Say you have the string "Hi"
. How do you get a value of 8
, 9
("H"
is the 8th letter of the alphabet, and "i"
is the 9th letter). Then say, add 1
to those integers and make it 9
, 10
which can then be made back into the string "Ij"
? Is it possible?
可以将文章内容翻译成中文,广告屏蔽插件可能会导致该功能失效(如失效,请关闭广告屏蔽插件后再试):
问题:
回答1:
use ord
to get the ASCII index, and chr
to bring it back.
'Hi'.chars.map{|x| (x.ord+1).chr}.join
回答2:
Note Cary Swoveland had already given a same answer in a comment to the question.
It is impossible to do that through the numbers 8 and 9 because these numbers do not contain information about the case of the letters. But if you do not insist on converting the string via the number 8 and 9, but instead more meaningful numbers like ASCII code, then you can do it like this:
"Hi".chars.map(&:next).join
# => "Ij"
回答3:
You can also create an enumerable of character ordinals from a string using the codepoints
method.
string = "Hi"
string.codepoints.map{|i| (i + 1).chr}.join
=> "Ij"
回答4:
Preserving case and assuming you want to wrap around at "Z":
upper = [*?A..?Z]
lower = [*?a..?z]
LOOKUP = (upper.zip(upper.rotate) + lower.zip(lower.rotate)).to_h
s.each_char.map { |c| LOOKUP[c] }.join
#=> "Ij"