Ruby 为什么嵌套在另一个each循环中的each循环中的rand不起作用

考虑以下阵列和范围：

friends = ["Joe", "Sam", "Tom"]
ary = [rand(1..5), rand(6..10), rand(11..20)]
range = (0..2)

我想创建一个代码，返回Joe，如下所示：

"Joe at the end of year 1 wrote 2 essays"
"Joe at the end of year 2 wrote 8 essays"
"Joe at the end of year 3 wrote 16 essays"

对于萨姆和汤姆来说，每年的论文数量也不尽相同
可以使用以下代码：

friends.each do |friend|
  "#{friend} at the end of year 1 wrote #{rand(1..5)} essays"
  "#{friend} at the end of year 2 wrote #{rand(6..10)} essays"
  "#{friend} at the end of year 3 wrote #{rand(11..20)} essays"
end

但是，此代码是重复和冗余的，不考虑

ari

的大小可能大于此处。因此，我想使用以下更紧凑的代码：

friends.each do |friend|
  range.each do |num|
    "#{friend} at the end of year #{num+1} wrote #{ary[num]} essays"
  end
end

---

但是这段代码会为每个朋友返回相同数量的文章，因此使用方法

rand

将是无用的。为什么呢？您建议什么解决方案？

您是否考虑过将范围存储在一个数组中，并根据需要从

rand

中提取

friends = ["Joe", "Sam", "Tom"]
ary =[(1..5), (6..10), (11..20)]
year = (1..3)
friends.each do |friend|
  year.each do |yr|
    p "#{friend} at the end of year #{yr} wrote #{rand(ary[yr - 1])} essays"
  end
end

例如，这会产生：

"Joe at the end of year 1 wrote 5 essays"
"Joe at the end of year 2 wrote 7 essays"
"Joe at the end of year 3 wrote 16 essays"
"Sam at the end of year 1 wrote 3 essays"
"Sam at the end of year 2 wrote 7 essays"
"Sam at the end of year 3 wrote 18 essays"
"Tom at the end of year 1 wrote 2 essays"
"Tom at the end of year 2 wrote 8 essays"
"Tom at the end of year 3 wrote 15 essays"

---

您是否考虑过将范围存储在数组中，并根据需要从

rand

绘图

friends = ["Joe", "Sam", "Tom"]
ary =[(1..5), (6..10), (11..20)]
year = (1..3)
friends.each do |friend|
  year.each do |yr|
    p "#{friend} at the end of year #{yr} wrote #{rand(ary[yr - 1])} essays"
  end
end

例如，这会产生：

"Joe at the end of year 1 wrote 5 essays"
"Joe at the end of year 2 wrote 7 essays"
"Joe at the end of year 3 wrote 16 essays"
"Sam at the end of year 1 wrote 3 essays"
"Sam at the end of year 2 wrote 7 essays"
"Sam at the end of year 3 wrote 18 essays"
"Tom at the end of year 1 wrote 2 essays"
"Tom at the end of year 2 wrote 8 essays"
"Tom at the end of year 3 wrote 15 essays"

---

除了@pjs之外，您还可以使用

each\u with\u index

方法

friends = ["Joe", "Sam", "Tom"]
ary =[(1..5), (6..10), (11..20)]
friends.each do |friend|
  ary.each_with_index do |value, year|
    p "#{friend} at the end of year #{year+1} wrote #{rand(value)} essays"
  end
end

另外，请回答您的问题：“…这样方法rand的使用就没用了”-当您创建一个数组时，其中的元素-方法，它们将返回在此数组中的工作结果，下次，您可以在控制台中使用

irb

：

2.3.0 :001 > ary = [rand(1..5), rand(6..10), rand(11..20)]
 => [2, 9, 12]

---

除了@pjs之外，您还可以使用

each\u with\u index

方法

friends = ["Joe", "Sam", "Tom"]
ary =[(1..5), (6..10), (11..20)]
friends.each do |friend|
  ary.each_with_index do |value, year|
    p "#{friend} at the end of year #{year+1} wrote #{rand(value)} essays"
  end
end

另外，请回答您的问题：“…这样方法rand的使用就没用了”-当您创建一个数组时，其中的元素-方法，它们将返回在此数组中的工作结果，下次，您可以在控制台中使用

irb

：

2.3.0 :001 > ary = [rand(1..5), rand(6..10), rand(11..20)]
 => [2, 9, 12]

---

each循环（或任何替代解决方案）应该在ary的每个元素上从头到尾迭代，而不重复。应遵守ary中元素的顺序，以便从第1年到第3年有一个书面文章的进度。@Asarluhi您尝试过运行提供的解决方案吗？我尊重你在问题中给出的相同进展。你是对的，我没有注意到你从ary中删除rand。非常感谢@pjs，它很有效！each循环（或任何替代解决方案）应该在ary的每个元素上从头到尾迭代，而不重复。应遵守ary中元素的顺序，以便从第1年到第3年有一个书面文章的进度。@Asarluhi您尝试过运行提供的解决方案吗？我尊重你在问题中给出的相同进展。你是对的，我没有注意到你从ary中删除rand。非常感谢@pjs，它很有效！