在Ruby中,如果字符串不在选项数组中,我将如何返回true?
# pseudo code
do_this if current_subdomain not_in ["www", "blog", "foo", "bar"]
......或者你知道一个更好的方法吗?
答案 0 :(得分:53)
do_this unless ["www", "blog", "foo", "bar"].include?(current_subdomain)
或
do_this if not ["www", "blog", "foo", "bar"].include?(current_subdomain)
我正在使用Array#include?方法。
然而,使用unless
是一个相当大的红宝石成语。
答案 1 :(得分:5)
除了使用数组外,您还可以这样做:
case current_subdomain
when 'www', 'blog', 'foo', 'bar'; do that
else do this
end
这实际上要快得多:
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)
答案 2 :(得分: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/)
=> []
答案 3 :(得分:2)
您可以尝试exclude?
方法而不是include?
示例:
do_this if ["www", "blog", "foo", "bar"].exclude?(current_subdomain)
希望这有帮助......谢谢
答案 4 :(得分:1)
我发现这种方法最容易理解!'foo'.in?(['bar', 'baz'])