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Without getting too much into biology, Proteins are made of Amino Acids. Each of the 20 Amino Acids that make up Proteins are represented by characters in a sequence. Each Amino Acid char has a different chemical formula, which I represent as strings. For example, "M" has a formula of "C5H11NO2S"
Given the 20 different formulas (and the varying frequency of each amino acid chars in a protein sequence) I want to compile all 20 of them into a single formula that will yield the total formula for the protein.
So first: multiply each formula by the frequency of its char in the sequence
Second : sum together all multiplied formulas into one formula.
To accomplish this, I first tried multiplying each amino acid char frequency in the sequence by the numbers in the chemical formula. I did this using .tally
sequence ="MGAAARTLRLALGLLLLATLLRPADACSCSPVHPQQAFCNADVVIRAKAVSEKEVDSGNDIYGNPIKRIQYEIKQIKMFKGPEKDIEFI"
sequence.chars.string.tally --> {"M"=>2, "G"=>5, "A"=>11, "R"=>5, "T"=>2, "L"=>9, "P"=>5, "D"=>5, "C"=>3, "S"=>4, "V"=>5, "H"=>1, "Q"=>4, "F"=>3, "N"=>3, "I"=>8, "K"=>7, "E"=>5, "Y"=>2}
Then, I listed all the amino acids chars and formulas into a hash
hash_of_formulas = {"A"=>"C3H7NO2", "R"=>"C6H14N4O2", "N"=>"C4H8N2O3", "D"=>"C4H7NO4", "C"=>"C3H7NO2S", "E"=>"C5H9NO4", "Q"=>"C5H10N2O3", "G"=>"C2H5NO2", "H"=>"C6H9N3O2", "I"=>"C6H13NO2", "L"=>"C6H13NO2", "K"=>"C6H14N2O2", "M"=>"C5H11NO2S", "F"=>"C9H11NO2", "P"=>"C5H9NO2", "S"=>"C3H7NO3", "T"=>"C4H9NO3", "W"=>"C11H12N2O2", "Y"=>"C9H11NO3", "V"=>"C5H11NO2"}
An example of what the process for my overall goal is:
In the sequence , "M" occurs twice so "C5H11NO2S" will become "C10H22N2O4S2". "C" has a formula of "C3H7NO2S" occurs 3 times: In the sequence so "C3H7NO2S" becomes "C9H21N3O6S3"
So, Summing together "C10H22N2O4S2" and "C9H21N3O6S3" will yield "C19H43N5O10S5"
How can I repeat the process of multiplying each formula by its frequency and then summing together all multiplied formulas?
I know that I could use regex for multiplying a formula by its frequency for an individual string using
formula_multiplied_by_frequency = "C5H11NO2S".gsub(/\d+/) { |x| x.to_i * 4}
But I'm not sure of any methods to use regex on strings embedded within hashes
If I understand correctly, you want the to provide the total formula for a given protein sequence. Here's how I'd do it:
NUCLEOTIDES = {"A"=>"C3H7NO2", "R"=>"C6H14N4O2", "N"=>"C4H8N2O3", "D"=>"C4H7NO4", "C"=>"C3H7NO2S", "E"=>"C5H9NO4", "Q"=>"C5H10N2O3", "G"=>"C2H5NO2", "H"=>"C6H9N3O2", "I"=>"C6H13NO2", "L"=>"C6H13NO2", "K"=>"C6H14N2O2", "M"=>"C5H11NO2S", "F"=>"C9H11NO2", "P"=>"C5H9NO2", "S"=>"C3H7NO3", "T"=>"C4H9NO3", "W"=>"C11H12N2O2", "Y"=>"C9H11NO3", "V"=>"C5H11NO2"}
NUCLEOTIDE_COMPOSITIONS = NUCLEOTIDES.each_with_object({}) { |(nucleotide, formula), compositions|
compositions[nucleotide] = formula.scan(/([A-Z][a-z]*)(\d*)/).map { |element, count| [element, count.empty? ? 1 : count.to_i] }.to_h
}
def formula(sequence)
sequence.each_char.with_object(Hash.new(0)) { |nucleotide, final_counts|
NUCLEOTIDE_COMPOSITIONS[nucleotide].each { |element, element_count|
final_counts[element] += element_count
}
}.map { |element, element_count|
"#{element}#{element_count.zero? ? "" : element_count}"
}.join
end
sequence = "MGAAARTLRLALGLLLLATLLRPADACSCSPVHPQQAFCNADVVIRAKAVSEKEVDSGNDIYGNPIKRIQYEIKQIKMFKGPEKDIEFI"
p formula(sequence)
# => "C434H888N51O213S"
You can't use regexp to multiply things. You can use it to parse a formula, but then it's on you and regular Ruby to do the math. The first job is to prepare a composition lookup by breaking down each nucleotide formula. Once we have a composition hash for each nucleotide, we can iterate over a nucleotide sequence, and add up all the elements of each nucleotide.
BTW, tally is not particularly useful here, since tally will need to iterate over the sequence, and then you have to iterate over tally anyway — and there is no aggregate operation going on that can't be done going over each letter independently.
