Sorting 彩虹的颜色

鉴于彩虹颜色的顺序与此完全相同：ROYGBIV R=红色，O=橙色，Y=黄色，G=绿色，B=蓝色，I=靛蓝，V=紫色，我如何获取列表：

def partialUnorderedList =["orange", "blue", "red","green"] 

并将其按ROYGBIV顺序分类，得出：

def  partialOrderedList = ["red", "orange", "green", "blue"]

---

如果您不介意引入另一个库，则可能非常适合解决此问题

def partialUnorderedList = ["orange", "blue", "red", "green"] 
Ordering roygbivOrdering = Ordering.explicit("red", "orange", "yellow", "green", "blue", "indigo", "violet")
def partialOrderedList = roygbivOrdering.sortedCopy(partialUnorderedList)

---

如果您不介意引入另一个库，则可能非常适合解决此问题

def partialUnorderedList = ["orange", "blue", "red", "green"] 
Ordering roygbivOrdering = Ordering.explicit("red", "orange", "yellow", "green", "blue", "indigo", "violet")
def partialOrderedList = roygbivOrdering.sortedCopy(partialUnorderedList)

---

一个简单的解决方案是使用闭包进行比较，如下所示：

def partialUnorderedList =["orange", "blue", "red","green"]

def order = "roygbiv".inject([:]){ map, letter -> map[letter] = map.size(); map}
def sortedList = partialUnorderedList.sort{ a, b ->
    order[a[0].toLowerCase()] <=> order[b[0].toLowerCase()]
}

这将为颜色的第一个字母创建索引的哈希映射。排序方法然后查找第一个方法的索引，并在此基础上进行比较

这仅在使用唯一的首字母时有效。要按全名排序，可以执行以下操作：

def partialUnorderedList =["orange", "blue", "red","green"]

def order = "red,orange,yellow,green,blue,indigo,violet".split(",").inject([:]){ map, name -> map[name] = map.size(); map}
def sortedList = partialUnorderedList.sort{ a, b ->
    order[a.toLowerCase()] <=> order[b.toLowerCase()]
}

---

一个简单的解决方案是使用闭包进行比较，如下所示：

def partialUnorderedList =["orange", "blue", "red","green"]

def order = "roygbiv".inject([:]){ map, letter -> map[letter] = map.size(); map}
def sortedList = partialUnorderedList.sort{ a, b ->
    order[a[0].toLowerCase()] <=> order[b[0].toLowerCase()]
}

这将为颜色的第一个字母创建索引的哈希映射。排序方法然后查找第一个方法的索引，并在此基础上进行比较

这仅在使用唯一的首字母时有效。要按全名排序，可以执行以下操作：

def partialUnorderedList =["orange", "blue", "red","green"]

def order = "red,orange,yellow,green,blue,indigo,violet".split(",").inject([:]){ map, name -> map[name] = map.size(); map}
def sortedList = partialUnorderedList.sort{ a, b ->
    order[a.toLowerCase()] <=> order[b.toLowerCase()]
}

---

以下是解决您问题的最简单解决方案：

def a =["red","orange","yellow","green","blue","indigo","violet"]
def input = ["orange", "blue", "red","green"]
list = []
for(i=0;i<input.size();i++)
{
   index =  a.findIndexValues{ it ==  input[i]}.join() as Integer
   list[index] =  input[i]
}
println list.minus(null)

关于其他投入：

[yellow, red, blue, orange, violet, indigo]

它输出：

[red, orange, yellow, blue, indigo, violet]

---

在上面的代码中使用了

以下是解决您问题的最简单解决方案：

def a =["red","orange","yellow","green","blue","indigo","violet"]
def input = ["orange", "blue", "red","green"]
list = []
for(i=0;i<input.size();i++)
{
   index =  a.findIndexValues{ it ==  input[i]}.join() as Integer
   list[index] =  input[i]
}
println list.minus(null)

关于其他投入：

[yellow, red, blue, orange, violet, indigo]

它输出：

[red, orange, yellow, blue, indigo, violet]

---

在上面的代码中使用了

以下是我对这个问题的看法：

def order = ["red", "orange", "yellow", "green", "blue", "indigo", "violet"]
def input = ["orange", "blue", "red", "green"]

def sorted = input.sort { order.indexOf it }

---

它使用输入列表中给定元素在引用列表中的索引位置作为排序标准。

以下是我对该问题的看法：

def order = ["red", "orange", "yellow", "green", "blue", "indigo", "violet"]
def input = ["orange", "blue", "red", "green"]

def sorted = input.sort { order.indexOf it }

---

它使用输入列表中给定元素在引用列表中的索引位置作为排序标准。

有人能告诉我indexOf是从哪里来的吗？@Ant感谢Christoph。这也是我提出的确切解决方案。另一个有趣的选择是使用枚举，而另一个选择是使用映射。有人愿意提供这些额外的分数吗“有人能告诉我indexOf是从哪里来的订单吗？”Ant感谢Christoph。这也是我提出的确切解决方案。另一个有趣的选择是使用枚举，而另一个选择是使用映射。有人愿意提供这些额外的分数吗