I have a challenge that calls for obfuscating numbers in a string, such as a SSN, for example: XXX-XX-4430. I've gotten pretty close:
def hide_all_ssns(string)
string.scan(/\w{3}-\w{2}-\w{4}/)
string.gsub('/\w{3}-\w{2}', 'XXX-XX')
end
but I get an error:
Error! hide_all_ssns obfuscates any SSNs in the string expected:
"XXX-XX-1422, XXX-XX-0744, XXX-XX-8762" got: "234-60-1422,
350-80-0744, 013-60-8762" (using ==)
I initially had the regular-expression (/\d{3}-\d{2}-\d{4}/) but thought that the problem was attempting to convert the integers in the string to X. Now I'm using \w, yet I am getting the same error.
Does anyone have any insight? I'm a newbie to coding and have exhausted Ruby-doc, as well as any blogs I can find on regex/gsub, but I am getting nowhere.
You're mis-using gsub (your regular expression needs to be between forward slashes), but I still thing gsub! is what you want...
def hide_all_ssns(string)
string.scan(/\w{3}-\w{2}-\w{4}/)
string.gsub!(/\w{3}-\w{2}/, 'XXX-XX')
end
Working example:
1.9.3p448 :063 > string = "123-45-6789"
=> "123-45-6789"
1.9.3p448 :064 > def hide_all_ssns(string)
1.9.3p448 :065?> string.scan(/\w{3}-\w{2}-\w{4}/)
1.9.3p448 :066?> string.gsub!(/\w{3}-\w{2}/, 'XXX-XX')
1.9.3p448 :067?> end
=> nil
1.9.3p448 :068 > hide_all_ssns(string)
=> "XXX-XX-6789"
1.9.3p448 :069 > string
=> "XXX-XX-6789"
Why does it have to be so hard? All U.S. social security numbers are the same format, right? So, work from that point. Here's some variations on a theme, ordered by escalating obscurity:
ssn = '123-45-6789' # => "123-45-6789"
ssn[0, 6] = 'XXX-XX' # => "XXX-XX"
ssn # => "XXX-XX-6789"
Or:
numbers = ssn.scan(/\d+/) # => ["123", "45", "6789"]
'XXX-XX-' + numbers.last # => "XXX-XX-6789"
Or:
ssn = '123-45-6789' # => "123-45-6789"
ssn[0, 6] = ssn[0, 6].gsub(/\d/, 'X') # => "XXX-XX"
ssn # => "XXX-XX-6789"
Or:
ssn[0,6] = ssn[0, 6].tr('0-9', 'X') # => "XXX-XX"
ssn # => "XXX-XX-6789"
Or:
numbers = ssn.split('-') # => ["123", "45", "6789"]
[*numbers[0, 2].map{ |s| 'X' * s.size }, numbers[-1]].join('-') # => "XXX-XX-6789"
Or:
ssn[/(\d+)-(\d+)-(\d+)/] # => "123-45-6789"
[$1, $2, $3] # => ["123", "45", "6789"]
[$3, *[$2, $1].map{ |s| s.gsub(/./, 'X') }].reverse.join('-') # => "XXX-XX-6789"
Of course, using one of these would cheating, since you're supposed to figure the challenge out by yourself, but they're good food for thought and a decent starting point for your own solution.
Short and simple... You could maybe try something like this:
crypted = ('X' * 6) + "4543-2329-1354-1111".to_s[14..18]
=> "XXXXXX-1111"
Related
I'm running Ruby 2.2.1.
The following code runs as expected as string hash keys are duped and frozen:
f = 'foo'
h = {f => 'bar'}
h.compare_by_identity
h[f] # => nil
h['foo'] # => nil
h[h.keys.first] # => "bar"
But I can't for the life of me figure out what is going on here:
h = {'foo' => 'bar'}
h.compare_by_identity
h.keys.first.frozen? # => true
'foo'.frozen? # => false
h.keys.first.object_id # => 20421220
'foo'.object_id # => 20067280
h['foo'] # => "bar"
h['foo'.dup] # => nil
It's interesting to note the the docs for #compare_by_identity started using #dup at 2.2.0. So it seems this behavior change is known.
2.1.7:
h1["a"] #=> nil # different objects.
2.2.0:
h1["a".dup] #=> nil # different objects.
However, the source is the same.
The same does not happen with other literals like arrays. Any ideas on why this behavior changed for string literals? The docs give no hints as to why.
Is there any easy way to convert strings like 3500goat to 3500Goat and goat350rat to Goat350rat?
I am trying to convert the first occurrence of alphabet in an alphanumeric string to uppercase. I was trying the code below using the method sub, but no luck.
stringtomigrate = 3500goat
stringtomigrate.sub!(/\D{0,1}/) do |w|
w.capitalize
This should work:
string.sub(/[a-zA-Z]/) { |s| s.upcase }
or a shorthand:
string.sub(/[a-zA-Z]/, &:upcase)
examples:
'3500goat'.sub(/[a-zA-Z]/, &:upcase)
# => "3500Goat"
'goat350rat'.sub(/[a-zA-Z]/, &:upcase)
# => "Goat350rat"
Try this
1.9.3-p545 :060 > require 'active_support/core_ext'
=> true
1.9.3-p545 :099 > "goat350rat to Goat350rat".sub(/[a-zA-Z]/){ |x| x.titleize}
=> "Goat350rat to Goat350rat"
I need to create regular expression for 2 and only 2 letters. I understood it has to be the following /[a-z]{2}/i, but it matches any string with 2 or more letters. Here is what I get:
my_reg_exp = /[a-z]{2}/i
my_reg_exp.match('aa') # => #<MatchData "aa">
my_reg_exp.match('AA') # => #<MatchData "AA">
my_reg_exp.match('a') # => nil
my_reg_exp.match('aaa') # => #<MatchData "aa">
Any suggestion?