EDIT: I probably made the regexp slightly more complicated that it needs to be, but it should parse stuff like CuSO4 correctly. I don't know if it's an accident or not that all nucleotides are only composed of elements with a single-character symbol... :P )
Givens
We are given a string representing a protein comprised of amino acids:
sequence = "MGAAARTLRLALGLLLLATLLRPADACSCSPVHPQQAFCNADVVIR" +
"AKAVSEKEVDSGNDIYGNPIKRIQYEIKQIKMFKGPEKDIEFI"
and a hash that contains the formulas of amino acids:
formulas = {
"A"=>"C3H7NO2", "R"=>"C6H14N4O2", "N"=>"C4H8N2O3", "D"=>"C4H7NO4",
"C"=>"C3H7NO2S", "E"=>"C5H9NO4", "Q"=>"C5H10N2O3", "G"=>"C2H5NO2",
"H"=>"C6H9N3O2", "I"=>"C6H13NO2", "L"=>"C6H13NO2", "K"=>"C6H14N2O2",
"M"=>"C5H11NO2S", "F"=>"C9H11NO2", "P"=>"C5H9NO2", "S"=>"C3H7NO3",
"T"=>"C4H9NO3", "W"=>"C11H12N2O2", "Y"=>"C9H11NO3", "V"=>"C5H11NO2"
}
Obtain counts of atoms in each amino acid
As a first step we can calculate the numbers of each atom in each amino acid:
counts = formulas.transform_values do |s|
s.scan(/[CHNOS]\d*/).
each_with_object({}) do |s,h|
h[s[0]] = s.size == 1 ? 1 : s[1..-1].to_i
end
end
#=> {"A"=>{"C"=>3, "H"=>7, "N"=>1, "O"=>2},
# "R"=>{"C"=>6, "H"=>14, "N"=>4, "O"=>2},
# ...
# "M"=>{"C"=>5, "H"=>11, "N"=>1, "O"=>2, "S"=>1}
# ...
# "V"=>{"C"=>5, "H"=>11, "N"=>1, "O"=>2}}
Compute formula for protein
Then it's simply:
def protein_formula(sequence, counts)
sequence.each_char.
with_object("C"=>0, "H"=>0, "N"=>0, "O"=>0, "S"=>0) do |c,h|
counts[c].each { |aa,cnt| h[aa] += cnt }
end.each_with_object('') { |(aa,nbr),s| s << "#{aa}#{nbr}" }
end
protein_formula(sequence, counts)
#=> "C434H888N120O213S5"
Another example:
protein_formula("MCMPCFTTDHQMARKCDDCCGGKGRGKCYGPQCLCR", count)
#=> "C158H326N52O83S11"
Explanation of calculation of counts
This calculation:
counts = formulas.transform_values do |s|
s.scan(/[CHNOS]\d*/).each_with_object({}) do |s,h|
h[s[0]] = s.size == 1 ? 1 : s[1..-1].to_i
end
end
uses the method Hash#transform_values. It will return a hash having the same keys as the hash formulas, with the values of those keys in formula modified by transform_values's block. For example, formulas["A"] ("C3H7NO2") is "transformed" to the hash {"C"=>3, "H"=>7, "N"=>1, "O"=>2} in the hash that is returned, counts.
transform_values passes each value of formulas to the block and sets the block variable equal to it. The first value passed is "C3H7NO2", so it sets:
s = "C3H7NO2"
We can write the block calculation more simply:
h = {}
s.scan(/[CHNOS]\d*/).each do |s|
h[s[0]] = s.size == 1 ? 1 : s[1..-1].to_i
end
h
(Once you understand this calculation, which I explain below, see Enumerable#each_with_object to understand why I used that method in my solution.)
After initializing h to an empty hash, the following calculations are performed:
h = {}
a = s.scan(/[CHNOS]\d*/)
#=> ["C3", "H7", "N", "O2"]
a is computed using String#scan with the regular expression /[CHNOS]\d*/. That regular expression, or regex, matches exactly one character in the character class [CHNOS] followed by zero of more (*) digits (\d). It therefore separates the string "C3H7NO2" into the substrings that are returned in the array shown under the calculation of a above . Continuing,
a.each do |s|
h[s[0]] = s.size == 1 ? 1 : s[1..-1].to_i
end
changes h to the following:
h #=> {"C"=>3, "H"=>7, "N"=>1, "O"=>2}
The block variable s is initially set equal to the first element of a that is passed to each's block:
s = "C3"
then we compute:
h[s[0]] = s.size == 1 ? 1 : s[1..-1].to_i
h["A"] = 2 == 1 ? 1 : "3".to_i
= false ? 1 : 3
3
This is repeated for each element of a.
