Arrays 将哈希转换为哈希数组
我有散列,它的所有值都是数组,如下所示:Arrays 将哈希转换为哈希数组,arrays,ruby,hashmap,Arrays,Ruby,Hashmap,我有散列,它的所有值都是数组,如下所示: list = { letter: ['a', 'b', 'c'], number: ['one', 'two', 'three'], fruit: ['apple', 'pear', 'kiwi'], car: ['vw', 'mb', 'bmw'], state: ['la', 'ny', 'fl'], color: ['red', 'wh
list = { letter: ['a', 'b', 'c'],
number: ['one', 'two', 'three'],
fruit: ['apple', 'pear', 'kiwi'],
car: ['vw', 'mb', 'bmw'],
state: ['la', 'ny', 'fl'],
color: ['red', 'white', 'black'],
tree: ['oak', 'pine', 'maple'],
animal: ['cat', 'dog', 'rat'],
clothes: ['tie', 'sock', 'glove'] }
事实上,这个散列可以有更多的键,值也可以更大,但每个值的大小总是相同的(在本例中为-3)
我想把这个散列转换成散列数组
每个散列将具有所有源散列键和各自的值
最后,我想:
list = [
{ letter: 'a', number: 'one', fruit: 'apple', car: 'vw', state: 'la',
color: 'red', tree: 'oak', animal: 'cat', clothes: 'tie' },
{ letter: 'b', number: 'two', fruit: 'pear', car: 'mb', state: 'ny',
color: 'white', tree: 'pine', animal: 'dog', clothes: 'sock' },
{ letter: 'c', number: 'three', fruit: 'kiwi', car: 'bmw', state: 'fl',
color: 'black', tree: 'elm', animal: 'rat', clothes: 'glove' }
]
最好的方法是什么?试试下面的代码
arr = []
3.times { |x| arr[x] = {}; list.each { |k,v| arr[x][k] = v[x] } }
arr.inspect
输出将是
=> [{:letter=>"a", :number=>"one", :fruit=>"apple", :car=>"vw", :state=>"la", :color=>"red", :tree=>"oak", :animal=>"cat", :clothes=>"tie"}, {:letter=>"b", :number=>"two", :fruit=>"pear", :car=>"mb", :state=>"ny", :color=>"white", :tree=>"pine", :animal=>"dog", :clothes=>"sock"}, {:letter=>"c", :number=>"three", :fruit=>"kiwi", :car=>"bmw", :state=>"fl", :color=>"black", :tree=>"maple", :animal=>"rat", :clothes=>"glove"}]
杠杆作用和
一应俱全:
list.map{|k,v| [k].product(v)}.transpose.map(&:to_h)
其思想是使用每个关键产品的值,映射、转置,然后通过转换为散列
[list.keys].product(list.values.transpose).map { |a| a.transpose.to_h }
#=> [{:letter=>"a", :number=>"one", :fruit=>"apple", :car=>"vw",
# :state=>"la", :color=>"red", :tree=>"oak", :animal=>"cat",
# :clothes=>"tie"},
# {:letter=>"b", :number=>"two", :fruit=>"pear", :car=>"mb",
# :state=>"ny", :color=>"white", :tree=>"pine", :animal=>"dog",
# :clothes=>"sock"},
# {:letter=>"c", :number=>"three", :fruit=>"kiwi", :car=>"bmw",
# :state=>"fl", :color=>"black", :tree=>"maple", :animal=>"rat",
# :clothes=>"glove"}]
假设list
定义如下
list = {
letter: ['a', 'b' ],
number: ['one', 'two' ],
fruit: ['apple', 'pear'],
car: ['vw', 'mb' ]
}
b = [list.keys]
#=> [[:letter, :number, :fruit, :car]]
c = list.values
#=> [["a", "b"], ["one", "two"], ["apple", "pear"], ["vw", "mb"]]
d = c.transpose
#=> [["a", "one", "apple", "vw"],
# ["b", "two", "pear", "mb"]]
e = b.product(d)
