在Ruby中，如何查找字符串是否不在數組中？

Question

在Ruby中，如果字符串不在選項數組中，我將如何返回true？

# pseudo code 
do_this if current_subdomain not_in ["www", "blog", "foo", "bar"] 

...或者您是否知道更好的書寫方式？

Answer 1

do_this unless ["www", "blog", "foo", "bar"].include?(current_subdomain) 

或

do_this if not ["www", "blog", "foo", "bar"].include?(current_subdomain) 

我使用的Array#include?方法。

但是使用unless是一個相當大的ruby成語。

Answer 2

do this if not ["www", "blog", "foo", "bar"].include?(current_subdomain) 

，或者您可以使用grep

>> d=["www", "blog", "foo", "bar"] 
>> d.grep(/^foo$/) 
=> ["foo"] 
>> d.grep(/abc/) 
=> [] 

Answer 3

除了使用數組，你也可以這樣做：

case current_subdomain 
when 'www', 'blog', 'foo', 'bar'; do that 
else do this 
end 

這實際上是要快得多：

require 'benchmark' 
n = 1000000 

def answer1 current_subdomain 
    case current_subdomain 
    when 'www', 'blog', 'foo', 'bar' 
    else nil 
    end 
end 

def answer2 current_subdomain 
    nil unless ["www", "blog", "foo", "bar"].include?(current_subdomain) 
end 

Benchmark.bmbm do |b| 
    b.report('answer1'){n.times{answer1('bar')}} 
    b.report('answer2'){n.times{answer2('bar')}} 
end 

Rehearsal ------------------------------------------- 
answer1 0.290000 0.000000 0.290000 ( 0.286367) 
answer2 1.170000 0.000000 1.170000 ( 1.175492) 
---------------------------------- total: 1.460000sec 

       user  system  total  real 
answer1 0.290000 0.000000 0.290000 ( 0.282610) 
answer2 1.180000 0.000000 1.180000 ( 1.186130) 


Benchmark.bmbm do |b| 
    b.report('answer1'){n.times{answer1('hello')}} 
    b.report('answer2'){n.times{answer2('hello')}} 
end 

Rehearsal ------------------------------------------- 
answer1 0.250000 0.000000 0.250000 ( 0.252618) 
answer2 1.100000 0.000000 1.100000 ( 1.091571) 
---------------------------------- total: 1.350000sec 

       user  system  total  real 
answer1 0.250000 0.000000 0.250000 ( 0.251833) 
answer2 1.090000 0.000000 1.090000 ( 1.090418) 

Answer 4

你可以試試exclude?方法，而不是include?

例子：

do_this if ["www", "blog", "foo", "bar"].exclude?(current_subdomain) 

希望這有助於...謝謝