Exclamation of construction of formula for the protein
We can simplify the following code1:
sequence.each_char.with_object("C"=>0, "H"=>0, "N"=>0, "O"=>0) do |c,h|
counts[c].each { |aa,cnt| h[aa] += cnt }
end.each_with_object('') { |(aa,nbr),s| s << "#{aa}#{nbr}" }
to more or less the following:
h = { "C"=>0, "H"=>0, "N"=>0, "O"=>0, "S"=>0 }
ch = sequence.chars
#=> ["M", "G", "A",..., "F", "I"]
ch.each do |c|
counts[c].each { |aa,cnt| h[aa] += cnt }
end
h #=> {"C"=>434, "H"=>888, "N"=>120, "O"=>213, "S"=>5}
When the first value of ch ("M") is passed to each's block (when h = { "C"=>0, "H"=>0, "N"=>0, "O"=>0, "S"=>0 }), the following calculations are performed:
c = "M"
g = counts[c]
#=> {"C"=>10, "H"=>22, "N"=>2, "O"=>4, "S"=>1}
g.each { |aa,cnt| h[aa] += cnt }
h #=> {"C"=>10, "H"=>22, "N"=>2, "O"=>4, "S"=>1}
Lastly, (when h #=> {"C"=>434, "H"=>888, "N"=>120, "O"=>213, "S"=>5})
s = ''
h.each { |aa,nbr| s << "#{aa}#{nbr}" }
s #=> "C434H888N120O213S5"
When aa = "C" and nbr = 434,
"#{aa}#{nbr}"
#=> "C434"
is appended to the string s.
1. (("C"=>0, "H"=>0, "N"=>0, "O"=>0) is shorthand for ({"C"=>0, "H"=>0, "N"=>0, "O"=>0}).
I've tried different ways and this is probably the closest that I got to it. I am trying to write a method that takes in an array of strings and returns it containing the strings that are at least 5 characters long and end with "y".
I'm a beginner and this is my second problem I've come across with, and I've tried multiple if statements and using a while loop, however I could not get to it and now this is where I am at. Thank you!
def phrases(arr1, arr2)
arr1 = ["funny", "tidy", "fish", "foogiliously"]
arr2 = ["happily", "lovely", "hello", "multivitaminly"]
if (arr1.length > 5 && arr1.length == "y")
return arr1
elsif (arr2.length > 5 && arr2.length == "y")
return arr2
end
end
puts phrases(["funny", "tidy", "fish", "foogiliously"])
puts phrases(["happily", "lovely", "hello", "multivitaminly"])
If I'm understanding your question correctly, you want to return a subset of the passed in array matching your conditions (length ≥ 5 and last character = 'y'). In that case:
def phrases(words)
words.grep(/.{4}y\z/)
end
What that regex does:
.{4} means 4 of any character
y is the letter y
\z is the end of the string, so we don't match in the middle of a long word
The docs for Enumerable#select are here (an Array is an Enumerable).
Output:
> phrases(["funny", "tidy", "fish", "foogiliously"])
=> ["funny", "foogiliously"]
> phrases(["happily", "lovely", "hello", "multivitaminly"])
=> ["happily", "lovely", "multivitaminly"]
If you only want word characters, rather than any character, you'd use this regex instead: /\A.{4,}y\z/. In that case, \A means the start of the string, and \w{4,} means at least 4 word characters.
If, when given an array and inclusion criterion, one wishes to construct an array that contains those elements of the first array that satisfy the inclusion criterion, one generally uses the method Array#select or Array#reject, whichever is more more convenient.
Suppose arr is a variable that holds the given array and include_element? is a method that takes one argument, an element of arr, and returns true or false, depending on whether the inclusion criterion is satisified for that element. For example, say the array comprises the integers 1 through 6 and the inclusion criterion is that the number is even (2, 4 and 6). We could write:
arr = [1,2,3,4,5,6]
def include_element?(e)
e.even?
end
include_element?(2)
#=> true
include_element?(3)
#=> false
arr.select { |e| include_element?(e) }
#=> [2, 4, 6]
The method include_element? is so short we probably would substitute it out and just write:
arr.select { |e| e.even? }
Array#select passes each element of its receiver, arr, to select's block, assigns the block variable e to that value and evaluates the expression in the block (which could be many lines, of course). Here that expresssion is just e.even?, which returns true or false. (See Integer#even? and Integer#odd?.)
If that expression evaluates as a truthy value, the element e is to be included in the array that is returned; if it evaluates as a falsy value, e is not to be included. Falsy values (logical false) are nil and false; truthy values (logical true) are all other Ruby objects, which of course includes true.
Notice that we could instead write:
arr.reject { |e| e.odd? }
Sometimes the inclusion criterion consists of a compound expression. For example, suppose the inclusion criterion were to keep elements of arr that are both even numbers and are at least 4. We would write:
arr.select { |e| e.even? && e >= 4 }
#=> [4, 6]
With other criteria we might write:
arr.select { |e| e.even? || e >= 4 }
#=> [2, 4, 5, 6]
or
arr.select { |e| e < 2 || (e > 3 && e < 6) }
#=> [1, 4, 5]
&& (logical 'and') and || (logical 'or') are operators (search "operator expressions"). As explained at the link, most Ruby operators are actually methods, but these two are among a few that are not.
Your problem now reduces to the following:
arr.select { |str| <length of str is at least 5> && <last character of str is 'y'> }
You should be able to supply code for the <...> bits.
You are trying to write a function that should work on a single array at a time I think. Also, you are taking in an array, and retaining only those elements that satisfy your conditions: at least 5 characters long, and ends with y. This is a filtering operation. Read about the methods available for ruby's Array class here
def phrases(array)
...
filtered_array
end
Now the condition you are using is this arr1.length > 5 && arr1.length == "y".
The first half should check if the string length is greater than 5, not the array length itself. The second half is an indexing operation, and your code for that is incorrect. basically you are checking if the last character in the string is y.
Usually strings are indexed in this manner: string[index]. In your case you can use string[-1]=='y' or string[string.length - 1]=='y'. This because arrays and strings are zero indexed in ruby. The first element has index of 0, the second has an index of 1, and the last one, therefore, will have an index of length-1. If you use negative indexes then the array is indexed from the end, so string[-1] is a quick way to get to the last element.
Considering this, the function will take the following structure:
def phrases(array)
filtered_array = [] # an empty array
loop through the input array
for each element check for the condition element.length > 5 && element[-1]=='y'
if true: push the element into the filtered_array
once the loop is done, return the filtered array
end
Read about ruby arrays, the methods push, filter and select in the above linked documentation to get a better idea. I'd also recommend the codeacademy ruby tutorial.
Edit: Both halves of the condition are incorrect. I had overlooked a mistake in my earlier answer. arr1.length refers to the length of the array. You want to check the length of each string in the array. So in your for loop you should check the length of the loop variable, if that is greater than 5.
You may want to spend some time reading about the methods in the core library, especially String#end_with? and Enumerable#select. You could then write a method that'd contain something like this:
['abc', 'qwerty', 'asdfghjk', 'y'].select{|s| s.length >= 5}.select{|s| s.end_with? 'y'}
#=> ["qwerty"]
I'm creating an array of permutated and unique letters in a string, only to sort them alphabetically and find the middle element in the set.
def middle_permutation(string)
length = string.length
permutation_set = string.split("").permutation(length).to_a.map{|item| item.join}.sort
permutation_set.length.even? ? permutation_set[(permutation_set.length)/2-1] : permutation_set[(permutation_set.length/2)+1]
end
For example:
middle_permutation("zxcvbnmasd") should equal "mzxvsndcba"
Even for small strings (N >=10), the calculations take pretty long to finish, and I can forget about anything double that; is there a quicker way?
I'm assuming the letters are unique, as in the OP's question.
Sort
Pluck the middle letter of the sorted string (rounded down). This is the first letter of the middle permutation.
If the original list had an even number of letters, the rest of the permutation is the reverse sort of the remaining letters.
If not, take the middle letter again. Now the rest of the result is the reverse sort of the remaining letters.
The method below returns the desired permutation directly, without iterating through permutations.
The asker has stated that the string contains no duplicated letters, which is a requirement for this method. I assume the characters of the string are sorted. If they are not, the creation of a sorted string would be the first step:
str = "ebadc".chars.sort.join
#=> "abcde"
Code
def mid_perm(str)
return mid_perm_even_length_strings(str) if str.size.even?
first_char_index = str.size/2
str[first_char_index] << mid_perm_even_length_strings(str[0,first_char_index] +
str[first_char_index+1..-1])
end
def mid_perm_even_length_strings(str)
first_char_index = str.size/2-1
str[first_char_index] + (str[0,first_char_index] + str[first_char_index+1..-1]).reverse
end
Examples
mid_perm 'abcd'
#=> "bdca"
mid_perm 'abcde'
#=> "cbeda"
mid_perm 'abcdefghijklmnopqrstuvwxyz'
#=> "mzyxwvutsrqponlkjihgfedcba"
Explanation
Let's start by defining a method to produce permutations of the letters of a string.
def perms(str)
str.chars.permutation(str.size).map(&:join)
end
Strings containing an even number of characters
Consider
a = perms "abcd"
#=> ["abcd", "abdc", "acbd", "acdb", "adbc", "adcb",
# "bacd", "badc", "bcad", "bcda", "bdac", "bdca",
# "cabd", "cadb", "cbad", "cbda", "cdab", "cdba",
# "dabc", "dacb", "dbac", "dbca", "dcab", "dcba"]
a contains 4! #=> 4*3*2 => 24 elements, 4 being the length of the string.
Notice that since the characters in perms' argument are sorted, the array returned is also sorted1.
a == a.sort #=>true
As a.size #=> 24, the "middle" element is either a[11] #=> "bdca" or a[12] #=> "cabd" (where 11 = (24-1)/2 and 12 = 24/2), depending on how we want to round. The question stipulates that, for even-length strings, we are to round down, so that would be "bdca".
Now let's slice a into str.size equal arrays, each containing a.size/str.size #=> 24/4 => 6 elements:
b = a.each_slice(a.size/str.size).to_a
#=> [["abcd", "abdc", "acbd", "acdb", "adbc", "adcb"],
# ["bacd", "badc", "bcad", "bcda", "bdac", "bdca"],
# ["cabd", "cadb", "cbad", "cbda", "cdab", "cdba"],
# ["dabc", "dacb", "dbac", "dbca", "dcab", "dcba"]]
The desired element is therefore
b[(a.size/str.size-1)/2-1][-1]
#=> "bdca"
This value can be computed more directly as follows.
first_char_index = str.size/2-1
#=> 1
first_char = str[first_char_index]
#=> "b"
remaining_chars = (str[0,first_char_index] + str[first_char_index+1..-1]).reverse
#=> "dca"
first_char + remaining_chars
#=> "bdca"
The same logic applies to all strings having an even number of characters. We therefore can write the method mid_perm_even_length_strings shown in the Code section above.
