Question

我希望能够比较两个共同的数组，而不是重复数据。

array1 = ['1. walking to the park to find some ice cream']
array2 = ['2. walking to the park to find ice cream and find more ice cream']

示例输出：

walking to the park find ice cream

Answer 1

以两个字符串开头
在空格上分割它们
对它们进行数组交集

a = '1. walking to the park to find some ice cream'
b = '2. walking to the park to find ice cream and find more ice cream'

a1 = a.split(" ")
b1 = b.split(" ")

common = a1 & b1
#=> ["walking", "to", "the", "park", "find", "ice", "cream"]

正则表达式将会变慢。这是数组联合和使用Regexp之间的快速比较：

require 'benchmark'

def arr_only(a,b)
  a1 = a.split(" ")
  b1 = b.split(" ")
  a1 & b1
end

def w_regex(a,b)
  match = Regexp.union(a.split(" "))
  b.scan(match).uniq
end

n = 100_000
Benchmark.bm do |x|
  x.report("arr:") { n.times {|i| arr_only(a, b) } }
  x.report("regex:"){ n.times {|i| w_regex(a, b) } }
end

#                   user     system      total        real
# arr:          1.030000   0.000000   1.030000 (  1.031033)
# regex:        4.970000   0.010000   4.980000 (  4.993263)

Answer 2

我会这样做：

a = '1. walking to the park to find some ice cream'
b = '2. walking to the park to find ice cream and find more ice cream'

match = Regexp.union(a.split(" "))
b.scan(match).uniq.join(" ")
# => "walking to the park find ice cream"

ruby比较两个共同的数组

2 个答案: