If I have a string like
6d7411014f
I want to read the the occurrence of first two integers and put the final number in a variable
Based on above example my variable would contain 67
more examples:
d550dfe10a
variable would be 55
What i've tried is \d but that gives me 6. how do I get the second number?
I'd use scan for this sort of thing:
n = my_string.scan(/\d/)[0,2].join.to_i
You'd have to decide what you want to do if there aren't two numbers though.
For example:
>> '6d7411014f'.scan(/\d/)[0,2].join.to_i
=> 67
>> 'd550dfe10a'.scan(/\d/)[0,2].join.to_i
=> 55
>> 'pancakes'.scan(/\d/)[0,2].join.to_i
=> 0
>> '6 pancakes'.scan(/\d/)[0,2].join.to_i
=> 6
References:
String#scan
Array#[]
Array#join
I really can't answer this exactly in Ruby, but a regex to do it is:
/^\D*(\d)\D*(\d)/
Then you have to concatenate $1 and $2 (or whatever they are called in Ruby).
Building off of sidyll's answer,
string = '6d7411014f'
matched_vals = string.match(/^\D*(\d)\D*(\d)/)
extracted_val = matched_vals[1].to_i * 10 + matched_vals[2].to_i
Related
I want to find a specific character in a given string of number for example if my input is:
1 4 5 7 9 12
Then for 4 the answer should be 1. My code is as follows:
secarr = second.split(" ")
answer = secarr.index(number) #here number is a variable which gets the character
puts answer
The above method works if I write "4" instead of number or any other specific character but does not work if I write a variable. Is there a method in ruby to do the same?
This is probably your variable number is an Integer, and secarr is an Array of Strings. Try to cast the number to string:
answer = secarr.index(number.to_s)
How can I generate an n-character pseudo random string containing only A-Z, 0-9 like SecureRandom.base64 without "+", "/", and "="? For example:
(0..n).map {(('1'..'9').to_a + ('A'..'Z').to_a)[rand(36)]}.join
Array.new(n){[*"A".."Z", *"0".."9"].sample}.join
An elegant way to do it in Rails 5 (I don't test it in another Rails versions):
SecureRandom.urlsafe_base64(n)
where n is the number of digits that you want.
ps: SecureRandom uses a array to mount your alphanumeric string, so keep in mind that n should be the amount of digits that you want + 1.
ex: if you want a 8 digit alphanumeric:
SecureRandom.urlsafe_base64(9)
Even brute force is pretty easy:
n = 20
c = [*?A..?Z + *?0..?9]
size = c.size
n.times.map { c[rand(size)] }.join
#=> "IE210UOTDSJDKM67XCG1"
or, without replacement:
c.sample(n).join
#=> "GN5ZC0HFDCO2G5M47VYW"
should that be desired. (I originally had c = [*(?A..?Z)] + [*(?0..?9)], but saw from #sawa's answer that that could be simplified quite a bit.)
To generate a random string from 10 to 20 characters including just from A to Z and numbers, both always:
require 'string_pattern'
puts "10-20:/XN/".gen
You can do simply like below:
[*'A'..'Z', *0..9].sample(10).join
Change the number 10 to any number to change the length of string
I have a string that looks like this:
Results 1 - 10 of 20
How would I find the number 10 and 20 of that sentence using regex in Ruby?
Something like:
first_number, second_number = compute_regex(my_string)...
Thanks
Like so:
first, second = *source.scan(/\d+/)[-2,2]
Explanation
\d+ matches any number
scan finds all matches of its regular expression argument in source
[-2,2] returns the last two numbers in an array: starts at index -2 from end, returns next 2
* splat operator unpacks these two matches into the variables first and second (NOTE: this operator is not necessary, you can remove this, and I like the concept)
Try this:
a = "Results 1 - 10 of 20"
first_number, second_number = a.match(/\w+ (\d) \- (\d+) of (\d+)/)[2..3].map(&:to_i)
The map piece is necessary because the regexp MatchData objects returned are strings.
I'm working with OPE IDs. One file has them with two trailing zeros, eg, [998700, 1001900]. The other file has them with one or two leading zeros for a total length of six, eg, [009987, 010019]. I want to convert every OPE ID (in both files) to an eight-digit string with exactly two leading zeros and however many zeros at the end to get it to be eight digits long.
Try this:
a = [ "00123123", "077934", "93422", "1231234", "12333" ]
a.map { |n| n.gsub(/^0*/, '00').ljust(8, '0') }
=> ["00123123", "00779340", "00934220", "001231234", "00123330"]
If you have your data parsed and stored as strings, it could be done like this, for example.
n = ["998700", "1001900", "009987", "0010019"]
puts n.map { |i|
i =~ /^0*([0-9]+?)0*$/
"00" + $1 + "0" * [0, 6 - $1.length].max
}
Output:
00998700
00100190
00998700
00100190
This example on codepad.
I'm note very sure though, that I got the description exactly right. Please check the comments and I correct in case it's not exactly what you were looking for.
With the help of the answers given by #detunized & #nimblegorilla, I came up with:
"998700"[0..-3].rjust(6, '0').to_sym
to make the first format I described (always with two trailing zeros) equal to the second.
I'm outputting a set of numbered files from a Ruby script. The numbers come from incrementing a counter, but to make them sort nicely in the directory, I'd like to use leading zeros in the filenames. In other words
file_001...
instead of
file_1
Is there a simple way to add leading zeros when converting a number to a string? (I know I can do "if less than 10.... if less than 100").
Use the % operator with a string:
irb(main):001:0> "%03d" % 5
=> "005"
The left-hand-side is a printf format string, and the right-hand side can be a list of values, so you could do something like:
irb(main):002:0> filename = "%s/%s.%04d.txt" % ["dirname", "filename", 23]
=> "dirname/filename.0023.txt"
Here's a printf format cheat sheet you might find useful in forming your format string. The printf format is originally from the C function printf, but similar formating functions are available in perl, ruby, python, java, php, etc.
If the maximum number of digits in the counter is known (e.g., n = 3 for counters 1..876), you can do
str = "file_" + i.to_s.rjust(n, "0")
Can't you just use string formatting of the value before you concat the filename?
"%03d" % number
Use String#next as the counter.
>> n = "000"
>> 3.times { puts "file_#{n.next!}" }
file_001
file_002
file_003
next is relatively 'clever', meaning you can even go for
>> n = "file_000"
>> 3.times { puts n.next! }
file_001
file_002
file_003
As stated by the other answers, "%03d" % number works pretty well, but it goes against the rubocop ruby style guide:
Favor the use of sprintf and its alias format over the fairly
cryptic String#% method
We can obtain the same result in a more readable way using the following:
format('%03d', number)
filenames = '000'.upto('100').map { |index| "file_#{index}" }
Outputs
[file_000, file_001, file_002, file_003, ..., file_098, file_099, file_100]