Linux 如何使用awk获得相同前4列的最小值

我需要一些帮助，以获得相同的前4列的最小值，同时保持那些没有重复的

输入文件：

20180509,1086776,0004,Car1,0.010000
20180509,1086776,0004,Car1,0.010002
20180509,1086776,0004,Car1,0.010002
20180509,1086776,0004,Car1,0.010002
20180509,1086776,0004,Car1,0.010001
20180509,1086776,0004,Car1,0.010002
20180509,1086776,0004,Car1,0.010003
20180509,1086776,0004,Car1,0.010004
20180509,1086776,0004,Car2,0.010004
20200509,1086776,0004,Car1,0.010005
20180509,0000006,0004,Car1,0.010005
20180509,1086776,0005,Car1,0.010005

输出：

20180509,1086776,0004,Car1,0.010000
20180509,1086776,0004,Car2,0.010004
20200509,1086776,0004,Car1,0.010005
20180509,0000006,0004,Car1,0.010005
20180509,1086776,0005,Car1,0.010005

输出文件的最后4行被保留，因为它具有前四列的唯一组合。有人能帮我吗？

---

解释也会有所帮助。谢谢

因此，我可以使用以下方法来实现这一点。我刚刚了解了如何放置一个键的多个列

awk -F, 'FNR==1 {rec=$0; m[$1,$2,$3,$4]=$5; next}
    {
        if (($1,$2,$3,$4) in m) {
            if ($5 < m[$1,$2,$3,$4]) {
                rec=$0
                m[$1,$2,$3,$4]=$5
            }
        }
        else {
            print rec
            rec=$0
            m[$1,$2,$3,$4]=$5
        }
    }
END {
    print rec
}' filename

awk-F，'FNR==1{rec=$0；m[$1，$2，$3，$4]=$5；next}
{
如果（$1，$2，$3，$4）单位为m）{
如果（$5

欢迎加入SO，请以代码的形式添加您的努力，这在SO上非常受鼓励（不是我的下一票），谢谢。按最后一个字段排序并选择唯一键，根据定义，它将为您提供每个唯一键的最小值。@Vinri Eclar-是否“相同的前4列”总是连续的？输出顺序需要如图所示吗？我试着检查其他代码，但它只涉及一列作为一个键，我不太擅长检查代码@Armali的前4列并不总是连续的。它们的顺序可能不同。最后，输出也不需要排序。