我有一个简单的数组:
arr = ["apples", "bananas", "coconuts", "watermelons"]
我还有一个函数f
,它将对单个字符串输入执行操作并返回一个值。这个操作非常昂贵,所以我想在哈希中记住结果。
我知道我可以用这样的东西制作所需的哈希:
h = {}
arr.each { |a| h[a] = f(a) }
我想要做的是不必初始化h,以便我可以写下这样的东西:
h = arr.(???) { |a| a => f(a) }
能做到吗?
答案 0 :(得分:116)
假设您有一个具有虚假名称的功能:“f”
def f(fruit)
fruit + "!"
end
arr = ["apples", "bananas", "coconuts", "watermelons"]
h = Hash[ *arr.collect { |v| [ v, f(v) ] }.flatten ]
会给你:
{"watermelons"=>"watermelons!", "bananas"=>"bananas!", "apples"=>"apples!", "coconuts"=>"coconuts!"}
更新:
正如评论中所提到的,Ruby 1.8.7为此引入了更好的语法:
h = Hash[arr.collect { |v| [v, f(v)] }]
答案 1 :(得分:50)
对某些给定的答案做了一些快速,肮脏的基准测试。 (这些发现可能与您基于Ruby版本,奇怪的缓存等完全相同,但一般结果将类似。)
arr
是ActiveRecord对象的集合。
Benchmark.measure {
100000.times {
Hash[arr.map{ |a| [a.id, a] }]
}
}
Benchmark @ real = 0.860651,@ cstime = 0.0,@ cutime = 0.0,@ stime = 0.0,@ utime = 0.8500000000000005,@ total = 0.8500000000000005
Benchmark.measure {
100000.times {
h = Hash[arr.collect { |v| [v.id, v] }]
}
}
Benchmark @ real = 0.74612,@ cstime = 0.0,@ cutime = 0.0,@ stime = 0.010000000000000009,@ utime = 0.740000000000002,@ total = 0.750000000000002
Benchmark.measure {
100000.times {
hash = {}
arr.each { |a| hash[a.id] = a }
}
}
Benchmark @ real = 0.627355,@ cstime = 0.0,@ cutime = 0.0,@ stime = 0.010000000000000009,@ utime = 0.6199999999999974,@ total = 0.6299999999999975
Benchmark.measure {
100000.times {
arr.each_with_object({}) { |v, h| h[v.id] = v }
}
}
Benchmark @ real = 1.650568,@ cstime = 0.0,@ cutime = 0.0,@ stime = 0.12999999999999998,@ utime = 1.51,@ total = 1.64
仅仅因为Ruby具有表现力和动态性,并不意味着你应该总是选择最漂亮的解决方案。基本的每个循环在创建哈希方面是最快的。
答案 2 :(得分:30)
h = arr.each_with_object({}) { |v,h| h[v] = f(v) }
答案 3 :(得分:11)
这是我可能会写的:
h = Hash[arr.zip(arr.map(&method(:f)))]
简单,清晰,明显,陈述性。你还想要什么?
答案 4 :(得分:6)
Ruby 2.6.0通过passing a block to the to_h
method启用了较短的语法:
<!DOCTYPE html>
<html>
<head>
<title>Page Title</title>
</head>
<body>
<button onclick="alert(UserInput('prompt with input', ''))">prompt with input</button><br/>
<button onclick="alert(UserInput('prompt with mandantory input', '', 0))">prompt with mandantory input</button>
</body>
</html>
答案 5 :(得分:5)
我正如本文http://robots.thoughtbot.com/iteration-as-an-anti-pattern#build-a-hash-from-an-array
中所描述的那样array = ["apples", "bananas", "coconuts", "watermelons"]
hash = array.inject({}) { |h,fruit| h.merge(fruit => f(fruit)) }
有关inject
方法的更多信息:http://ruby-doc.org/core-2.0.0/Enumerable.html#method-i-inject
答案 6 :(得分:1)
另一个,稍微更清楚恕我直言 -
Hash[*array.reduce([]) { |memo, fruit| memo << fruit << f(fruit) }]
将长度用作f() -
2.1.5 :026 > array = ["apples", "bananas", "coconuts", "watermelons"]
=> ["apples", "bananas", "coconuts", "watermelons"]
2.1.5 :027 > Hash[*array.reduce([]) { |memo, fruit| memo << fruit << fruit.length }]
=> {"apples"=>6, "bananas"=>7, "coconuts"=>8, "watermelons"=>11}
2.1.5 :028 >
答案 7 :(得分:0)
除了弗拉多·辛格(Vlado Cingel)的答案(我还不能添加评论,所以我添加了答案)。
Inject也可以这种方式使用:该块必须返回累加器。只有块中的赋值会返回赋值,并报告错误。
array = ["apples", "bananas", "coconuts", "watermelons"]
hash = array.inject({}) { |h,fruit| h[fruit]= f(fruit); h }