我希望能够对输入csv文件进行排序，该文件由额外列中创建的值逗号分隔。以下是输入csv文件的示例
Timestamp,Email,Name,Year,Make,Model,Car_ID,Judge_ID,Judge_Name,Racer_Turbo,Racer_Supercharged,Racer_Performance,Racer_Horsepower,Car_Overall,Engine_Modifications,Engine_Performance,Engine_Chrome,Engine_Detailing,Engine_Cleanliness,Body_Frame_Undercarriage,Body_Frame_Suspension,Body_Frame_Chrome,Body_Frame_Detailing,Body_Frame_Cleanliness,Mods_Paint,Mods_Body,Mods_Wrap,Mods_Rims,Mods_Interior,Mods_Other,Mods_ICE,Mods_Aftermarket,Mods_WIP,Mods_Overall
8/5/2018 14:10,honoland13@japanpost.jp,Hernando,2015,Acura,TLX,48,J04,Bob,0,0,2,2,4,4,0,2,4,4,2,4,2,2,2,2,2,0,4,4,4,6,2,0,4
8/5/2018 15:11,nlighterness2q@umn.edu,Noel,2015,Jeep,Wrangler,124,J02,Carl,0,6,4,2,4,6,6,4,4,4,6,6,6,6,6,4,6,6,6,6,6,4,6,4,6
8/5/2018 17:10,eguest47@microsoft.com,Edan,2015,Lexus,Is250,222,J05,Adrian,0,0,0,0,0,0,0,0,6,6,6,0,0,6,6,6,0,0,0,0,0,0,0,0,4
8/5/2018 17:34,hchilley40@fema.gov,Hieronymus,1993,Honda,Civic eG,207,J06,Aaron,0,0,2,2,2,2,2,2,0,4,2,2,2,2,2,2,4,2,2,0,0,0,2,2,0
8/5/2018 14:30,nnowick3d@tuttocitta.it,Nickolas,2016,Ford,Mystang,167,J02,Carl,0,0,2,2,0,2,2,0,0,0,0,2,0,2,2,2,0,0,2,0,0,0,0,0,2
8/5/2018 16:12,mdearl39@amazon.co.uk,Martin,2013,Hyundai,Gen coupe,159,J04,Bob,0,0,2,0,0,0,2,0,0,0,0,2,0,2,2,0,2,0,2,0,0,0,0,0,0
8/5/2018 17:00,alynamg@blogtalkradio.com,Aldridge,2009,Infiniti,G37,20,J06,Aaron,2,0,2,2,0,0,2,0,0,2,2,2,2,2,2,2,2,2,4,2,2,0,2,0
我的代码目前所做的是筛选csv文件，并选择car_id列、year列、make列和model列。然后，它遍历从racer_turbo到最后一列的每一列，对于每一行，它将这些列中的值相加为一个总值，并将其与其他值(id、make、model等(一起打印。打印时，还有一个排名列在其他5列之前。下面是我的代码。
BEGIN {
FS = ",";
OFS = "t";
print "Ranking", "Car_ID", "Year", "Make", "Model", "Total";
}
{
rank;
total = 0;
if(NR > 1) {
for(i = 8; i < NF; i++) {
total += $i;
}
print ++rank,$7, $4, $5, $6, total;
}
rows[$5][total][$0]
}
END {
print "n";
print "Ranking", "Car_ID", "Year", "Make", "Model", "Total";
ranking;
PROCINFO["sorted_in"] = "@ind_str_asc"
for (m in rows) {
n = asorti(rows[m], t, "@ind_num_desc");
n = (n>3) ? 3 : n
for(i = 1; i <= n; i++) for(s in rows[m][t[i]]) {
$0 = s;
$1 = ++r;
print ++ranking, $7, $4, $5, $6, total;
}
}
}
我想在END块中再次打印输出，但是，使用在代码的前一块中创建的total列，按照每个品牌的前三名对汽车进行排名。然而，我现在运行的代码的输出如下
Ranking Car_ID Year Make Model Total
1 48 2015 Acura TLX 58
2 124 2015 Jeep Wrangler 118
3 222 2015 Lexus Is250 36
4 207 1993 Honda Civic eG 40
5 167 2016 Ford Mystang 18
6 159 2013 Hyundai Gen coupe 14
7 20 2009 Infiniti G37 36
...
Ranking Car_ID Year Make Model Total
1 113 2012 Acura Tsx sportwagon 10
2 112 2008 Acura TL 10
3 50 2015 Acura TLX 10
4 15 2014 Audi S4 10
5 18 2015 Audi S3 10
6 116 2008 Audi A4 10
7 2 2016 Bmw M2 10
8 172 2014 Bmw 4 10
9 28 1995 Bmw 318xi 10
...
看看在第二个打印部分的合计列中，它如何显示每个打印的汽车的合计为10，而不是与第一个打印部分中每个汽车的值相同，并且显示每个品牌的最高3个合计。
以下是预期输出
Ranking Car_ID Year Make Model Total
1 48 2015 Acura TLX 58
2 124 2015 Jeep Wrangler 118
3 222 2015 Lexus Is250 36
4 207 1993 Honda Civic eG 40
5 167 2016 Ford Mystang 18
6 159 2013 Hyundai Gen coupe 14
7 20 2009 Infiniti G37 36
8 178 2009 Honda Oddesy 66
...
Ranking Car_ID Year Make Model Total
1 112 2008 Acura TL 110
2 50 2015 Acura TLX 102
3 127 2013 Acura Tsx 86
4 15 2014 Audi S4 120
5 18 2015 Audi S3 38
6 116 2008 Audi A4 28
7 2 2016 Bmw M2 24
8 172 2014 Bmw 4 22
9 111 2007 Bmw 328i 10
10 218 2010 Chevy Camaro 64
11 170 2014 Chevy Cruze 50
12 0 2015 Chevy Camaro 0
...
用我目前的代码可以挽救吗？或者，更好的方法是创建一个单独的awk文件，对生成的输出进行排序，并生成另一个按前3名排序的文件？
我正在运行GNU AWK v4.0.2。
假设Car_ID(以下简称为id(在行之间是唯一的，请尝试：
BEGIN {
FS = ","
OFS = "t"
print "Ranking", "Car_ID", "Year", "Make", "Model", "Total"
}
{
rank
total = 0
if (NR > 1) {
for (i = 8; i < NF; i++) {
total += $i
}
print ++rank, $7, $4, $5, $6, total
ttl[$5][$7] = total
row[$7] = $0
}
}
END {
print "n"
print "Ranking", "Car_ID", "Year", "Make", "Model", "Total"
ranking
id
PROCINFO["sorted_in"] = "@ind_str_asc"
for (m in ttl) {
n = asorti(ttl[m], t, "@val_num_desc")
n = (n>3) ? 3 : n
for (i = 1; i <= n; i++) {
id = t[i]
total = ttl[m][id]
$0 = row[id]
print ++ranking, $7, $4, $5, $6, total
}
}
}
我稍微修改了数据结构，将id指定为主键。然后创建一个二维数组ttl，该数组包含值total由CCD_ 6和CCD_。在END循环中，我们可以检索使用CCD_ 9
附带说明一下，您的原始数据结构使用total作为索引。如果具有相同值的多行恰好具有相同值则total的任何一个索引都将被覆盖。