shell regex: Extract prices - regex

Question

Given the following list of prices, I am trying to figure out how to normalize/extract only the digits.
INPUT DESIRED_OUTPUT
CA$1399.00 1399.00
$1399.11 1399.11
$1,399.22< 1399.22
Z$1 399.33 1399.33
$1399.44# 1399.44
C$ 1399.55 1399.55
1,399.66 1399.66
1399.77 1399.77
,1399.88 1399.88
25 1399.88 1399.88
399.99 399.99
88.88 99.99 99.99 (if >2 matches on one line, only the last one matters)
.1399.88 DO NOT MATCH (not a price; too many ".")
666.000 DO NOT MATCH (not a price: too many 0's)
I suppose it is a good idea to begin is with what they all have in common:
Prices always contain .NN, but never contain .NNN
Upon further inspection, other rules become apparent:
.NN must be preceded by one or more digits.
NNN.NN can be preceded by either ,, , or a simple digit, but nothing else.
Anything following .NN and preceding *N.NN marks the end of the match.
Finally, the regex needs to consider commas in things like 1,399.66 (1399.66) to determine whether it is a price, but then strip them. 1, 399.66, for instance does not equal 1399.66: it should be 399.66.
I am looking at sed, grep, and awk for a portable and efficient solution. How should I go about approaching this problem?
I found a similar question, but I have no idea how to try the following regex with sed:
^\d+(,\d{1,2})?$
EDIT: Yes, my input format is can be a little weird, because it is the result of the concatenation of scraped pages.

Answer 1

You can use the following shell script:
#/bin/sh
grep -v '\.\d\+\.' | # get rid of lines with multiple dots within the same number
grep -v '\.\d\d\d\+' | # get rid of lines with more than 2 digits after .
sed -e 's/\(.*\.[0-9][0-9]\).*$/\1/' | # remove anything after last .NN
sed -e 's/^.* \([0-9][0-9][0-9][0-9]\)\./\1./' | # "* NNNN." => "NNNN."
sed -e 's/^.* \([0-9][0-9]\)\./\1./' | # "* NN." => "NN."
sed -e 's/^.* \([0-9]\)\./\1./' | # "* N." => "N."
sed -e 's/^\(.*\)[ ,]\(\([0-9]\)\{3,\}\)\./\1\2./g' | # "*,NNN." or "* NNN." => "*NNN."
sed -e 's/^\(.*\)[ ,]\(\([0-9]\)\{6,\}\)\./\1\2./g' | # "*,NNNNNN." or "* NNNNNN." => "*NNNNNN."
sed -e 's/^\(.*\)[ ,]\(\([0-9]\)\{9,\}\)\./\1\2./g' | # "*,NNNNNNNNN." or "* NNNNNNNNN." => "*NNNNNNNNN."
grep -o '\d\+\.\d\d' # print only the price
In case of numbers that are separated by space or , in groups of 3 digits, this solution works up to 9 digits before the .. If you need to extract bigger prices, just add more lines, increasing the number in the regex by 3. ;-)
Put it in a file called extract_prices, make it executable (chmod +x extract_prices) and run it: ./extract_prices < my_list.txt
Tested on OS X using the following input:
CA$1399.00
$1399.11
$1,399.22<
Z$1 399.33
Z$12 777 666.34 # <-- additonal monster price
$1399.44#
C$ 1399.55
1,399.66
1399.77
,1399.88
25 1399.88
399.99
88.88 99.99
.1399.88
666.000
Which generates the following output:
1399.00
1399.11
1399.22
1399.33
12777666.34
1399.44
1399.55
1399.66
1399.77
1399.88
1399.88
399.99
99.99

Answer 2

A solution with awk that splits on all characters that are not numbers or decimal point and prints the last field that matches a price. The leading sed script handles the exception case #3 where we have a space instead of a comma marking the thousands spot.
sed -e 's/ / x /g; :a; s/\(\$[1-9][0-9]*\) /\1/; ta' | awk -F '[^0-9.]' -v p='[0-9]+\\.[0-9][0-9]' '$0 ~ p { gsub(/,/, ""); for (i=NF; i>0; i--) if ($i ~ "^" p "$") { print $i; next } }'
Notes:
1) The sed script uses a test to iterate; therefore, it can handle millions, billions, etc.
2) The sed script also handles the multiple space condition such that $1[ ][ ]1000.00 does not become $11000.00 in the end.
3) Commas are simply stripped/ignored... if there is an issue with comma separation of numbers, the issue can be resolved by getting rid of the gsub in the awk script and fixing the filter in the leading sed script
Here is a more complicated version that builds on the idea in note #3 to make commas and spaces part of the number only if the space or comma is at a thousands separator.
sed -e ':a; s/\(\$[1-9][0-9]*\) \([0-9][0-9][0-9][ .]\)/\1\2/; ta; :b; s/\([1-9][0-9]*\),\([0-9][0-9][0-9][,.]\)/\1\2/; tb;' | awk -F '[^0-9.]' -v p='[0-9]+\\.[0-9][0-9]' '$0 ~ p { for (i=NF; i>0; i--) if ($i ~ "^" p "$") { print $i; next } }'
If chance of success is high on each line, then getting rid of "p" would make for a more efficient script.
sed -e ':a; s/\(\$[1-9][0-9]*\) \([0-9][0-9][0-9][ .]\)/\1\2/; ta; :b; s/\([1-9][0-9]*\),\([0-9][0-9][0-9][,.]\)/\1\2/; tb;' | awk -F '[^0-9.]' '{ for (i=NF; i>0; i--) if ($i ~ /^[0-9]+\.[0-9][0-9]$/) { print $i; next } }'
Finally, for safety, we can check in the sed filter to make sure we have a valid space or comma delimited number before we do either substitution.
sed -e ':a; /\$[1-9][0-9]\?[0-9]\?\( [0-9][0-9][0-9]\)\+\.[0-9][0-9]/ s/\(\$[1-9][0-9]*\) \([0-9][0-9][0-9][ .]\)/\1\2/; ta; :b; /[1-9][0-9]\?[0-9]\?\(,[0-9][0-9][0-9]\)\+\.[0-9][0-9]/ s/\([1-9][0-9]*\),\([0-9][0-9][0-9][,.]\)/\1\2/; tb;' | awk -F '[^0-9.]' '{ for (i=NF; i>0; i--) if ($i ~ /^[0-9]+\.[0-9][0-9]$/) { print $i; next } }'

Answer 3

This might work for you (GNU sed):
sed -r '/\n/!s/([^0-9]*\b(([0-9])[ ,]([0-9]{3})|([0-9]+))(\.[0-9]{2})\b)+/\n\3\4\5\6\n/;/^[0-9]+\.[0-9]{2}\b/P;D' file
This works with the data provided but some of the specification is a bit sketchy.