How to parse an input sentence? - prolog

I am making a program which takes user input and handles that input based on the user's choice of options. Say I input a string like "hello 4 is a number, and 5 is as well". How can I take the numbers from the string and put them in variables? (In my example I am looking to use the 4 and 5.)

So, i write some code to solve your problem. First and easyest case: you pass the sentence as a list, like this [hello,4,is,5,and]. In this case, you need to check if an element of the list is a number with number/1:
parseSentence([],[]).
parseSentence([H|T],L):-
\+(number(H)),
parseSentence(T,L).
parseSentence([H|T],[H|T1]):-
number(H),
parseSentence(T,T1).
?- parseSentence([hello,4,is,5,and],L).
L = [4, 5].
false
Second and most interesting case: you pass the sentence as a string, like this: "hello 4 is a number, and 5 is as well". Here the code (a bit redundant to make it more clear):
parseSentenceMod([],[]).
parseSentenceMod([H|T],[A|T1]):-
atom_number(H,A),
parseSentenceMod(T,T1).
parseSentenceMod([H|T],L):-
\+(atom_number(H,_)),
parseSentenceMod(T,L).
findAtom([],[]).
findAtom([H|T],[H1|T1]):-
atom_string(H1,H),
findAtom(T,T1).
parseString(S,P,A,N):-
split_string(S," ","",P),
findAtom(P,A),
parseSentenceMod(A,N).
?- parseString("hello 4 is a number, and 5 is as well",S,A,N).
A = [hello, '4', (is), a, 'number,', and, '5', (is), (as), well],
N = [4, 5],
S = ["hello", "4", "is", "a", "number,", "and", "5", "is", "as", "well"]
false
atom_number/2 checks if the atom is a number, atom_string/2 creates an atom given a string and split_string/4 splits the string in substrings separated by, in this case, a blank space. Hope it helps.

Related

Ruby regex: union 2K values in one regex,

I code a process to process bunch of text files and capture its name if any of 2000 literals exists in it (1 or many). So I'm thinking to combine that many values into one regex, do you think it's doable, I did test for 100 and looks like it's OK. Tx all
Code below depics my flow and sample code, just without looping.
# 1. read regex value list as file [alpha,fox, delta] # 2000 values
# 2. read file into s #5000 files
# 3. find if any of #1 values exists in each #2 file. *with regex tweaks to match format dbname.dob.table
s = '1 dbName.dbo.ALPHA 2 DBNAME.bcd.ALPHA 3 dbName..ALPHA 4 ALPHA 5x dbName.alphA 6x alpha.XX 7x ###dbName.###a.alpha --alpha
dbName..FOX dbName.dbo.DELTA clarity.aba..fox '
value1 = '(?<=^|\s)(?:dbName\.[a-z]*\.)?(?:alpha)(?=\s|$)'
value2 = '(?<=^|\s)(?:dbName\.[a-z]*\.)?(?:fox)(?=\s|$)'
##...
value2000 = '(?<=^|\s)(?:dbName\.[a-z]*\.)?(?:delta)(?=\s|$)'
regex = /#{value1}|#{value2}|#{value2000}/i ## can I union 2000 regex's ???
puts 'reg1: ' + regex.to_s
puts 'result: ' + s.scan(regex).to_s
if s.scan(regex) then puts '...Match!!!d' end
Declaring 2000 variables is highly unnecessary; you should define all values in a single array, then somehow loop through them.
Also, the regular expression is highly repetitive - e.g. the use of (?:dbName\.[a-z]*\.) 2000 times. This can be simplified by grouping all of your values within the non-capture group as follows:
values = %w(alpha fox delta)
regex = /(?<=^|\s)(?:dbName\.[a-z]*\.)?(?:#{Regexp.union(values)})(?=\s|$)/
This is the result:
/(?<=^|\s)(?:dbName\.[a-z]*\.)?(?:(?-mix:alpha|fox|delta))(?=\s|$)/
If you extend that values array to contain 2000 strings, the other code does not need to change.
Provided two conditions are met, I would do it as follows, which I think would be far more efficient than using a gigantic regular expression, which, by its nature, requires that a linear search of the "bad words" be performed for each word in the string, until a match is found or it is determined that there are no matches.
We are given a file whose path is contained in a variable fname and an array of bad words:
arr = ["alpha", "fox", "delta", "charlie", "mabel"]
The first condition that I spoke of above is that, by way of example, "ALPHA" and "Alpha" match "alpha", but "aLPha" does not (or some variant of that).
The second condition is that there is a regular expression with a capture group that would capture a bad word if a bad word were present at the given location in a match. For example:
regex = (?<=^|\s)(?:dbName\.[a-z]*\.)?(\p{Alpha}+)(?=\s|$)
Wherever there is a match, the capture group (\p{Alpha}+) would capture a string of one or more alphanumeric characters whose value is assigned to the global variable $1. We will then check to see if the value of $1 is a bad word. (The regular expression might have other capture groups as well, in which case we might be looking for $2 or $3, say, or a named capture group.)
If there were more than one such regular expression to check for, the code below could be executed for each of them until a match is found or it is determined that there are no more matches.
The first step is to convert the array of bad words to a set:
require 'set'
bad_words = arr.flat_map { |w| [w, w.capitalize, w.upcase] }.to_set
#=> #<Set: {"alpha", "Alpha", "ALPHA", "fox", "Fox", "FOX",
# "delta", "Delta", "DELTA", "charlie", "Charlie", "CHARLIE",
# "mabel", "Mabel", "MABEL"}>
This allows very fast word lookups--much faster than stepping through an array. We may then search the file as follows.
rv = IO.foreach(fname).any? do |line|
line.gsub(regex).any? { bad_words.include?($1) }
end
IO::foreach without a block is seen to return an enumerator. We can then chain that to any? to determine if there is a line that contains a match of the regular expression and the value of its capture group is contained in the set bad_words. If such a line is found the search terminates and true is returned; else, false is returned.
It is seen that String#gsub without a block returns an enumerator, which here I've chained to any?. This form of gsub has nothing to do with string replacements; it just generates matches. Those matches are passed to the block, but we are only interested in the contents of the capture group, which are held by $1. Hence the expression bad_words.include?($1).

