Arrays 展平内部阵列

我正在尝试创建一个输出以下内容的区域：

[@month，@monthly\u count]

，以便完整的输出如下：

[[“一月”，0]，“二月”，0]，“三月”，0]，“四月”，2]，“五月”，3]，“六月”，19]，“七月”，0]，“八月”，0]，“九月”，0]，“十月”，0]，“十一月”，0]，“十二月”，0]

我的代码是：

@months = [["January"],["February"],["March"],["April"],["May"],["June"],["July"],["August"],["September"],["October"],["November"],["December"]]
@monthly_count = [[0], [0], [0], [2], [3], [19], [0], [0], [0], [0], [0], [0]] 
@monthly_activity_count = Array.new(12){Array.new}
          i = 0
    12.times do |i|
      @monthly_activity_count[i] << @months[i]
      @monthly_activity_count[i] << @monthly_count[i]
      @monthly_activity_count[i].flatten
      i += 1
    end

我试图在迭代器中使用

array.flatte

来展平每个单独的数组，同时保持每个月的数组边界，但这不起作用。如何正确制作阵列？

展平（级别）

适合您

[[["January"], [0]], [["February"], [0]], [["March"], [0]], [["April"], [2]], [["May"], [3]], [["June"], [19]], [["July"], [0]], [["August"], [0]], [["September"], [0]], [["October"], [0]], [["November"], [0]], [["December"], [0]]].flatten(2)

---

有关详细信息

请尝试执行

展平在您的代码中

12.times do |i|
    @monthly_activity_count[i] << @months[i]
    @monthly_activity_count[i] << @monthly_count[i]
    @monthly_activity_count[i].flatten!
    i += 1
  end

12.0倍|
@每月活动计数[i]如果我了解您想要了解的内容，您可以从稍微简单一点的设置开始：

months = ["January", "February", "March", "April", "May", "June", "July", "August", "September", "October", "November", "December"]

在此基础上，您可以使用以下任一行生成所需的数组（它们完全等效）：

这两行都将遍历months数组，将月份名称及其索引传递到块中。代码块（被
{
和
}
包围的部分）返回一个仅包含月份名称和索引的数组-所有这些小数组都被分组到一个包含数组中，因为您使用的是
映射
（或
收集
）

然而，我无法想象为什么需要这样一个结构。如果你知道你想要的月份的索引，你可以这样得到它的名称：

months[index]

如果您知道月份名称并想知道其索引，您可以这样找到：

month_name = "March"
index = months.index { |month| month == month_name }

# only using some of the array to make the example easier to read
@months = [["January"],["February"],["March"],["April"],["May"]]
@monthly_count = [[0], [0], [0], [2], [3]] 


## First Method

zipped = @months.zip(@monthly_count)
=> [[["January"], [0]], [["February"], [0]], [["March"], [0]], [["April"], [2]], [["May"], [3]]]

@monthly_activity_count = zipped.each { |pair| pair.flatten! }
=> [["January", 0], ["February", 0], ["March", 0], ["April", 2], ["May", 3]]

# could also do it as a one liner
@months.zip(@monthly_count).each { |pair| pair.flatten! }
# the flatten! in the line above is not bad, just a warning to help you understand 
# the result. The difference between `flatten` and `flatten!` is that `flatten`
# will create a new array to hold the result, whereas `flatten!` will 
# modify the array you call it on.


## Second Method

@months.flatten.zip(@monthly_count.flatten)
=> [["January", 0], ["February", 0], ["March", 0], ["April", 2], ["May", 3]]


## Ideal Method

# if you can get your months and monthly_count data as simple arrays, eg ["January", 
# "February", ...] then you can remove the flattens in the previous line, giving:
@months.zip(@monthly_count)

months中的
index
。index
将迭代所有月份，并将每个月份传递到代码块中。它希望您提供一个代码块，当您找到想要索引的对象时，该代码块将返回
true
。因此，代码块（
month==month\u name
）设置为将传入的月份名称与存储在
month\u name
变量中的名称相匹配。有关Ruby代码块的更多信息，请参阅。
您可以这样做：

month_name = "March"
index = months.index { |month| month == month_name }

# only using some of the array to make the example easier to read
@months = [["January"],["February"],["March"],["April"],["May"]]
@monthly_count = [[0], [0], [0], [2], [3]] 


