How to extract data from text columns - ruby

Question

I have two addresses side-by-side in a multi-line string:
Adresse de prise en charge : Adresse d'arrivée :
rue des capucines rue des tilleuls
92210 Saint Cloud 67000 Strasbourg
Tél.: Tél.:
I need to extract the addresses on the left and right with a regexp, and assign them to variables. I need to match:
address1: "rue des capucines 92210 Saint Cloud"
address2: "rue des tilleuls 67000 Strasbourg"
I thought of separating them with spaces, but I cant find any regexp to count the spaces. I tried:
en\s*charge\s*:\s*((.|\n)*)\s*
and similar, but that gives me both addresses, and is not what I'm looking for. Any help will be deeply appreciated.

Answer 1

I'd do something like this:
str = <<EOT
Adresse de prise en charge : Adresse d'arrivée :
rue des capucines rue des tilleuls
92210 Saint Cloud 67000 Strasbourg
Tél.: Tél.:
EOT
left_addr = []
right_addr = []
lines = str.squeeze("\n").gsub(':', '').lines.map(&:strip) # => ["Adresse de prise en charge Adresse d'arrivée", "rue des capucines rue des tilleuls", "92210 Saint Cloud 67000 Strasbourg", "Tél. Tél."]
center_line_pos = lines.max.length / 2 # => 35
lines.each do |l|
left_addr << l[0 .. (center_line_pos - 1)].strip
right_addr << l[center_line_pos .. -1].strip
end
At this point left_addr and right_addr look like:
left_addr # => ["Adresse de prise en charge", "rue des capucines", "92210 Saint Cloud", "Tél."]
right_addr # => ["Adresse d'arrivée", "rue des tilleuls", "67000 Strasbourg", "Tél."]
And here's what they contain:
puts left_addr
puts '------'
puts right_addr
# >> Adresse de prise en charge
# >> rue des capucines
# >> 92210 Saint Cloud
# >> Tél.
# >> ------
# >> Adresse d'arrivée
# >> rue des tilleuls
# >> 67000 Strasbourg
# >> Tél.
If you need the results all in one line without the 'Tel:':
puts left_addr[0..-2].join(' ').squeeze(' ')
puts '------'
puts right_addr[0..-2].join(' ').squeeze(' ')
# >> Adresse de prise en charge rue des capucines 92210 Saint Cloud
# >> ------
# >> Adresse d'arrivée rue des tilleuls 67000 Strasbourg
Here's a breakdown of what is going on:
str.squeeze("\n") # => " Adresse de prise en charge : Adresse d'arrivée :\n rue des capucines rue des tilleuls\n 92210 Saint Cloud 67000 Strasbourg\n Tél.: Tél.:\n"
.gsub(':', '') # => " Adresse de prise en charge Adresse d'arrivée \n rue des capucines rue des tilleuls\n 92210 Saint Cloud 67000 Strasbourg\n Tél. Tél.\n"
.lines # => [" Adresse de prise en charge Adresse d'arrivée \n", " rue des capucines rue des tilleuls\n", " 92210 Saint Cloud 67000 Strasbourg\n", " Tél. Tél.\n"]
.map(&:strip) # => ["Adresse de prise en charge Adresse d'arrivée", "rue des capucines rue des tilleuls", "92210 Saint Cloud 67000 Strasbourg", "Tél. Tél."]

Answer 2

Assuming that each address section in each line is indented as much as or further than the corresponding "Adresse" in the first line, the following can extract not only two addresses aligned sidewards, but n addresses in general.
lines = string.split(/#{$/}+/)
# => [
# => "Adresse de prise en charge : Adresse d'arrivée :",
# => " rue des capucines rue des tilleuls",
# => " 92210 Saint Cloud 67000 Strasbourg",
# => " Tél.: Tél.:"
# => ]
break_points = []
lines.first.scan(/\bAdresse\b/){break_points.push($~.begin(0))}
ranges = break_points.push(0).each_cons(2).map{|s, e| s..(e - 1)}
# => [0..53, 54..-1]
address1, address2 =
lines[1..-2]
.map{|s| ranges.map{|r| s[r]}}
.transpose
.map{|a| a.join(" ").strip.squeeze(" ")}
# => [
# => "rue des capucines 92210 Saint Cloud",
# => "rue des tilleuls 67000 Strasbourg"
# => ]

Answer 3

str =
" Adresse de prise en charge : Adresse d'arrivée :
rue des capucines rue des tilleuls
92210 Saint Cloud 67000 Strasbourg
Tél.: Tél.:"
adr_prise, adr_arr = str.lines[3].strip.split(/ {2,}/) #split on 2+ spaces
code_prise, cite_prise, code_arr, cite_arr = str.lines[6].strip.split(/ {2,}/)