Regular expression that only leaves the second number

I'm trying to make a regular expression that leaves only the second number.
{"1"=>"2", "3"=>"6"}
For example, that leaves only the numbers:
2, 6
I tried this:
(([0-9]+)[=>]*([0-9]+))
{"1"=>"2", "3"=>"6"}.values
will give you:
["2", "6"]

In ruby, how do I use string.scan(/regex/) method for numbers from 1 to 12?

That's what I am doing:
c.scan(/[1-9]|1[0-2]/)
For some reason, it returns only numbers from 1 to 9, ignoring the second part. I tried experimenting a little bit, it seems that the method will search for 10-12 only if 1 is excluded from [1-9] part, e.g., c.scan(/[2-9]|1[0-2]/) will do. What is the reason?
P.S. I know that this method lacks lookbehinds and will search for numbers and "part of numbers" as well
Change the order of your patterns and add word boundaries if necessary.
c.scan(/\b(?:1[0-2]|[1-9])\b/)
The pattern before | is used first. So in our case, it matches all the numbers from 10 to 12. After that the next pattern, that is the one after | is used and now it matches all the remaining numbers ranges from 1 to 9. Note that this would match 9 in 59 also. So i suggest you to put your pattern inside a capturing or non-capturing group and add word boundary \b (matches between a word character and a non-word character) before and after to that group .
DEMO
| matches left to right, and the first part of the right side (1) is always matched by the left side. Reverse them:
c.scan(/1[0-2]|[1-9]/)
Here's another way you might consider extracting numbers between 1 and 12 (assuming that's what you want to do):
c = '14 0 11x 15 003 y12'
c.scan(/\d+/).map(&:to_i).select { |n| (1..12).cover?(n) }
#=> [11, 3, 12]
I've returned an array of integers, rather than strings, thinking that probably would be more useful, but if you want strings:
c.scan(/\d+/).map { |s| s.to_i.to_s }
.select { |s| ['10', '11', '12', *'1'..'9'].include?(s) }
#=> ["11", "3", "12"]
I see several advantages to this approach, versus using a single regex:
it's easy to understand;
the regex is simple;
it's easy to modify if the permissible values change; and
it can be broken into three pieces to facilitate testing.

Overlapping string matching using regular expressions

Imagine we have some sequence of letters in the form of a string, call it
str = "gcggcataa"
The regular expression
r = /(...)/
matches any three characters, and when I execute the code
str.scan(r)
I get the following output:
["gcg", "gca", "taa"]
However, what if I wanted to scan through and instead of the distinct, non-overlapping strings as above but instead wanted to get this output:
["gcg", "cgg", "ggc", "gca", "cat", "ata", "taa"]
What regular expression would allow this?
I know I could do this with a loop but I don't want to do that
str = "gcggcataa"
str.chars.each_cons(3).map(&:join) # => ["gcg", "cgg", "ggc", "gca", "cat", "ata", "taa"]