#=> [[[:letter, :number, :fruit, :car], ["a", "one", "apple", "vw"]],
# [[:letter, :number, :fruit, :car], ["b", "two", "pear", "mb"]]]
e.map { |a| a.transpose.to_h }
#=> [{:letter=>"a", :number=>"one", :fruit=>"apple", :car=>"vw"},
# {:letter=>"b", :number=>"two", :fruit=>"pear", :car=>"mb"}]
f = a.transpose
#=> [[:letter, "a"], [:number, "one"], [:fruit, "apple"], [:car, "vw"]]
f.to_h
#=> {:letter=>"a", :number=>"one", :fruit=>"apple", :car=>"vw"}
步骤如下
list = {
letter: ['a', 'b' ],
number: ['one', 'two' ],
fruit: ['apple', 'pear'],
car: ['vw', 'mb' ]
}
b = [list.keys]
#=> [[:letter, :number, :fruit, :car]]
c = list.values
#=> [["a", "b"], ["one", "two"], ["apple", "pear"], ["vw", "mb"]]
d = c.transpose
#=> [["a", "one", "apple", "vw"],
# ["b", "two", "pear", "mb"]]
e = b.product(d)
#=> [[[:letter, :number, :fruit, :car], ["a", "one", "apple", "vw"]],
# [[:letter, :number, :fruit, :car], ["b", "two", "pear", "mb"]]]
e.map { |a| a.transpose.to_h }
#=> [{:letter=>"a", :number=>"one", :fruit=>"apple", :car=>"vw"},
# {:letter=>"b", :number=>"two", :fruit=>"pear", :car=>"mb"}]
f = a.transpose
#=> [[:letter, "a"], [:number, "one"], [:fruit, "apple"], [:car, "vw"]]
f.to_h
#=> {:letter=>"a", :number=>"one", :fruit=>"apple", :car=>"vw"}
让我们更仔细地看最后一步map
将e
的第一个元素传递给块,并将块变量a
设置为其值:
a = e.first
#=> [[:letter, :number, :fruit, :car],
# ["a", "one", "apple", "vw"]]
区块计算如下所示
list = {
letter: ['a', 'b' ],
number: ['one', 'two' ],
fruit: ['apple', 'pear'],
car: ['vw', 'mb' ]
}
b = [list.keys]
#=> [[:letter, :number, :fruit, :car]]
c = list.values
#=> [["a", "b"], ["one", "two"], ["apple", "pear"], ["vw", "mb"]]
d = c.transpose
#=> [["a", "one", "apple", "vw"],
# ["b", "two", "pear", "mb"]]
e = b.product(d)
#=> [[[:letter, :number, :fruit, :car], ["a", "one", "apple", "vw"]],
# [[:letter, :number, :fruit, :car], ["b", "two", "pear", "mb"]]]
e.map { |a| a.transpose.to_h }
#=> [{:letter=>"a", :number=>"one", :fruit=>"apple", :car=>"vw"},
# {:letter=>"b", :number=>"two", :fruit=>"pear", :car=>"mb"}]
f = a.transpose
#=> [[:letter, "a"], [:number, "one"], [:fruit, "apple"], [:car, "vw"]]
f.to_h
#=> {:letter=>"a", :number=>"one", :fruit=>"apple", :car=>"vw"}
e.map{a | a.transpose.to_h}
的其余计算类似。仅用于:
list.values.transpose.each_与_对象([]){v,a | a类似keys=list.keys;list.values.transpose.map{v |[keys,v].transpose.to_h}
不一样,更像是,为什么我不直接映射值…:p我会更新,谢谢3
从哪里来?在他的例子中,3是通过检查list
来获取的。他可以通过一些必需的逻辑从list
动态获取。如果需要,我也会提供:)1.map
比更惯用>收集
2.每对
都是多余的。这类似于@Marcin的答案。主要区别在于我使用了和使用了,而不是,后者是阴阳。@mechnicov,谢谢你的夸奖,也感谢你提到了混淆列表的问题。我解决了这个问题。