将两个范围连接成二维数组 Ruby
Joining two ranges into 2d array Ruby
如何在 ruby 中将两个范围连接成一个二维数组?使用 zip 无法提供我需要的结果。
(0..2) and (0..2)
# should become => [[0,0],[0,1],[0,2], [1,0],[1,1],[1,2], [2,0],[2,1],[2,2]]
range_a = (0..2)
range_b = (5..8)
def custom_join(a, b)
a.inject([]){|carry, a_val| carry += b.collect{|b_val| [a_val, b_val]}}
end
p custom_join(range_a, range_b)
输出:
[[0, 5], [0, 6], [0, 7], [0, 8], [1, 5], [1, 6], [1, 7], [1, 8], [2, 5], [2, 6], [2, 7], [2, 8]]
Ruby 有一个内置方法:repeated_permutation。
(0..2).to_a.repeated_permutation(2).to_a
直接解决方案:
range_a = (0..2)
range_b = (5..8)
def custom_join(a, b)
[].tap{|result| a.map{|i| b.map{|j| result << [i, j]; } } }
end
p custom_join(range_a, range_b)
输出:
[[0, 5], [0, 6], [0, 7], [0, 8], [1, 5], [1, 6], [1, 7], [1, 8], [2, 5], [2, 6], [2, 7], [2, 8]]
简单地说,这样做就可以了:
a = (0...2).to_a
b = (0..2).to_a
result = []
a.each { |ae| b.each { |be| result << [ae, be] } }
p result
# => [[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2]]
我很困惑。这是问题发布后的一天,没有人提出明显的建议:Array#product:
[*0..2].product [*1..3]
#=> [[0, 1], [0, 2], [0, 3], [1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3]]
如何在 ruby 中将两个范围连接成一个二维数组?使用 zip 无法提供我需要的结果。
(0..2) and (0..2)
# should become => [[0,0],[0,1],[0,2], [1,0],[1,1],[1,2], [2,0],[2,1],[2,2]]
range_a = (0..2)
range_b = (5..8)
def custom_join(a, b)
a.inject([]){|carry, a_val| carry += b.collect{|b_val| [a_val, b_val]}}
end
p custom_join(range_a, range_b)
输出:
[[0, 5], [0, 6], [0, 7], [0, 8], [1, 5], [1, 6], [1, 7], [1, 8], [2, 5], [2, 6], [2, 7], [2, 8]]
Ruby 有一个内置方法:repeated_permutation。
(0..2).to_a.repeated_permutation(2).to_a
直接解决方案:
range_a = (0..2)
range_b = (5..8)
def custom_join(a, b)
[].tap{|result| a.map{|i| b.map{|j| result << [i, j]; } } }
end
p custom_join(range_a, range_b)
输出:
[[0, 5], [0, 6], [0, 7], [0, 8], [1, 5], [1, 6], [1, 7], [1, 8], [2, 5], [2, 6], [2, 7], [2, 8]]
简单地说,这样做就可以了:
a = (0...2).to_a
b = (0..2).to_a
result = []
a.each { |ae| b.each { |be| result << [ae, be] } }
p result
# => [[0, 0], [0, 1], [0, 2], [1, 0], [1, 1], [1, 2]]
我很困惑。这是问题发布后的一天,没有人提出明显的建议:Array#product:
[*0..2].product [*1..3]
#=> [[0, 1], [0, 2], [0, 3], [1, 1], [1, 2], [1, 3], [2, 1], [2, 2], [2, 3]]