gawk如何实现具有相同字段的多行则输出最后字段数值最大的行

愤怒的CMD

有文本a.txt，内容如下

AAA,BBB,CC,D,EEEEEEEE,950
 SSS,GGG,CC,F,EEEEEEEE,1000
 SSS,GGG,CC,F,EEEEEEEE,800
 PPP,YYY,FF,E,DDDDDDDD,1250
 KKK,JJJ,HH,D,TTTTTTTT,400
 KKK,JJJ,HH,D,TTTTTTTT,1200
 KKK,JJJ,HH,D,TTTTTTTT,800COPY

文件中存在以下情况，2条以上的记录前5项相同，最后1项不同，如：
SSS,GGG,CC,F,EEEEEEEE,1000
SSS,GGG,CC,F,EEEEEEEE,800

KKK,JJJ,HH,D,TTTTTTTT,400
KKK,JJJ,HH,D,TTTTTTTT,1200
KKK,JJJ,HH,D,TTTTTTTT,800
希望输出结果：
AAA,BBB,CC,D,EEEEEEEE,950
SSS,GGG,CC,F,EEEEEEEE,1000
PPP,YYY,FF,E,DDDDDDDD,1250
KKK,JJJ,HH,D,TTTTTTTT,1200

即：
1、前5项没有相同记录的，直接输出；
2、前5项有相同记录的，输出第6项数字最大的记录。

希望高手指点，谢谢！

pcl_test

//&cls&cscript -nologo -e:jscript "%~f0"<"1.txt"&pause&exit
var map={}, s='';
while(!WSH.StdIn.AtEndOfStream){
    var str = WSH.StdIn.ReadLine();
    var a = str.replace(/[^\,]+$/, '').replace(/^\s*/, '');
    var b = str.replace(/^.+\,/g, '').replace(/\s*$/, '');
    if(!map[a]){
        map[a]=b;
     }else{
        if(map[a]>=b)map[a]=b;
     }
}
for(var m in map)s+=m+map[m]+'\r\n'
WSH.Echo(s);COPY

愤怒的CMD

谢谢pcl_test ，我最想知道的gawk怎样解决这个问题

bailong360

@echo off
<%0 more +4 >$.awk
gawk -f $.awk a.txt
del $.awk&exit
BEGIN {FS=","}
{
    if (data[$1","$2","$3","$4","$5]==0) 
        data[$1","$2","$3","$4","$5]=$6
     else 
        data[$1","$2","$3","$4","$5]=$6>data[$1","$2","$3","$4","$5]?$6:data[$1","$2","$3","$4","$5]
    
}
END {
    for (line in data)
        printf("%s,%d\n",line,data[line])>"$new.txt"
}COPY

拙作,楼下既然有更好的代码了,就放在这里当作纪念吧

CrLf

回复 4# bailong360


    直接用 data[$1,$2,$3,$4,$5] 就好，这里的 , 相当于一个文件分割符

bailong360

回复 5# CrLf


    我原先也是这样的,后来发现输出以后没有逗号,所以改成了这个样子

看了下兄弟楼下的代码,感叹自己的思路实在是僵硬啊..

CrLf

gawk -F"," "$6>data[$1,$2,$3,$4,$5]{data[$1,$2,$3,$4,$5]=$0} END{for (line in data)print data[line]>\"$new.txt\"}" a.txt
代码有误，更正见 10 楼

yiwuyun

佩服。好像这样也可

gawk -F"," "$6>data[$1$2]{data[$1$2]=$0} END{for (line in data)print data[line]}" a.txtCOPY

改成这样：

gawk -F"," "$6>a[split(data[$1$2],a,\",\")]{data[$1$2]=$0} END{for (line in data)print data[line]}" 2.txtCOPY

愤怒的CMD

感谢大家的帮助，今天没法评分了，明天补上

CrLf

卧槽，大伙没发现我写错了吗，用 $6 和 $0 来比较的...
修改下，还是像 4 楼那样用逗号分隔吧

gawk -F"," -v d="," "$6>data[$1 d $2 d $3 d $4 d $5]{data[$1 d $2 d $3 d $4 d $5]=$6} END{for (line in data)print line d data[line]}" a.txtCOPY

或用 gsub

gawk -F"," "$6>data[$1,$2,$3,$4,$5]{data[$1,$2,$3,$4,$5]=$6} END{for (line in data){l=line;gsub(/\x1c/,\",\",l);print l \",\" data[line]}}" a.txtCOPY

分隔符不同，排序居然还不一样

yiwuyun

$6和$0比较，就是$6和$6比较。逗号分隔符的计算就是以最后一个为准。认识错了。
gawk -F"," "$6>int(data[$1$2]){data[$1$2]=$0} END{for(line in data)print data[line]}" a.txt

愤怒的CMD

回复 10# CrLf


感谢帮助，我对这个还是了解的太少了

愤怒的CMD

回复 8# yiwuyun

谢谢帮助，我觉得还是data[$1$2$3$4$5]比较好，否则第三到5列要是有不一样的就无法区分了

CrLf

回复 8# yiwuyun


    a[split(data[$1$2],a,\",\")] 好精巧！可以去掉 \",\"，因为你已经设了 FS 了
    但 $1$2 是没有分隔符的，怎么区分 SSS,GGG 和 SSSG,GG 呢？

yiwuyun

嗯。多谢指正，考虑不周，实际应用应多取几段，或者再多加几根下划线。这样重复的概率就小多了。