groovy学习(一)列表

numbers = [11, 12, 13, 14]
println(numbers[0])
println(numbers[3])
println(numbers[-1])//最左边的元素
println(numbers[-3])//最左边的第3元素
println(numbers[0..2])//第0, 1, 2元素
println(numbers[1..<3])//第1,2个元素

numbers[1] = 22//给第1个元素赋值
println(numbers)

numbers[1] = [33, 44]//将第1个元素替换为列表,numbers变成列表的列表
println(numbers)

numbers << 15
println(numbers) //添加到最后

numbers = [11, 12, 13, 14]
numbers += [15, 16] //连接列表
println(numbers)

numbers = [11, 12, 13, 14]
numbers.add(15) //注意这里返回的是true,表示添加成功
println(numbers)

numbers = [11, 12, 13, 14]
numbers.add(2, 15) //第2个位置添加15,第2个以及以后的元素住后移
println(numbers)

numbers = [11, 12, 13, 14]
println(numbers.get(1))//返回第1个位置的元素

numbers = [11, 12, 13, 14]
println(numbers.isEmpty())//是否为空

numbers = [11, 12, 13, 14]
println(numbers.size())//长度

println([11, 12, [13, 14]].flatten()) //嵌套的列表转化成一个同有嵌套的列表

numbers = [11, 12, 13, 14]
println(numbers.getAt(1))//返回第一个元素

numbers = [11, 12, 13, 14]
println(numbers.getAt(1..2))//返回第1到2个元素

numbers = [11, 12, 13, 14]
println(numbers.getAt([0, 2, 3]))//返回第0, 2, 3个元素

numbers = [11, 12, 13, 14]
println(numbers.intersect([13, 14, 15, 11]))//取交集

numbers = [11, 12, 13, 14]
println(numbers.pop())//返回最后一个元素,相当于栈顶

numbers = [11, 12, 13, 14]
println(numbers.reverse()) //逆序

numbers = [14, 13, 11, 12]
println(numbers.sort()) //排序

numbers = [14, 13, 11, 12]
println(numbers.remove(3))//返回删除的第3个元素
println(numbers)

numbers = ['Ken', 'John', 'Sally', 'Jon']
println(numbers.remove('Ken')) //Ken在列表中返回true,不在返回false
println(numbers)

转载于:https://www.cnblogs.com/zhouyang209117/p/4899385.html