## First Method

zipped = @months.zip(@monthly_count)
=> [[["January"], [0]], [["February"], [0]], [["March"], [0]], [["April"], [2]], [["May"], [3]]]

@monthly_activity_count = zipped.each { |pair| pair.flatten! }
=> [["January", 0], ["February", 0], ["March", 0], ["April", 2], ["May", 3]]

# could also do it as a one liner
@months.zip(@monthly_count).each { |pair| pair.flatten! }
# the flatten! in the line above is not bad, just a warning to help you understand 
# the result. The difference between `flatten` and `flatten!` is that `flatten`
# will create a new array to hold the result, whereas `flatten!` will 
# modify the array you call it on.


## Second Method

@months.flatten.zip(@monthly_count.flatten)
=> [["January", 0], ["February", 0], ["March", 0], ["April", 2], ["May", 3]]


## Ideal Method

# if you can get your months and monthly_count data as simple arrays, eg ["January", 
# "February", ...] then you can remove the flattens in the previous line, giving:
@months.zip(@monthly_count)

有关
zip
方法的文档，请参阅

有关ruby中！（bang）方法的解释，请参阅。
这并没有改变迭代器中的任何内容，如
@monthly\u count[i]。展平（2）
，并给了我错误的输出
[“一月”，0，“二月”，0，“三月”，0，“四月”，2，“五月”，3，“六月”，19，“七月”，0，“八月”，0，“九月”，0，“十月”，0，“十一月”，0，“12月”，0]
当我做了
@每月\u计数时。展平（2）
在迭代器外。我会在迭代器外调整dif级别，但我真的希望在迭代器内有一些东西。你的代码有更多的上下文吗？在这部分代码之前，
@monthly\u activity\u count
和
@months
设置为什么？@bencopock在上面用适当的上下文编辑过，谢谢你的建议What Is
@monthly_count
？您的预期输出是什么？@virgeasault:
@monthly_count
不是一个整数数组，它是一个数组数组，其中每个内部数组只包含一个整数。同样，
@monthly
也是一个数组数组，其中每个内部数组只包含一个字符串。如果
@monthly_count
是一个整数数组，
@months
是一个字符串数组，你不需要首先将结果展平。这是有效的！使用
！
符号在这里是不受欢迎的吗？我很犹豫，因为有人教我在CSS中不要太多使用它，不确定这是否与“强制”相同“是的，它是强制的，但这并不意味着你永远不必使用它。基本上，它会将flatte的值赋回到调用实例中。+1
# only using some of the array to make the example easier to read
@months = [["January"],["February"],["March"],["April"],["May"]]
@monthly_count = [[0], [0], [0], [2], [3]] 


## First Method

zipped = @months.zip(@monthly_count)
=> [[["January"], [0]], [["February"], [0]], [["March"], [0]], [["April"], [2]], [["May"], [3]]]

@monthly_activity_count = zipped.each { |pair| pair.flatten! }
=> [["January", 0], ["February", 0], ["March", 0], ["April", 2], ["May", 3]]

# could also do it as a one liner
@months.zip(@monthly_count).each { |pair| pair.flatten! }
# the flatten! in the line above is not bad, just a warning to help you understand 
# the result. The difference between `flatten` and `flatten!` is that `flatten`
# will create a new array to hold the result, whereas `flatten!` will 
# modify the array you call it on.


## Second Method

@months.flatten.zip(@monthly_count.flatten)
=> [["January", 0], ["February", 0], ["March", 0], ["April", 2], ["May", 3]]


## Ideal Method

# if you can get your months and monthly_count data as simple arrays, eg ["January", 
# "February", ...] then you can remove the flattens in the previous line, giving:
@months.zip(@monthly_count)