如何使用以下两个数组构建数组:
名称 = [a, b, c]
how_many_of_each [3, 5, 2]
要得到
my_array = [a,a,a,b,b,b,b,b,c,c]
使用zip
、flat_map
和数组乘法:
irb(main):001:0> value = [:a, :b, :c]
=> [:a, :b, :c]
irb(main):002:0> times = [3, 5, 2]
=> [3, 5, 2]
irb(main):003:0> value.zip(times).flat_map { |v, t| [v] * t }
=> [:a, :a, :a, :b, :b, :b, :b, :b, :c, :c]
name.zip(how_many_of_each).inject([]) do |memo, (x, y)|
y.times { memo << x}
memo
end
=> [:a, :a, :a, :b, :b, :b, :b, :b, :c, :c]
编辑:哦,好吧,有更好的,见@David Grayson。
这将以一种易于理解的方式完成:
my_array = []
name.count.times do |i|
how_many_of_each[i].times { my_array << name[i] }
end
您可以选择您想要的(只需交换评论#
):
class Array
def multiply_times(how_many)
r = []
#how_many.length.times { |i| how_many[i].times { r << self[i] } }
self.each_with_index { |e, i| how_many[i].times { r << e } }
r
end
end
p ['a', 'b', 'c'].multiply_times([3, 5, 2])
#=> ["a", "a", "a", "b", "b", "b", "b", "b", "c", "c"]
array = ["a", "b", "c"]
how_many = [2, 2, 2]
result = []
array.each_with_index do |item, index|
how_many[index].times { result << item }
end
print result # => ["a", "a", "b", "b", "c", "c"]