比较多个数组并删除重复项的最简单方法是什么?
所以(在这种情况下是数组内的数组)...
a = [[2, 1], [3, 3], [7, 2], [5, 6]]
b = [[2, 1], [6, 7], [9, 9], [4, 3]]
c = [[2, 1], [1, 1], [2, 2], [9, 9]]
d = [[2, 1], [9, 9], [2, 2], [3, 1]]
...会出来(优先给数组a,然后是b,然后是c,然后是d)
a = [[2, 1], [3, 3], [7, 2], [5, 6]]
b = [[6, 7], [9, 9], [4, 3]]
c = [[1, 1], [2, 2]]
d = [[3, 1]]
最佳答案
就是set difference或减法,你可以这样写。运算符重载可以是一种幸福:)
a
就是这样。
a
[[2, 1], [3, 3], [7, 2], [5, 6]]
b = b - a
[[6, 7], [9, 9], [4, 3]]
c = c - b - a # or c - (a + b)
[[1, 1], [2, 2]]
d = d - c - b - a # or d - (a + b + c)
[[3, 1]]
关于ruby - 在 Ruby 中比较数组并删除重复项?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/3365326/