sub(/^\+/,""}1' file
23x^3+12x^2+33
3x^2+4x^1+19
1x^1+2
2x^2+1x^1
And an explanation:
$ awk '
BEGIN {
OFS="+" # separate with a + (negative values
} # would be dealt with in gsub
{
for(i=1;i<=NF;i++) # process all components
$i=$i (NF-i?"x^" NF-i:"") # add x and exponent
gsub(/(^|\+)0(x\^[0-9]+)?/,"") # clean 0s and leftover +s
sub(/^\+/,"") # remore leading + if first component was 0
}1' file # output

Answer 11

This might work for you (GNU sed);)
sed -r ':a;/^\S+$/!bb;s/0x\^[^+]+\+//g;s/\^1\+/+/;s/\+0$//;b;:b;h;s/\S+$//;s/\S+\s+/a/g;s/^/cba/;:c;s/(.)(.)\2\2\2\2\2\2\2\2\2\2/\1\1\2/;tc;s/([a-z])\1\1\1\1\1\1\1\1\1/9/;s/([a-z])\1\1\1\1\1\1\1\1/8/;s/([a-z])\1\1\1\1\1\1\1/7/;s/([a-z])\1\1\1\1\1\1/6/;s/([a-z])\1\1\1\1\1/5/;s/([a-z])\1\1\1\1/4/;s/([a-z])\1\1\1/3/;s/([a-z])\1\1/2/;s/([a-z])\1/1/;s/[a-z]/0/g;s/^0+//;G;s/(.*)\n(\S+)\s+/\2x^\1+/;ba' file
This is not a serious solution!
Shows how sed can count, kudos goes to Greg Ubben back in 1989 when he wrote wc in sed!