为什么一个嵌套在另一个每个循环的每个循环中兰德不起作用

Question

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) 

我想创造出返回乔东西代码如下：

"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" 

并且Sam和Tom每年都有不同数量的散文。
它可以使用下面的代码：

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 

但是这个代码是重复和多余的，没有考虑到的ary大小可以比这里更大。所以我想用下面的更紧凑的代码：

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

但这个代码将返回每个朋友的相同数量的论文，使之使用方法rand的将是无用的。这是为什么？你会建议什么解决方案？

Answer 1

您是否考虑将范围存储在数组中，并根据需要从rand进行绘制？

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 

这就产生，例如：

"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" 

Answer 2

除了@pjs，您可以使用each_with_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，回答你的问题：” ..使rand方法的使用将是无用的“ - 当你创建一个数组时，其中的方法 - 这些方法的元素，他们将返回他们在这个数组中的工作结果，下一次，你可以在你的控制台中尝试使用irb：

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