For example (for a 12-character string)
mid_perm_even_length_strings 'abcdefghijkl'
#=> "flkjihgedcba"
Strings containing an odd number of characters
Now consider
str = "abcde"
a = perms str
#=> ["abcde", "abced", "abdce", "abdec", "abecd", "abedc",
# "acbde", "acbed", "acdbe", "acdeb", "acebd", "acedb",
# "adbce", "adbec", "adcbe", "adceb", "adebc", "adecb",
# "aebcd", "aebdc", "aecbd", "aecdb", "aedbc", "aedcb",
# "bacde", "baced", "badce", "badec", "baecd",..., "bedca",
# "cabde", "cabed", "cadbe", "cadeb", "caebd", "caedb",
# "cbade", "cbaed", "cbdae", "cbdea", "cbead", "cbeda",
# "cdabe", "cdaeb", "cdbae", "cdbea", "cdeab", "cdeba",
# "ceabd", "ceadb", "cebad", "cebda", "cedab", "cedba",
# "dabce", "dabec", "dacbe", "daceb", "daebc",..., "decba",
# "eabcd", "eabdc", "eacbd", "eacdb", "eadbc",..., "edcba"]
Here the permutation contains 5! #=> 100 elements, in 5 blocks of 20. (Again, a.each_cons(2).all? { |s1,s2| s1 < s2 } #=> true.)
The middle element of a is clearly the middle element of the block of elements that begin with
str[str.size/2] #=> "c"
That block would be the array
b = a.each_slice(a.size/str.size).to_a[str.size/2]
#=> ["cabde", "cabed", "cadbe", "cadeb", "caebd", "caedb",
# "cbade", "cbaed", "cbdae", "cbdea", "cbead", "cbeda",
# "cdabe", "cdaeb", "cdbae", "cdbea", "cdeab", "cdeba",
# "ceabd", "ceadb", "cebad", "cebda", "cedab", "cedba"]
which would be 'c' plus the middle element of the array
["abde", "abed", "adbe", "adeb", "aebd", "aedb",
"bade", "baed", "bdae", "bdea", "bead", "beda",
"dabe", "daeb", "dbae", "dbea", "deab", "deba",
"eabd", "eadb", "ebad", "ebda", "edab", "edba"]
That array is merely the permutations of the string "abde". Since that string contains an even number characters, its middle element is
mid_perm_even_length_strings 'abde'
#=> "beda"
It follows that the middle element of the permutations of the letters of "abcde" is therefore
'c' + 'abde'
#=> "cabde"
This clearly applies to all strings containing an odd number of characters.
1. The doc for Array#permutation states, "The implementation makes no guarantees about the order in which the permutations are yielded.". We therefore might need to tack .sort to the end of the operative line of perms, but with Ruby v2.4 (and I suspect, earlier versions) that is, in fact not necessary here.
I was able to compact it like this:
def middle_permutation(string)
list = string.chars.permutation.map(&:join).sort
list[list.length / 2 - (list.length.even? ? 1 : 0)]
end
Which yields:
middle_permutation('zxcvbnmasd')
# => "mzxvsndcba"
You don't need to generate all permutations. Just find overall number of permutations as PN = N! where N is string (of different chars) length and calculate only needed PN/2-th permutation by its number - for example, using this approach
public static int[] perm(int n, int k)
{
int i, ind, m=k;
int[] permuted = new int[n];
int[] elems = new int[n];
for(i=0;i<n;i++) elems[i]=i;
for(i=0;i<n;i++)
{
ind=m%(n-i);
m=m/(n-i);
permuted[i]=elems[ind];
elems[ind]=elems[n-i-1];
}
return permuted;
}
So it turns out there are two tracks to this, odd strings and even strings.
For odd strings, you take out the middle character Element of the sorted array and the one before it, in that order. When you do that you have two remaining arrays, the one the right and left, both alphabetically sorted. You tack on elements of the right array, starting with the last element, then do the same for the one on the left.
For even strings, Do the same but only take one character in the first step: the (N/2) element.
Here's my solution:
def middle_permutation(string)
string_array = string.chars.sort
mid_string = []
length = string.length
if length.even?
mid_string << string_array[length/2-1]
string_array.delete_at(length/2-1)
(mid_string << string_array.reverse).flatten.join
else
mid_string << string_array[(length/2)-1..length/2].reverse
string_array.slice!((length/2)-1, 2)
(mid_string << string_array.reverse).flatten.join
end
end
I was going through the exercises in Ruby Koans and I was struck by the following Ruby quirk that I found really unexplainable:
array = [:peanut, :butter, :and, :jelly]
array[0] #=> :peanut #OK!
array[0,1] #=> [:peanut] #OK!
array[0,2] #=> [:peanut, :butter] #OK!
array[0,0] #=> [] #OK!
array[2] #=> :and #OK!
array[2,2] #=> [:and, :jelly] #OK!
array[2,20] #=> [:and, :jelly] #OK!
array[4] #=> nil #OK!
array[4,0] #=> [] #HUH?? Why's that?
array[4,100] #=> [] #Still HUH, but consistent with previous one
array[5] #=> nil #consistent with array[4] #=> nil
array[5,0] #=> nil #WOW. Now I don't understand anything anymore...
So why is array[5,0] not equal to array[4,0]? Is there any reason why array slicing behaves this weird when you start at the (length+1)th position??
Slicing and indexing are two different operations, and inferring the behaviour of one from the other is where your problem lies.
The first argument in slice identifies not the element but the places between elements, defining spans (and not elements themselves):
:peanut :butter :and :jelly
0 1 2 3 4
4 is still within the array, just barely; if you request 0 elements, you get the empty end of the array. But there is no index 5, so you can't slice from there.
When you do index (like array[4]), you are pointing at elements themselves, so the indices only go from 0 to 3.
this has to do with the fact that slice returns an array, relevant source documentation from Array#slice:
* call-seq:
* array[index] -> obj or nil
* array[start, length] -> an_array or nil
* array[range] -> an_array or nil
* array.slice(index) -> obj or nil
* array.slice(start, length) -> an_array or nil
* array.slice(range) -> an_array or nil
which suggests to me that if you give the start that is out of bounds, it will return nil, thus in your example array[4,0] asks for the 4th element that exists, but asks to return an array of zero elements. While array[5,0] asks for an index out of bounds so it returns nil. This perhaps makes more sense if you remember that the slice method is returning a new array, not altering the original data structure.
EDIT:
After reviewing the comments I decided to edit this answer. Slice calls the following code snippet when the arg value is two:
if (argc == 2) {
if (SYMBOL_P(argv[0])) {
rb_raise(rb_eTypeError, "Symbol as array index");
}
beg = NUM2LONG(argv[0]);
len = NUM2LONG(argv[1]);
if (beg < 0) {
beg += RARRAY(ary)->len;
}
return rb_ary_subseq(ary, beg, len);
}
if you look in the array.c class where the rb_ary_subseq method is defined, you see that it is returning nil if the length is out of bounds, not the index:
if (beg > RARRAY_LEN(ary)) return Qnil;
In this case this is what is happening when 4 is passed in, it checks that there are 4 elements and thus does not trigger the nil return. It then goes on and returns an empty array if the second arg is set to zero. while if 5 is passed in, there are not 5 elements in the array, so it returns nil before the zero arg is evaluated. code here at line 944.
I believe this to be a bug, or at least unpredictable and not the 'Principle of Least Surprise'. When I get a few minutes I will a least submit a failing test patch to ruby core.
At least note that the behavior is consistent. From 5 on up everything acts the same; the weirdness only occurs at [4,N].
Maybe this pattern helps, or maybe I'm just tired and it doesn't help at all.
array[0,4] => [:peanut, :butter, :and, :jelly]
array[1,3] => [:butter, :and, :jelly]
array[2,2] => [:and, :jelly]
array[3,1] => [:jelly]
array[4,0] => []
At [4,0], we catch the end of the array. I'd actually find it rather odd, as far as beauty in patterns go, if the last one returned nil. Because of a context like this, 4 is an acceptable option for the first parameter so that the empty array can be returned. Once we hit 5 and up, though, the method likely exits immediately by nature of being totally and completely out of bounds.
This makes sense when you consider than an array slice can be a valid lvalue, not just an rvalue:
array = [:peanut, :butter, :and, :jelly]
# replace 0 elements starting at index 5 (insert at end or array):
array[4,0] = [:sandwich]
# replace 0 elements starting at index 0 (insert at head of array):
array[0,0] = [:make, :me, :a]
# array is [:make, :me, :a, :peanut, :butter, :and, :jelly, :sandwich]
# this is just like replacing existing elements:
array[3, 4] = [:grilled, :cheese]
# array is [:make, :me, :a, :grilled, :cheese, :sandwich]
This wouldn't be possible if array[4,0] returned nil instead of []. However, array[5,0] returns nil because it's out of bounds (inserting after the 4th element of a 4-element array is meaningful, but inserting after the 5th element of a 4 element array is not).
Read the slice syntax array[x,y] as "starting after x elements in array, select up to y elements". This is only meaningful if array has at least x elements.
This does make sense
You need to be able to assign to those slices, so they are defined in such a way that the beginning and the end of the string have working zero-length expressions.
array[4, 0] = :sandwich
array[0, 0] = :crunchy
=> [:crunchy, :peanut, :butter, :and, :jelly, :sandwich]
I found explanation by Gary Wright very helpful as well.
http://www.ruby-forum.com/topic/1393096#990065
The answer by Gary Wright is -
http://www.ruby-doc.org/core/classes/Array.html
The docs certainly could be more clear but the actual behavior is
self-consistent and useful.
Note: I'm assuming 1.9.X version of String.
It helps to consider the numbering in the following way:
-4 -3 -2 -1 <-- numbering for single argument indexing
0 1 2 3
+---+---+---+---+
| a | b | c | d |
+---+---+---+---+
0 1 2 3 4 <-- numbering for two argument indexing or start of range
-4 -3 -2 -1
The common (and understandable) mistake is too assume that the semantics
of the single argument index are the same as the semantics of the
first argument in the two argument scenario (or range). They are not
the same thing in practice and the documentation doesn't reflect this.
The error though is definitely in the documentation and not in the
implementation:
single argument: the index represents a single character position
within the string. The result is either the single character string
found at the index or nil because there is no character at the given
index.
s = ""
s[0] # nil because no character at that position
s = "abcd"
s[0] # "a"
s[-4] # "a"
s[-5] # nil, no characters before the first one
two integer arguments: the arguments identify a portion of the string to
extract or to replace. In particular, zero-width portions of the string
can also be identified so that text can be inserted before or after
existing characters including at the front or end of the string. In this
case, the first argument does not identify a character position but
instead identifies the space between characters as shown in the diagram
above. The second argument is the length, which can be 0.
s = "abcd" # each example below assumes s is reset to "abcd"
To insert text before 'a': s[0,0] = "X" # "Xabcd"
To insert text after 'd': s[4,0] = "Z" # "abcdZ"
To replace first two characters: s[0,2] = "AB" # "ABcd"
To replace last two characters: s[-2,2] = "CD" # "abCD"
To replace middle two characters: s[1..3] = "XX" # "aXXd"
The behavior of a range is pretty interesting. The starting point is the
same as the first argument when two arguments are provided (as described
above) but the end point of the range can be the 'character position' as
with single indexing or the "edge position" as with two integer
arguments. The difference is determined by whether the double-dot range
or triple-dot range is used:
s = "abcd"
s[1..1] # "b"
s[1..1] = "X" # "aXcd"
s[1...1] # ""
s[1...1] = "X" # "aXbcd", the range specifies a zero-width portion of
the string
s[1..3] # "bcd"
s[1..3] = "X" # "aX", positions 1, 2, and 3 are replaced.
s[1...3] # "bc"
s[1...3] = "X" # "aXd", positions 1, 2, but not quite 3 are replaced.
If you go back through these examples and insist and using the single
index semantics for the double or range indexing examples you'll just
get confused. You've got to use the alternate numbering I show in the
ascii diagram to model the actual behavior.
I agree that this seems like strange behavior, but even the official documentation on Array#slice demonstrates the same behavior as in your example, in the "special cases" below:
a = [ "a", "b", "c", "d", "e" ]
a[2] + a[0] + a[1] #=> "cab"
a[6] #=> nil
a[1, 2] #=> [ "b", "c" ]
a[1..3] #=> [ "b", "c", "d" ]
a[4..7] #=> [ "e" ]
a[6..10] #=> nil
a[-3, 3] #=> [ "c", "d", "e" ]
# special cases
a[5] #=> nil
a[5, 1] #=> []
a[5..10] #=> []
Unfortunately, even their description of Array#slice doesn't seem to offer any insight as to why it works this way:
Element Reference—Returns the element at index, or returns a subarray starting at start and continuing for length elements, or returns a subarray specified by range. Negative indices count backward from the end of the array (-1 is the last element). Returns nil if the index (or starting index) are out of range.
An explanation provided by Jim Weirich
One way to think about it is that index position 4 is at the very edge
of the array. When asking for a slice, you return as much of the
array that is left. So consider the array[2,10], array[3,10] and
array[4,10] ... each returns the remaining bits of the end of the
array: 2 elements, 1 element and 0 elements respectively. However,
position 5 is clearly outside the array and not at the edge, so
array[5,10] returns nil.
Consider the following array:
>> array=["a","b","c"]
=> ["a", "b", "c"]
You can insert an item to the begining (head) of the array by assigning it to a[0,0]. To put the element between "a" and "b", use a[1,0]. Basically, in the notation a[i,n], i represents an index and n a number of elements. When n=0, it defines a position between the elements of the array.
Now if you think about the end of the array, how can you append an item to its end using the notation described above? Simple, assign the value to a[3,0]. This is the tail of the array.
So, if you try to access the element at a[3,0], you will get []. In this case you are still in the range of the array. But if you try to access a[4,0], you'll get nil as return value, since you're not within the range of the array anymore.
Read more about it at http://mybrainstormings.wordpress.com/2012/09/10/arrays-in-ruby/ .
tl;dr: in the source code in array.c, different functions are called depending on whether you pass 1 or 2 arguments in to Array#slice resulting in the unexpected return values.
(First off, I'd like to point out that I don't code in C, but have been using Ruby for years. So if you're not familiar with C, but you take a few minutes to familiarize yourself with the basics of functions and variables it's really not that hard to follow the Ruby source code, as demonstrated below. This answer is based on Ruby v2.3, but is more or less the same back to v1.9.)