You can add the anchors like this:
my_reg_exp = /^[a-z]{2}$/i
Test:
my_reg_exp.match('aaa')
#=> nil
my_reg_exp.match('aa')
#=> #<MatchData "aa">
Hao's solution matches isn't locale sensitive. If this is important for your use case:
/\a[[:alpha:]]{2}\z/
2.0.0-p451 :005 > 'aba' =~ /\A[[:alpha:]]{2}\Z/
=> nil
2.0.0-p451 :006 > 'ab' =~ /\A[[:alpha:]]{2}\Z/
=> 0
2.0.0-p451 :007 > 'xy' =~ /\A[[:alpha:]]{2}\Z/
=> 0
2.0.0-p451 :008 > 'zxy' =~ /\A[[:alpha:]]{2}\Z/
=> nil
Per usual, if you need further assistance, leave a comment.
You can use /\b[a-z]{2}\b/i to match a two-letter string. /b Matches a word-break.
This means you can scan a string to find all occurrences:
'Foo is a bar'.scan(/\b[a-z]{2}\b/i) #=> ["is"]
Or find the first match in a string using:
'a bc def'[/\b[a-z]{2}\b/i] # => "bc"
I have an Excel column containing part numbers. Here is a sample
As you can see, it can be many different datatypes: Float, Int, and String. I am using roo gem to read the file. The problem is that roo interprets integer cells as Float, adding a trailing zero to them (16431 => 16431.0). I want to trim this trailing zero. I cannot use to_i because it will trim all the trailing numbers of the cells that require a decimal in them (the first row in the above example) and will cut everything after a string char in the String rows (the last row in the above example).
Currently, I have a a method that checks the last two characters of the cell and trims them if they are ".0"
def trim(row)
if row[0].to_s[-2..-1] == ".0"
row[0] = row[0].to_s[0..-3]
end
end
This works, but it feels terrible and hacky. What is the proper way of getting my Excel file contents into a Ruby data structure?
def trim num
i, f = num.to_i, num.to_f
i == f ? i : f
end
trim(2.5) # => 2.5
trim(23) # => 23
or, from string:
def convert x
Float(x)
i, f = x.to_i, x.to_f
i == f ? i : f
rescue ArgumentError
x
end
convert("fjf") # => "fjf"
convert("2.5") # => 2.5
convert("23") # => 23
convert("2.0") # => 2
convert("1.00") # => 1
convert("1.10") # => 1.1
For those using Rails, ActionView has the number_with_precision method that takes a strip_insignificant_zeros: true argument to handle this.
number_with_precision(13.00, precision: 2, strip_insignificant_zeros: true)
# => 13
number_with_precision(13.25, precision: 2, strip_insignificant_zeros: true)
# => 13.25
See the number_with_precision documentation for more information.
This should cover your needs in most cases: some_value.gsub(/(\.)0+$/, '').
It trims all trailing zeroes and a decimal point followed only by zeroes. Otherwise, it leaves the string alone.
It's also very performant, as it is entirely string-based, requiring no floating point or integer conversions, assuming your input value is already a string:
Loading development environment (Rails 3.2.19)
irb(main):001:0> '123.0'.gsub(/(\.)0+$/, '')
=> "123"
irb(main):002:0> '123.000'.gsub(/(\.)0+$/, '')
=> "123"
irb(main):003:0> '123.560'.gsub(/(\.)0+$/, '')
=> "123.560"
irb(main):004:0> '123.'.gsub(/(\.)0+$/, '')
=> "123."
irb(main):005:0> '123'.gsub(/(\.)0+$/, '')
=> "123"
irb(main):006:0> '100'.gsub(/(\.)0+$/, '')
=> "100"
irb(main):007:0> '127.0.0.1'.gsub(/(\.)0+$/, '')
=> "127.0.0.1"
irb(main):008:0> '123xzy45'.gsub(/(\.)0+$/, '')
=> "123xzy45"
irb(main):009:0> '123xzy45.0'.gsub(/(\.)0+$/, '')
=> "123xzy45"
irb(main):010:0> 'Bobby McGee'.gsub(/(\.)0+$/, '')
=> "Bobby McGee"
irb(main):011:0>
Numeric values are returned as type :float
def convert_cell(cell)
if cell.is_a?(Float)
i = cell.to_i
cell == i.to_f ? i : cell
else
cell
end
end
convert_cell("foobar") # => "foobar"
convert_cell(123) # => 123
convert_cell(123.4) # => 123.4
How do I a string against a regex such that it will return true if the whole string matches (not a substring)?
eg:
test( \ee\ , "street" ) #=> returns false
test( \ee\ , "ee" ) #=> returns true!
Thank you.
You can match the beginning of the string with \A and the end with \Z. In ruby ^ and $ match also the beginning and end of the line, respectively:
>> "a\na" =~ /^a$/
=> 0
>> "a\na" =~ /\Aa\Z/
=> nil
>> "a\na" =~ /\Aa\na\Z/
=> 0
This seems to work for me, although it does look ugly (probably a more attractive way it can be done):
!(string =~ /^ee$/).nil?
Of course everything inside // above can be any regex you want.
Example:
>> string = "street"
=> "street"
>> !(string =~ /^ee$/).nil?
=> false
>> string = "ee"
=> "ee"
>> !(string =~ /^ee$/).nil?
=> true
Note: Tested in Rails console with ruby (1.8.7) and rails (3.1.1)
So, what you are asking is how to test whether the two strings are equal, right? Just use string equality! This passes every single one of the examples that both you and Tomas cited:
'ee' == 'street' # => false
'ee' == 'ee' # => true
"a\na" == 'a' # => false
"a\na" == "a\na" # => true