Question

我有两个不同的数组，我需要一个方法来回复我：

array_1 = [1,2,3]
array_2 = ["a","b","c","d","e","f","g"]

def letter_to_number()
 #what goes here?
end

letter_to_number("g") #return 1
letter_to_number("a") #return 1
letter_to_number("b") #return 2
letter_to_number("f") #return 3

我对如何做到这一点的任何方式都不满意。我觉得可能有一种更简单的方法。

Answer 1

如果我理解正确，你的意思是得到这样的数字：

array1(repeat):  1 2 3 1 2 3 1 2 3
array2:          a b c d e f g

您可以在array_2中获取字母索引，按%的长度计算模数（array_1），并获取array_1中的值使用此索引：

array_1[array_2.index("g") % array_1.length]
# => 1

我将留下如何为您制作这样的方法。

Answer 2

def letter_to_number(array_1, array_2, letter)
  array_1[array_2.index(letter) % array_1.size]
end

array_1 = [1,2,3]
array_2 = ["a","b","c","d","e","f","g"]

letter_to_number(array_1, array_2, "g") #=> 1
letter_to_number(array_1, array_2, "a") #=> 1
letter_to_number(array_1, array_2, "b") #=> 2
letter_to_number(array_1, array_2, "f") #=> 3


array_1 = [1,2,3,4]
array_2 = ["a","b","c","d","e","f","g"]

letter_to_number(array_1, array_2, "g") #=> 3
letter_to_number(array_1, array_2, "a") #=> 1
letter_to_number(array_1, array_2, "b") #=> 2
letter_to_number(array_1, array_2, "f") #=> 2

Answer 3

上面的答案是正确的，除非需要处理一个角落案例，如下所示。

def letter_to_number(array_1, array_2, char)
  index = array_2.index(char) 
  (index)? array_1[index % array_1.length] : -1
end

Ruby循环数组相互之间

3 个答案: