Turn number into a fraction in words ruby? - ruby

Is there a way in ruby to turn a Float into a fraction in words?
Such that:
fraction_to_words(1/3r)
=> "one third"
fraction_to_words(5/7r)
=> "five sevenths"
fraction_to_words(2/5r)
=> "two fifths"

Your question is the same as converting any number to a string. Turning a float to a rational is easy (though may give weird results depending on floating point shenanigans). Your problem is the next step - there is no built-in for converting numbers to words. If you do things yourself this might get long:
x = 0.75.to_r
denoms = ['unused', 'whole', 'half', 'third', 'quarter', 'fifth']
numerator = ['zero', 'one', 'two', 'three', 'four', 'five']
rational_name = numerator[x.numerator].to_s() + " " +denoms[x.denominator].to_s
if x.numerator > 1
rational_name += 's'
end
You will have to do a lot of work yourself, or find something someone already did to convert integers to words, such as in Using Ruby convert numbers to words?. There is apparently a module to do this as well: https://github.com/radar/humanize
You would still need to provide your own words for the "special" denominators (two->half, three->third, etc..), and add 'th' (2 elevenths) etc...
For example using that humanize library (never used it, so I can't comment on how robust it is):
rational_name = x.numerator.humanize..to_s() + " "
if x.denominator < 6
rational_name += denoms[x.denominator].to_s()
else
rational_name += x.denominator.humanize + 'th'
end
if x.numerator > 1
rational_name += 's'
end
I missed halves here for example. (has 'halfs' right now for 1.5). Anyway, this is one way to get started.

Related

How would I find an unknown pattern in an array of bytes?

I am building a tool to help me reverse engineer database files. I am targeting my tool towards fixed record length flat files.
What I know:
1) Each record has an index(ID).
2) Each record is separated by a delimiter.
3) Each record is fixed width.
4) Each column in each record is separated by at least one x00 byte.
5) The file header is at the beginning (I say this because the header does not contain the delimiter..)
Delimiters I have found in other files are: ( xFAxFA, xFExFE, xFDxFD ) But this is kind of irrelevant considering that I may use the tool on a different database in the future. So I will need something that will be able to pick out a 'pattern' despite how many bytes it is made of. Probably no more than 6 bytes? It would probably eat up too much data if it was more. But, my experience doing this is limited.
So I guess my question is, how would I find UNKNOWN delimiters in a large file? I feel that given, 'what I know' I should be able to program something, I just dont know where to begin...
# Really loose pseudo code
def begin_some_how
# THIS IS THE PART I NEED HELP WITH...
# find all non-zero non-ascii sets of 2 or more bytes that repeat more than twice.
end
def check_possible_record_lengths
possible_delimiter = begin_some_how
# test if any of the above are always the same number of bytes apart from each other(except one instance, the header...)
possible_records = file.split(possible_delimiter)
rec_length_count = possible_records.map{ |record| record.length}.uniq.count
if rec_length_count == 2 # The header will most likely not be the same size.
puts "Success! We found the fixed record delimiter: #{possible_delimiter}
else
puts "Wrong delimiter found"
end
end
possible = [",", "."]
result = [0, ""]
possible.each do |delimiter|
sizes = file.split( delimiter ).map{ |record| record.size }
next if sizes.size < 2
average = 0.0 + sizes.inject{|sum,x| sum + x }
average /= sizes.size #This should be the record length if this is the right delimiter
deviation = 0.0 + sizes.inject{|sum,x| sum + (x-average)**2 }
matching_value = average / (deviation**2)
if matching_value > result[0] then
result[0] = matching_value
result[1] = delimiter
end
end
Take advantage of the fact that the records have constant size. Take every possible delimiter and check how much each record deviates from the usual record length. If the header is small enough compared rest of the file this should work.

Can someone explain how this ruby basic-calculator code handles addition and subtraction?

So I was working on a Codewars problem here, and found some code posted to Github that works out-of-the-box. Problem is, I don't understand how part of it works. Here are the Codewars directions:
Description:
Create a simple calculator that given a string of operators (+ - * and /) and numbers separated by spaces returns the value of that expression
Example:
Calculator.new.evaluate("2 / 2 + 3 * 4 - 6") # => 7
Remember about the order of operations! Multiplications and divisions have a higher priority and should be performed left-to-right. Additions and subtractions have a lower priority and should also be performed left-to-right.
Here's the code:
class Calculator
def evaluate(string)
operator_stack = []
number_stack = []
string.split(" ").each do |token|
if /\d/.match(token)
number_stack << token.to_i
elsif operator_stack.length > 0 && /[*]|[\/]/.match(operator_stack[-1])
x, y = number_stack.pop, number_stack.pop
temp_result = y.send(operator_stack.pop, x)
number_stack << temp_result
operator_stack << token
else
operator_stack << token
end
end
while(number_stack.length > 0 && operator_stack.length > 0)
x, y = number_stack.shift, number_stack.shift
temp_result = x.send(operator_stack.shift,y)
number_stack.unshift(temp_result)
end
return number_stack[0]
end
end
Now I've learned enough Ruby that I can read through and understand what the various functions do, but when it comes to the mathematical operations the code does, I don't see where or how it handles addition and subtraction. There is some regex that's used to match for multiplication and division present in this line:
elsif operator_stack.length > 0 && /[*]|[\/]/.match(operator_stack[-1])
But since I don't see the plus or minus sign anywhere in the code, I don't get how it performs those operations. Can anyone help?
BTW, I'm done with the Codewars problem and have moved on. I also discovered you can solve this calculator problem with "instance_eval string", which blew my mind when I first saw it. But, it makes sense after reading through what I found here. I should have guessed that there was a one-liner that would work as a basic calculator :)
I would still like to know how this code handles addition and subtraction. Can anyone enlighten me?
The actual operations are performed in these lines:
temp_result = y.send(operator_stack.pop, x)
and later
temp_result = x.send(operator_stack.shift,y)
which says "send the operator_stack.shift/pop message with parameter y to objectx, which is basically the same as doing x <operator> y where <operator> is the operator on top of operator_stack

How do I convert a spreadsheet "letternamed" column coordinate to an integer?

In spreadsheets I have cells named like "F14", "BE5" or "ALL1". I have the first part, the column coordinate, in a variable and I want to convert it to a 0-based integer column index.
How do I do it, preferably in an elegant way, in Ruby?
I can do it using a brute-force method: I can imagine loopping through all letters, converting them to ASCII and adding to a result, but I feel there should be something more elegant/straightforward.
Edit: Example: To simplify I do only speak about the column coordinate (letters). Therefore in the first case (F14) I have "F" as the input and I expect the result to be 5. In the second case I have "BE" as input and I expect getting 56, for "ALL" I want to get 999.
Not sure if this is any clearer than the code you already have, but it does have the advantage of handling an arbitrary number of letters:
class String
def upcase_letters
self.upcase.split(//)
end
end
module Enumerable
def reverse_with_index
self.map.with_index.to_a.reverse
end
def sum
self.reduce(0, :+)
end
end
def indexFromColumnName(column_str)
start = 'A'.ord - 1
column_str.upcase_letters.map do |c|
c.ord - start
end.reverse_with_index.map do |value, digit_position|
value * (26 ** digit_position)
end.sum - 1
end
I've added some methods to String and Enumerable because I thought it made the code more readable, but you could inline these or define them elsewhere if you don't like that sort of thing.
We can use modulo and the length of the input. The last character will
be used to calculate the exact "position", and the remainders to count
how many "laps" we did in the alphabet, e.g.
def column_to_integer(column_name)
letters = /[A-Z]+/.match(column_name).to_s.split("")
laps = (letters.length - 1) * 26
position = ((letters.last.ord - 'A'.ord) % 26)
laps + position
end
Using decimal representation (ord) and the math tricks seems a neat
solution at first, but it has some pain points regarding the
implementation. We have magic numbers, 26, and constants 'A'.ord all
over.
One solution is to give our code better knowlegde about our domain, i.e.
the alphabet. In that case, we can switch the modulo with the position of
the last character in the alphabet (because it's already sorted in a zero-based array), e.g.
ALPHABET = ('A'..'Z').to_a
def column_to_integer(column_name)
letters = /[A-Z]+/.match(column_name).to_s.split("")
laps = (letters.length - 1) * ALPHABET.size
position = ALPHABET.index(letters.last)
laps + position
end
The final result:
> column_to_integer('F5')
=> 5
> column_to_integer('AK14')
=> 36
HTH. Best!
I have found particularly neat way to do this conversion:
def index_from_column_name(colname)
s=colname.size
(colname.to_i(36)-(36**s-1).div(3.5)).to_s(36).to_i(26)+(26**s-1)/25-1
end
Explanation why it works
(warning spoiler ;) ahead). Basically we are doing this
(colname.to_i(36)-('A'*colname.size).to_i(36)).to_s(36).to_i(26)+('1'*colname.size).to_i(26)-1
which in plain English means, that we are interpreting colname as 26-base number. Before we can do it we need to interpret all A's as 1, B's as 2 etc. If only this is needed than it would be even simpler, namely
(colname.to_i(36) - '9'*colname.size).to_i(36)).to_s(36).to_i(26)-1
unfortunately there are Z characters present which would need to be interpreted as 10(base 26) so we need a little trick. We shift every digit 1 more then needed and than add it at the end (to every digit in original colname)
`

String to BigNum and back again (in Ruby) to allow circular shift

As a personal challenge I'm trying to implement the SIMON block cipher in Ruby. I'm running into some issues finding the best way to work with the data. The full code related to this question is located at: https://github.com/Rami114/Personal/blob/master/Simon/Simon.rb
SIMON requires both XOR, shift and circular shift operations, the last of which is forcing me to work with BigNums so I can perform the left circular shift with math rather than a more complex/slower double loop on byte arrays.
Is there a better way to convert a string to a BigNum and back again.
String -> BigNum (where N is 64 and pt is a string of plaintext)
pt = pt.chars.each_slice(N/8).map {|x| x.join.unpack('b*')[0].to_i(2)}.to_a
So I break the string into individual characters, slice into N-sized arrays (the word size in SIMON) and unpack each set into a BigNum. That appears to work fine and I can convert it back.
Now my SIMON code is currently broken, but that's more the math I think/hope and not the code. The conversion back is (where ct is an array of bignums representing the ciphertext):
ct.map { |x| [x.to_s(2).rjust(128,'0')].pack('b*') }.join
I seem to have to right-justify pad the string as bignums are of undefined width so I have no leading 0s. Unfortunately the pack requires the defined with to have sensible output.
Is this a valid method of conversion? Is there a better way? I'm not sure on either count and hoping someone here can help out.
E: For #torimus, the circular shift implementation I'm using (From link above)
def self.lcs (bytes, block_size, shift)
((bytes << shift) | (bytes >> (block_size - shift))) & ((1<< block_size)-1)
end
If you would be equally happy with unpack('B*') with msb first binary numbers (which you could well be if all your processing is circular), then you could also use .unpack('Q>') instead of .unpack('B*')[0].to_i(2) for generating pt:
pt = "qwertyuiopasdfghjklzxcvbnmQWERTYUIOPASDFGHJKLZXCVBNM1234567890!#"
# Your version (with 'B' == msb first) for comparison:
pt_nums = pt.chars.each_slice(N/8).map {|x| x.join.unpack('B*')[0].to_i(2)}.to_a
=> [8176115190769218921, 8030025283835160424, 7668342063789995618, 7957105551900562521,
6145530372635706438, 5136437062280042563, 6215616529169527604, 3834312847369707840]
# unpack to 64-bit unsigned integers directly
pt_nums = pt.unpack('Q>8')
=> [8176115190769218921, 8030025283835160424, 7668342063789995618, 7957105551900562521,
6145530372635706438, 5136437062280042563, 6215616529169527604, 3834312847369707840]
There are no native 128-bit pack/unpacks to return in the other direction, but you can use Fixnum to solve this too:
split128 = 1 << 64
ct = pt # Just to show round-trip
ct.map { |x| [ x / split128, x % split128 ].pack('Q>2') }.join
=> "\x00\x00\x00\x00\x00\x00\x00\x00qwertyui . . . " # truncated
This avoids a lot of the temporary stages on your code, but at the expense of using a different byte coding - I don't know enough about SIMON to comment whether this is adaptable to your needs.

Ruby, Count syllables

I am using ruby to calculate the Gunning Fog Index of some content that I have, I can successfully implement the algorithm described here:
Gunning Fog Index
I am using the below method to count the number of syllables in each word:
Tokenizer = /([aeiouy]{1,3})/
def count_syllables(word)
len = 0
if word[-3..-1] == 'ing' then
len += 1
word = word[0...-3]
end
got = word.scan(Tokenizer)
len += got.size()
if got.size() > 1 and got[-1] == ['e'] and
word[-1].chr() == 'e' and
word[-2].chr() != 'l' then
len -= 1
end
return len
end
It sometimes picks up words with only 2 syllables as having 3 syllables. Can anyone give any advice or is aware of a better method?
text = "The word logorrhoea is often used pejoratively to describe prose that is highly abstract and contains little concrete language. Since abstract writing is hard to visualize, it often seems as though it makes no sense and all the words are excessive. Writers in academic fields that concern themselves mostly with the abstract, such as philosophy and especially postmodernism, often fail to include extensive concrete examples of their ideas, and so a superficial examination of their work might lead one to believe that it is all nonsense."
# used to get rid of any puncuation
text = text.gsub!(/\W+/, ' ')
word_array = text.split(' ')
word_array.each do |word|
puts word if count_syllables(word) > 2
end
"themselves" is being counted as 3 but it's only 2
The function I give you before is based upon these simple rules outlined here:
Each vowel (a, e, i, o, u, y) in a
word counts as one syllable subject to
the following sub-rules:
Ignore final -ES, -ED, -E (except
for -LE)
Words of three letters or
less count as one syllable
Consecutive vowels count as one
syllable.
Here's the code:
def new_count(word)
word.downcase!
return 1 if word.length <= 3
word.sub!(/(?:[^laeiouy]es|ed|[^laeiouy]e)$/, '')
word.sub!(/^y/, '')
word.scan(/[aeiouy]{1,2}/).size
end
Obviously, this isn't perfect either, but all you'll ever get with something like this is a heuristic.
EDIT:
I changed the code slightly to handle a leading 'y' and fixed the regex to handle 'les' endings better (such as in "candles").
Here's a comparison using the text in the question:
# used to get rid of any puncuation
text = text.gsub!(/\W+/, ' ')
words = text.split(' ')
words.each do |word|
old = count_syllables(word.dup)
new = new_count(word.dup)
puts "#{word}: \t#{old}\t#{new}" if old != new
end
The output is:
logorrhoea: 3 4
used: 2 1
makes: 2 1
themselves: 3 2
So it appears to be an improvement.
One thing you ought to do is teach your algorithm about diphthongs. If I'm reading your code correctly, it would incorrectly flag "aid" as having two syllables.
You can also add "es" and the like to your special-case endings (you already have "ing") and just not count it as a syllable, but that might still result in some miscounts.
Finally, for best accuracy, you should convert your input to a spelling scheme or alphabet that has a definite relationship to the word's pronunciation. With your "themselves" example, the algorithm has no reliable way to know that the "e" "ves" is dropped. However, if you respelled it as "themselvz", or taught the algorithm the IPA and fed it [ðəmsɛlvz], it becomes very clear that the word is only pronounced with two syllables. That, of course, assumes you have control over the input, and is probably more work than just counting the syllables yourself.
To begin with it seems you should decrement len for the suffixes that should be excluded.
len-=1 if /.*[ing,es,ed]$/.match(word)
You could also check out Lingua::EN::Readability.
It can also calculate several readability measures, such as a Fog Index and a Flesch-Kincaid level.
PS. I think I know where you got the function from. DS.
There is also a rubygem called Odyssey that calculates Gunning Fog, along with some of the other popular ones (Flesch-Kincaid, SMOG, etc.)

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