Scenario #1
array.length == 4; array.slice(4) #=> nil
If you look at the source code for Array#slice (rb_ary_aref), you see that when only one argument is passed in (lines 1277-1289), rb_ary_entry is called, passing in the index value (which can be positive or negative).
rb_ary_entry then calculates the position of the requested element from the beginning of the array (in other words, if a negative index is passed in, it computes the positive equivalent) and then calls rb_ary_elt to get the requested element.
As expected, rb_ary_elt returns nil when the length of the array len is less than or equal to the index (here called offset).
1189: if (offset < 0 || len <= offset) {
1190: return Qnil;
1191: }
Scenario #2
array.length == 4; array.slice(4, 0) #=> []
However when 2 arguments are passed in (i.e. the starting index beg, and length of the slice len), rb_ary_subseq is called.
In rb_ary_subseq, if the starting index beg is greater than the array length alen, nil is returned:
1208: long alen = RARRAY_LEN(ary);
1209:
1210: if (beg > alen) return Qnil;
Otherwise the length of the resulting slice len is calculated, and if it's determined to be zero, an empty array is returned:
1213: if (alen < len || alen < beg + len) {
1214: len = alen - beg;
1215: }
1216: klass = rb_obj_class(ary);
1217: if (len == 0) return ary_new(klass, 0);
So since the starting index of 4 is not greater than array.length, an empty array is returned instead of the nil value that one might expect.
Question answered?
If the actual question here isn't "What code causes this to happen?", but rather, "Why did Matz do it this way?", well you'll just have to buy him a cup of coffee at the next RubyConf and ask him.
In Ruby, I would like to take an array of numbers, select 2 different numbers, add those 2 numbers together and see weather there equal to a variable x.y'd a variable x. Here is the code I used
def arrayIsEqual? (numArray, x)
return true if numArray.sample + numArray.sample == x
return false if numArray.empty? || numArray.count == 1
end
for example
numArray = [4,2,7,5]
x = 11
arrayIsEqual (numArray, n) should return true, since 4 + 7 = n(11)
How do I get this to work?
I don't want it to be 2 random numbers, just any 2 different numbers that add up to n
It looks like you're trying to see if there are any two numbers in the array that add up to the specified value x. However, your code just picks two numbers at random and checks if those numbers add up.
Ruby has the Array#combination method, which generates all combinations of a given length:
def contains_pair_for_sum?(arr, n)
!!arr.uniq.combination(2).detect { |a, b| a + b == n }
end
A few things to note:
First, we named it according to Ruby conventions: each word is separated_by_underscores. The ? on the end means that the method is a predicate method and returns a true or false value.
Inside the method, a few things happen. Let's look at that line, piece by piece.
arr: We take the array that was passed in.
<...>.uniq: We only look at the unique elements (because the OP wants to pick two different numbers).
<...>.combination(2): We ask for all combinations from the array of length 2. If the array was [4, 5, 6], we'd get [[4, 5], [4, 6], [5, 6]].
<...>.detect { |a, b| a + b == n }: We look for the first combination that adds up to n. If we found one, that's the result of that method. Otherwise, we get nil.
!!<...>: Finally, we take the result we got from detect and negate it twice. The first negation produces a Boolean value (true if the value we got was nil, or false if it's anything else); the second negation produces a Boolean value that's identical to the truth value of the first negation. This is a Ruby idiom to coerce a result into being either true or false.
Let's see it in action:
array = [4, 5, 9, 7, 8]
contains_pair_for_sum?(array, 11)
# => true (because [4, 7] sums to 11)
contains_pair_for_sum?(array, 17)
# => true (because [9, 8] sums to 17)
contains_pair_for_sum?(array, 100)
# => false (no pair matched)
I understand that your question is "is there any pair of numbers in my array equals x", in which case this will do what you need:
def has_pair_equal?(num_array, x)
(0..num_array.length-1).any? do |i|
num_array[i+1..-1].any? { |n| n + num_array[i] == x }
end
end
This checks all sums of pairs of numbers in the array, and checks if their sum is x. sample randomly picks an item from the array, which means that what your code does is "return true sometimes if there is a pair of numbers in my array equals x"
def array_is_equal? (num_array, x)
equality = 0
num_array.each do |a|
equality += 1 if a == x
return true if equality == 2
end
return false
end
Use lowercase and underscores for variables in Ruby. The convention is different here than in some other languages.
One liner
x=[4,2,7,5]; x.each_with_index.any? {|y,i| x.each_with_index.any? {|z,j| unless i==j; z+y==11; end } }
And as a function
def pair_sum_match?(arr, x)
arr.each_with_index.any? do |y,i|
arr.each_with_index.any? do |z,j|
unless i==j
z+y==x
end
end
end
end
Updated: Added each_with_index to avoid self inclusion on checks. It's a lot longer now :-/
Just iterate over it once and use the target number to see if it matches. 100 times faster then most of the answers here
numbers = ( -10..10 ).to_a
numbers.unshift( numbers.first + -1 ) # if you do -20 or 20
numbers.push( numbers.last + 1 )
target = 5
searched = { }
matches = { }
numbers.each do |number|
if searched[ target - number + 1 ] == true
matches[ "#{ number }_plus_#{ target - number }" ] = target
end
searched[ number + 1 ] = true
end
ap matches