Answer 4

Assumptions
I have assumed that the first and last lines are not wanted and the street names are separated by at least two spaces, and the same for the postal code/city strings. This permits the street name (and postal code/city pair) for "prise en charge" to end below "Adresse d'arrivée :".
Code
def parse_text(text)
text.split(/\n+\s+/)[1..-2].
map { |s| s.gsub(/\d+\K\s+/,' ').split(/\s{2,}/) }.
transpose.
map { |a| a.join(' ') }
end
Examples
Example 1
text = <<BITTER_END
Adresse de prise en charge : Adresse d'arrivée :
rue des capucines rue des tilleuls
92210 Saint Cloud 67000 Strasbourg
Tél.: Tél.:
BITTER_END
parse_text(text)
#=> ["rue des capucines 9210 Saint Cloud",
# "rue des tileuls 670 Strasbourg"]
Example 2
text = <<_
Adresse 1 : Adresse 2 : Adresse 3 :
rue nom le plus long du monde par un mile rue gargouilles rue des tilleuls
92210 Saint Cloud 31400 Nice 67000 Strasbourg
France France France
Tél.: Tél.: Tél.:
_
parse_text(text)
#=> ["rue nom le plus long du monde par un mile 92210 Saint Cloud France",
# "rue gargouilles 31400 Nice France",
# "rue des tilleuls 67000 Strasbourg France"]
Explanation
The steps for text given in the question:
Split into lines, removing blank lines and leading spaces:
a1 = text.split(/\n+\s+/)
#=> ["Adresse de prise en charge : Adresse d'arrivée :",
# "rue des capucines rue des tilleuls",
# "92210 Saint Cloud 67000 Strasbourg",
# "Tél.: Tél.:\n"]
Remove first and last lines:
a2 = a1[1..-2]
#=> ["rue des capucines rue des tilleuls",
# "92210 Saint Cloud 67000 Strasbourg"]
Remove extra spaces between the postal codes and cities and split each line on two or more spaces:
r = /
\d+ # match one or more digits
\K # forget everything matched so far
\s+ # match one of more spaces
/x # extended/free-spacing regex definition mode
a3 = a2.map { |s| s.gsub(/\d+\K\s+/,' ').split(/\s{2,}/) }
#=> [["rue des capucines", "rue des tilleuls"],
# ["92210 Saint Cloud", "67000 Strasbourg"]]
Group by column:
a4 = a3.transpose
#=> [["rue des capucines", "92210 Saint Cloud"],
# ["rue des tilleuls", "67000 Strasbourg"]]
Join strings:
a4.map { |a| a.join(' ') }
#=> ["rue des capucines 92210 Saint Cloud",
# "rue des tilleuls 67000 Strasbourg"]

Answer 5

Inspired by #steenslag's very pragmatic answer, here's a pretty dense one-liner just for fun.
# Assume the input data is in the variable `text`
left_addr, right_addr = text.lines.values_at(3, 6).map do |line|
line.scan(/(?:\d+ +)?\S+(?: \S+)*/)
.map {|part| part.squeeze(' ') }
end
.transpose
.map {|addr| addr.join(' ') }
puts left_addr
# => rue des capucines 92210 Saint Cloud
puts right_addr
# => rue des tilleuls 67000 Strasbourg
Like #steenslag's answer, this assumes that the desired data is always on lines 3 and 6. It also assumes that on line 6 both columns will have a postal code and city and that the postal code will always start with a digit.
Because it's a pretty dense one-liner and because it makes a lot of assumptions, I don't think this is the best answer and I'm marking it Community Wiki.
If I have time I'll come back and unpack this later.

Answer 6

Assuming that the "center line position" is known, this would work:
left_lines, right_lines = str.scan(/^(.{50})(.*)$/).transpose
The regular expression captures 50 characters at the beginning of each line plus the remaining characters until the line's end.
scan returns a nested array: (I'm using placeholders because the actual lines are too long)
[
['1st left line', '1st right line'],
['2nd left line', '2nd right line'],
...
]
transpose converts it to:
[
['1st left line', '2nd left line', ...], # <- assigned to left_lines
['1st right line', '2nd right line', ...] # <- assigned to right_lines
]
The lines (excluding the first and last line) have to be joined and spaces have to be removed: (see strip and squeeze)
left_lines[1..-2].join(' ').strip.squeeze(' ')
#=> "rue des capucines 92210 Saint Cloud"
Same for right_lines:
right_lines[1..-2].join(' ').strip.squeeze(' ')
#=> "rue des tilleuls 67000 Strasbourg"