Matching repeated pattern in string

I have street names and numbers in a file, like so:
Sokolov 19, 20, 23 ,25
Hertzl 80,82,84,86
Hertzl 80a,82b,84e,90
Aba Hillel Silver 2,3,5,6,
Weizman 8
Ahad Ha'am 9 13 29
I parse the lines one by one with regex. I want a regex that will find and match:
The name of the street,
The street numbers with its possible a,b,c,d attached.
I've come up with this mean while:
/(\D{2,})\s+(\d{1,3}[a-d|א-ד]?)(?:[,\s]{1,3})?/
It finds the street name and first number. I need to find all the numbers.
I don't want to use two separate regex's if possible, and I prefer not to use Ruby's scan but just have it in one regex.
You can use regex to find all the numbers, with their separators:
re = /\A(.+?)\s+((?:\d+[a-z]*[,\s]+)*\d+[a-z]*)/
txt = "Sokolov 19, 20, 23 ,25
Hertzl 80,82,84,86
Hertzl 80a,82b,84e,90
Aba Hillel Silver 2,3,5,6,
Weizman 8
Ahad Ha'am 9 13 29"
matches = txt.lines.map{ |line| line.match(re).to_a[1..-1] }
p matches
#=> [["Sokolov", "19, 20, 23 ,25"],
#=> ["Hertzl", "80,82,84,86"],
#=> ["Hertzl", "80a,82b,84e,90"],
#=> ["Aba Hillel Silver", "2,3,5,6"],
#=> ["Weizman", "8"],
#=> ["Ahad Ha'am", "9 13 29"]]
The above regex says:
\A Starting at the front of the string
(…) Capture the result
.+? Find one or more characters, as few as possible that make the rest of this pattern match.
\s+ Followed by one or more whitespace characters (which we don't capture)
(…) Capture the result
(?:…)* Find zero or more of what's in here, but don't capture them
\d+ One or more digits (0–9)
[a-z]* Zero or more lowercase letters
[,\s]+ One or more commas and/or whitespace characters
\d+ Followed by one or more digits
[a-z]* And zero or more lowercase letters
However, if you want to break the number up into pieces you will need to use scan or split or the equivalent.
result = matches.map{ |name,numbers| [name,numbers.scan(/[^,\s]+/)] }
p result
#=> [["Sokolov", ["19", "20", "23", "25"]],
#=> ["Hertzl", ["80", "82", "84", "86"]],
#=> ["Hertzl", ["80a", "82b", "84e", "90"]],
#=> ["Aba Hillel Silver", ["2", "3", "5", "6"]],
#=> ["Weizman", ["8"]],
#=> ["Ahad Ha'am", ["9", "13", "29"]]]
This is because regex captures inside a repeating group do not capture each repetition. For example:
re = /((\d+) )+/
txt = "hello 11 2 3 44 5 6 77 world"
p txt.match(re)
#=> #<MatchData "11 2 3 44 5 6 77 " 1:"77 " 2:"77">
The whole regex matches the whole string, but each capture only saves the last-seen instance. In this case, the outer capture only gets "77 " and the inner capture only gets "77".
Why do you prefer not to use scan? This is what it is made for.
If you want your 3rd example to work, you need to have the [a-d] change to include the e in the range. After changing that you can use (\D{2,})\s+(\d{1,3}[a-e]?(?:[,\s]{1,3})*)*. Using the examples you gave I did some testing using Rubular.
Using some more groupings you can have the repetition on those last few conditions (which seem to be pretty tricky. This way the spacing and comma at the end will get caught in the repetition after consuming the space initially.
The only way around the limitation that you can only capture the last instance of a repeated expression is to write your regex for a single instance and let the regex machine do the repeating for you, as occurs with the global substitute options, admittedly similar to scan. Unfortunately, in that case, you have to match for either the street name or the street number and then have no way to easily associate the captured numbers with the captured names.
Regex is great at what it does, but when you try to extend its application beyond it's natural limitations, it's not pretty. ;-)
I want a regex that will find and match....
Do the street names also contain digits (0-9), other characters beside an apostrophe?
Are the street numbers based off arbitrary data? Is it always just an optional a, b, c, or d?
Are you needing a minimum and maximum limitation of string length?
Here are some possible options:
If you are unsure about what the street name contains, but know your street number pattern will be numbers with an optional letter, commas or spaces.
/^(.*?)\s+(\d+(?:[a-z]?[, ]+\d+)*)(?=,|$)/
See working demo
If the street names contain only letters with optional apostrophe's and the street numbers contain numbers with an optional letter, comma.
/^([a-zA-Z' ]+)\s+(\d+(?:[a-z]?[, ]+\d+)*)(?=,|$)/
See working demo
If your street name and street number pattern are always consistant, you could easily do.
/^([a-zA-Z' ]+)\s+([0-9a-z, ]+)$/
See working demo

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