我想我明白了:

a = [3, 2, 4, 6, 3, 8]
a.delete(3)
#=> 3
a
#=> [2, 4, 6, 8]

A .delete_at(3) 3在这里是位置。

你可以简单地运行:

[2,4,6,3,8].delete(3)

首次发生的无损清除:

a = [2, 4, 6, 3, 8]
n = a.index 3
a.take(n)+a.drop(n+1)

因此，当3出现多次，而你只想删除第一次出现的3时，你可以简单地执行如下操作。

arr = [2, 4, 6, 3, 8, 10, 3, 12]

arr.delete_at arr.index 3

#This will modify arr as [2, 4, 6, 8, 10, 3, 12] where first occurrence of 3 is deleted. Returns the element deleted. In this case => 3.

另一个选择:

a = [2,4,6,3,8]

a -= [3]

结果是

=> [2, 4, 6, 8]