嵌套的每个语句都不起作用

Question

我正在尝试在 Ruby 中编写一个程序来查找给定数字的质因数，而不使用 Ruby 的 .Prime class。我写的代码对我来说似乎是合乎逻辑的（如果是迂回的话），但出于某种原因，最后一行代码似乎不起作用，我也不知道为什么。该代码仍然只是 returns 给定数字的所有因数，而不是素数。

def prime(n)
    array = []
    r = Range.new(2, n - 1)
    r.each { |x| array << x if n % x == 0 }
    array.each { |x| puts x if (2..x - 1).each { |p| x % p != 0 }}
end

Answer 1

您想检查 (2..x - 1) 中的 .all? 个数字，而不是 .each:

def prime(n)
  array = []
  r = Range.new(2, n - 1)
  r.each { |x| array << x if n % x == 0 }
  array.each { |x| puts x if (2..x - 1).all? { |p| x % p != 0 }}
end

顺便说一句，我也建议给你的变量起一个有意义的名字，因为现在你的代码真的很难阅读，这样的实现更清晰：

def prime?(num)
  return false if num < 2
  (2..Math.sqrt(num).to_i).each do |i|
    return false if num % i == 0
  end
  true
end

def prime_factors(num)
  (2..num - 1).inject([]) do |factors, i|
    (num % i == 0) ? factors << i : factors
  end.select { |factor| prime?(factor) }
end

p prime_factors(44)

Answer 2

将这个单一功能拆分为三个 factors n、prime n 和 prime_factors n，然后将它们组合在一起似乎更合乎逻辑。这种设计更加模块化，更适合测试各个部分。例如如果所有的逻辑都放在一起，你怎么能肯定你的程序正确地找到了素数？

def factors n 
   (2..(n/2).floor).select {|x| n % x == 0 } 
end

def prime? n
   return false unless n >= 2
   (2..Math.sqrt(n).floor).all? {|x| n % x != 0 }
end

def prime_factors n
   factors(n).select {|x| prime? x }
end

Answer 3

def prime_factors(n)
    # It's clear what factors will hold.
    # It helps you when you or someone else read it out 
    # after a while.
    factors = []

    # Keep the r. No problem.
    r = Range.new(2, n-1)

    # Array#select is nice for this task.
    # select is good in case where you need to filter out certain
    # values out of a list. It nicely separates the selection task.
    factors = r.select { |x| n % x == 0 }

    # It's better not to go for double block on a single line.
    # It will become more readable this way.
    # Also it's better to handle how the result is displayed outside
    # the function and just to return the computed values.
    factors.select do |x| 
        (2..x - 1).all? { |p| x % p != 0 }
    end
end

嵌套的每个语句都不起作用

Nested each statements aren't working

ruby

each