去重结构数组

Question

我有一个结构数组，我想删除所有重复元素但保留数组中的最后一个元素。类似于 hashmap 的东西，我可以将每次匹配的最后一个结构更新到新数组

我有一个像这样的结构

type samplestruct struct {
    value1           string    
    value2           string   
    value3           string   
    value4           string
    value5           string
}

在我的结构数组中，如果任何结构的 value1、value2 和 value3 相同，则删除所有重复项并保留最后一个结构。

func unique(sample []samplestruct) []samplestruct {
    var unique []samplestruct

    for _, v := range sample {
        skip := false
        for _, u := range unique {
            if v.value1 == u.value1 && v.value2 == u.value2 && v.value3 == u.value3 {
                skip = true
                break
            }
        }
        if !skip {
            unique = append(unique, v)
        }
    }
    return unique

}

此代码返回与提供的条件匹配的第一个结构，但我想要与条件匹配的最后一个结构

给定输入 -

[
samplestruct{"ram","rahim","india","34","india"},
samplestruct{"ram","rahim","india","38","America"},
samplestruct{"ram","rahim","india","40","Jamica"},
samplestruct{"amit","rawat","bangladesh","35","hawai"},
samplestruct{"amit","rawat","bangladesh","36","india"}
]

预期产出 -

[
samplestruct{"ram","rahim","india","40","Jamica"},
samplestruct{"amit","rawat","bangladesh","36","india"}
]

Answer 1

问题中的代码几乎就在那里。当在 中找到匹配元素时unique，用当前值覆盖该元素：

func unique(sample []samplestruct) []samplestruct {
    var unique []samplestruct
sampleLoop:
    for _, v := range sample {
        for i, u := range unique {
            if v.value1 == u.value1 && v.value2 == u.value2 && v.value3 == u.value3 {
                unique[i] = v
                continue sampleLoop
            }
        }
        unique = append(unique, v)
    }
    return unique
}

根据数据集的大小和幸存元素的数量，其他答案中显示的基于地图的方法可能更合适。这是地图方法的正确实现：

func unique(sample []samplestruct) []samplestruct {
    var unique []samplestruct
    type key struct{ value1, value2, value3 string }
    m := make(map[key]int)
    for _, v := range sample {
        k := key{v.value1, v.value2, v.value3}
        if i, ok := m[k]; ok {
            // Overwrite previous value per requirement in
            // question to keep last matching value.
            unique[i] = v
        } else {
            // Unique key found. Record position and collect
            // in result.
            m[k] = len(unique)
            unique = append(unique, v)
        }
    }
    return unique
}