英文:
Any way to optimize this character counter i wrote in ruby for String class
问题
# 字符计数器
class String
def count_lcases
count(('a'..'z').to_a.join(''))
end
def count_upcases
count(('A'..'Z').to_a.join(''))
end
def count_num
count((0..9).to_a.join(''))
end
def count_spl_chars
length - count_lcases - count_upcases - count_num
end
end
input = ARGV[0]
if ARGV.empty?
puts '请提供输入'
exit
end
puts '小写字符 = %d' % [input.count_lcases]
puts '大写字符 = %d' % [input.count_upcases]
puts '数字字符 = %d' % [input.count_num]
puts '特殊字符 = %d' % [input.count_spl_chars]
对于你提出的优化问题，可以考虑使用正则表达式来一次性匹配并计数不同类型的字符。不过，这个示例中的方法也是有效的。
英文:
# Character Counter
class String
def count_lcases
count(('a'..'z').to_a.join(''))
end
def count_upcases
count(('A'..'Z').to_a.join(''))
end
def count_num
count((0..9).to_a.join(''))
end
def count_spl_chars
length - count_lcases - count_upcases - count_num
end
end
input = ARGV[0]
if ARGV.empty?
puts 'Please provide an input'
exit
end
puts 'Lowercase characters = %d' % [input.count_lcases]
puts 'Uppercase characters = %d' % [input.count_upcases]
puts 'Numeric characters = %d' % [input.count_num]
puts 'Special characters = %d' % [input.count_spl_chars]
I used ranges to count characters but count function is called 3 times.
I can always use loops and count it one by one.I was wondering is there any way to optimize this?...
答案1
得分: 3
如果您使用Ruby 2.7，您可以使用tally；字符串的字符只需迭代一次。
def classify_char(c)
case c
when /[a-z]/ then :lcase
when /[A-Z]/ then :ucase
when /\d/ then :digit
else :other
end
end
p "asg3456 ERTYaeth".chars.map{|c| classify_char(c) }.tally
# => {:lcase=>7, :digit=>4, :other=>2, :ucase=>4}
英文:
If you are using Ruby 2.7 you could use tally; the string's chars are just iterated one time.
def classify_char(c)
case c
when /[a-z]/ then :lcase
when /[A-Z]/ then :ucase
when /\d/ then :digit
else :other
end
end
p "asg3456 ERTYaeth".chars.map{|c| classify_char(c) }.tally
# => {:lcase=>7, :digit=>4, :other=>2, :ucase=>4}
答案2
得分: 0
如果 Ruby 版本在 2.3 到 2.7 之间，这段代码将有效：
CHAR_CLASSES = {
lcase: ?a..?z,
ucase: ?A..?Z,
digit: ?0..?9,
}
p "asg3456 ERTYaeth".each_char.with_object(Hash.new(0)) { |c, o|
o[CHAR_CLASSES.find { |label, group| group === c }&.first || :other] += 1
}
对于小于 2.3 的情况，
p "asg3456 ERTYaeth".each_char.with_object(Hash.new(0)) { |c, o|
p = CHAR_CLASSES.find { |label, group| group === c }
o[p ? p.first : :other] += 1
}
英文:
If Ruby 2.3...2.7, this will work:
CHAR_CLASSES = {
lcase: ?a..?z,
ucase: ?A..?Z,
digit: ?0..?9,
}
p "asg3456 ERTYaeth".each_char.with_object(Hash.new(0)) { |c, o|
o[CHAR_CLASSES.find { |label, group| group === c }&.first || :other] += 1
}
For < 2.3,
p "asg3456 ERTYaeth".each_char.with_object(Hash.new(0)) { |c, o|
p = CHAR_CLASSES.find { |label, group| group === c }
o += 1
}
通过集体智慧和协作来改善编程学习和解决问题的方式。致力于成为全球开发者共同参与的知识库，让每个人都能够通过互相帮助和分享经验来进步。
评论