我有两个二维阵列,比较二维阵列
a = [[17360, "Z51.89"],
[17361, "S93.601A"],
[17362, "H66.91"],
[17363, "H25.12"],
[17364, "Z01.01"],
[17365, "Z00.121"],
[17366, "Z00.129"],
[17367, "K57.90"],
[17368, "I63.9"]]
和
b = [[17360, "I87.2"],
[17361, "s93.601"],
[17362, "h66.91"],
[17363, "h25.12"],
[17364, "Z51.89"],
[17365, "z00.121"],
[17366, "z00.129"],
[17367, "k55.9"],
[17368, "I63.9"]]
我想在两个阵列来计算类似的行,而不管字符案件,即"h25.12"
将等于"H25.12"
。
我试过,
count = a.count - (a - b).count
但(a - b)
收益
[[17360, "Z51.89"],
[17361, "S93.601A"],
[17362, "H66.91"],
[17363, "H25.12"],
[17364, "Z01.01"],
[17365, "Z00.121"],
[17366, "Z00.129"],
[17367, "K57.90"]]
我需要算作5
因为有5个类似的行,当我们不考虑字符大小写。
为什么选择近距离投票? – 2015-03-25 06:39:08
这些不是二维数组,Ruby没有这种东西(除非你想要计算Matrix),那些数组就是数组。 – 2015-03-25 06:47:33
好的!这意味着按照惯例,Ruby没有“二维数组”? – 2015-03-25 07:03:13