groovy学习之列表操作--操作列表元素

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    def "过滤列表"() {
        def ages = [20, 36, 42, 56]
        def midage = 21..50
        expect:
        ages.grep(midage) == [36, 42]
        println ages.grep(midage)
    }

    def "list of lists,demo flatten"() {

        def list = [[1, 2], [3, 4, 5]].flatten()
        expect:
        list == [1, 2, 3, 4, 5]
        println list
    }

    def "demo list intersect("() {
        def list1 = [1, 2, 3]
        def list2 = [2, 3, 4]
        expect: "求交集"

        list1.intersect(list2) == [2, 3]
    }

    def "demo list disjoint("() {
        expect: "不交"
        assert [1, 2, 3].disjoint([4, 5, 6])
    }

    def "demo list pop,like stack("() {
        def list = [1, 2, 3]
        def popped = list.pop()
        expect:
        popped == 3
        list == [1, 2]
    }

    def "列表排序和反转"() {
        assert [1, 2].reverse() == [2, 1]
        assert [3, 1, 2].sort() == [1, 2, 3]

        def list = [[1, 0], [0, 1, 2]]
        when: "根据第一个元素排序，默认升序 asc"
        list = list.sort { a, b -> a[0] <=> b[0] }
        then:
        list == [[0, 1, 2], [1, 0]]

        when: "根据size大小来排序"
        list.sort { item -> item.size() }
        then:
        list == [[1, 0], [0, 1, 2]]

    }

    def "两种方法，列表元素去重"() {
        def x = [1, 1, 1]
        expect:
        assert [1] == new HashSet(x).toList()
        assert [1] == x.unique()
    }

    def "列表去null"() {
        def x = [1, null, 1]
        expect:
        assert [1, 1] == x.findAll { it != null }
        assert [1, 1] == x.grep { it }
    }

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