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问题:
I would like to generate a sequence of letters i.e. "A", "DE" "GJE", etc. that correspond to a number. The first 26 are pretty easy so 3 returns "C", 26 returns "Z", and 27 would return "AA", 28 "AB", and so on.
The thing I can't quite figure out is how to do this so it will handle any number passed in. So if I pass in 4123 I should get back some combination of 3 letters since (26 * 26 * 26) allows for up to +17,000 combinations.
Any suggestions?
回答1:
class Numeric
Alph = ("a".."z").to_a
def alph
s, q = "", self
(q, r = (q - 1).divmod(26)); s.prepend(Alph[r]) until q.zero?
s
end
end
3.alph
# => "c"
26.alph
# => "z"
27.alph
# => "aa"
4123.alph
# => "fbo"
回答2:
A tweak on @sawa original answer for Ruby 2.0 since I couldn't get his to work as is:
class Numeric
Alpha26 = ("a".."z").to_a
def to_s26
return "" if self < 1
s, q = "", self
loop do
q, r = (q - 1).divmod(26)
s.prepend(Alpha26[r])
break if q.zero?
end
s
end
end
and here it is going in reverse from string to integer:
class String
Alpha26 = ("a".."z").to_a
def to_i26
result = 0
downcased = downcase
(1..length).each do |i|
char = downcased[-i]
result += 26**(i-1) * (Alpha26.index(char) + 1)
end
result
end
end
Usage:
1234567890.to_s26
# => "cywoqvj"
"cywoqvj".to_i26
# => 1234567890
1234567890.to_s26.to_i26
# => 1234567890
"".to_i26
# => 0
0.to_s26
# => ""
回答3:
Strings do have a succ
method so they are usable in a Range. The successor to "Z" happens to be "AA", so this works:
h = {}
('A'..'ZZZ').each_with_index{|w, i| h[i+1] = w }
p h[27] #=> "AA"
回答4:
I liked this answer from: https://stackoverflow.com/a/17785576/514483
number.to_s(26).tr("0123456789abcdefghijklmnopq", "ABCDEFGHIJKLMNOPQRSTUVWXYZ")
回答5:
Using the base conversion method found here. I also changed it for the lack of "0" that we have in this numbering system. End cases have been addressed.
def baseAZ(num)
# temp variable for converting base
temp = num
# the base 26 (az) number
az = ''
while temp > 0
# get the remainder and convert to a letter
num26 = temp % 26
temp /= 26
# offset for lack of "0"
temp -= 1 if num26 == 0
az = (num26).to_s(26).tr('0-9a-p', 'ZA-Y') + az
end
return az
end
irb I/O:
>> baseAZ(1)
=> "A"
>> baseAZ(26^2 + 1)
=> "Y"
>> baseAZ(26*26 + 1)
=> "ZA"
>> baseAZ(26*26*26 + 1)
=> "YZA"
>> baseAZ(26*26*26 + 26*26 + 1)
=> "ZZA"
回答6:
def letter_sequence(n)
n.to_s(26).each_char.map {|i| ('A'..'Z').to_a[i.to_i(26)]}.join
end
回答7:
Here is a short performant recursion based solution
class Numeric
# 'z' is placed in the begining of the array because 26 % 26 is 0 not 26
Alph = ['z'] + ("a".."y").to_a
def to_alph
# if self is 0 or negative return a blank string.
# this is also used to end the recursive chain of to_alph calls
# so don't replace this with raising an error or returning anything else
return '' if self < 1
# (otherwise) return two concatenated strings:
# the right side is the letter for self modules 26
# the left side is comprised of:
# 1. minus one, because this is not a zero-based numbering system.
# therefore, round numbers (26, 52...) should return one digit 'z'
# instead of two digits 'aa' or 'ba'.
# 2. divide by 26 and call to_alph on that.
# this repeats recursively for every digit of the final string,
# plus once more that will return '' to end the recursion chain.
return ((self - 1) / 26).to_alph + Alph[self % 26]
end
end
回答8:
Based on sawa's answer, I wanted a method that worked independently, albeit recursively, to achieve the desired outcome:
def num_to_col(num)
raise("invalid value #{num} for num") unless num > 0
result, remainder = num.divmod(26)
if remainder == 0
result -= 1
remainder = 26
end
final_letter = ('a'..'z').to_a[remainder-1]
result > 0 ? previous_letters = num_to_col(result) : previous_letters = ''
"#{previous_letters}#{final_letter}".upcase
end