TypeError-在Ruby on Rails数组中没有将String隐式转换为Integer

Question

我正在尝试制作一个输出以下内容的区域： [@month, @monthly_count]以便完整的输出如下所示： [["January", 0], ["February", 0], ["March", 0], ["April", 2], ["May", 3], ["June", 19], ["July", 0], ["August", 0], ["September", 0], ["October", 0], ["November", 0], ["December", 0]]

我当前收到TypeError - no implicit conversion of String into Integer:错误TypeError - no implicit conversion of String into Integer:在下面的@monthly_value_created[i] << @months[i]中。

这是我的代码：

@months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]
@monthly_idea_value = [[0], [0], [0], [3], [35], [744], [0], [0], [0], [0], [0], [0]] 
#Array of arrays of integers isn't best, but unless it's causing the error I'm not worried about it right now.
#Bc the .flatten! method at the end fixes the output

      @monthly_value_created = Array.new(12){Array.new}
      i = 0
      12.times do |i|
        @monthly_value_created[i] << @months[i]
        @monthly_value_created[i] << @monthly_idea_value[i]
        @monthly_value_created.flatten!
        i += 1
      end

如何避免此错误？

Answer 1

您可以使用Array＃zip

@months.zip(@monthly_idea_value).map &:flatten

如果@monthly_idea_value是整数数组而不是数组数组，则不需要映射和展平。

Answer 2

可能会更有效：

@months.map.each_with_index do |month, index|
  [month, @monthly_idea_value[index][0]]
end

Answer 3

因此，您的代码中有两个错误。

首先i自己增加#times已经通过了变量。

您犯的第二个错误是您弄平了错误的东西（您只想弄平子数组而不是弄平您构建的数组结构。

@months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]
@monthly_idea_value = [[0], [0], [0], [3], [35], [744], [0], [0], [0], [0], [0], [0]]

@monthly_value_created = Array.new(12){Array.new}
12.times do |i|
  @monthly_value_created[i] << @months[i]
  @monthly_value_created[i] << @monthly_idea_value[i]
  @monthly_value_created[i].flatten!
end

=> [["January", 0], ["February", 0], ["March", 0], ["April", 3], ["May", 35], ["June", 744], ["July", 0], ["August", 0], ["September", 0], ["October", 0], ["November", 0], []]

仍然fylooi或Santhosh的方法更好，也更红宝石：)