获取列的和+；awk+；双引号+；迭代

这是我的档案：

$ cat -v test6 | head
"Rec_Open_Date"|"MSISDN"|"IMEI"|"Data_Volume_Bytes"|"Device_Manufacturer"|"Device_Model"|"Product_Description"|"Data_Volume_MB"|">20MB/30"|">200MB/30"|">2048MB/30"|">5120MB/30"|">10240MB/30"
"2015-10-06"|"427"|"060"|"137765"|"Samsung Korea"|"Samsung SM-G900I"|"$39 Plan"|"0.131383"|"0"|"0"|"0"|"0"|"0"
"2015-10-06"|"592"|"620"|"0"|"Apple Inc"|"Apple iPhone 6 (A1586)"|"PREPAY  STD - TRIAL - #16"|"0"|"0"|"0"|"0"|"0"|"0"
"2015-10-06"|"007"|"290"|"0"|"Apple Inc"|"Apple iPhone 6 (A1586)"|"PREPAY PLUS - $0 -"|"0"|"0"|"0"|"0"|"0"|"0"
"2015-10-06"|"592"|"050"|"48836832"|"Apple Inc"|"Apple iPhone 5S (A1530)"|"Talk and Text Connect Flexi Plan"|"46.5744"|"1"|"1"|"0"|"0"|"0"
"2015-10-06"|"409"|"720"|"113755347"|"Samsung Korea"|"Samsung SM-G360G"|"$29 CARRYOVER PLAN"|"108.486"|"1"|"1"|"1"|"0"|"0"
"2015-10-06"|"742"|"620"|"19840943"|"Apple Inc"|"Apple iPhone S (A1530)"|"PREPAY STD - $0 - #2"|"18.9218"|"1"|"1"|"0"|"0"|"0"
"2015-10-06"|"387"|"180"|"0"|"HUAWEI Technologies Co Ltd"|"HUAWEI HUAWEI G526-L11"|"PREPAY STD - $1 - #4"|"0"|"0"|"0"|"0"|"0"|"0"
"2015-10-06"|"731"|"570"|"2258243"|"Samsung Korea"|"Samsung SM-N910U"|"Business Freedom"|"2.15363"|"1"|"0"|"0"|"0"|"0"
"2015-10-06"|"556"|"910"|"13332272"|"Samsung Korea"|"Samsung GT-I9505"|"$49 Plan"|"12.7146"|"1"|"1"|"0"|"0"|"0"

这是我获得1列sume的方法，我必须删除“usinggsub”

$ awk -F'|' 'NR>1{n=$9; gsub(/"/,"",n); sum+=n} END {print sum}' test6
684

我想做的是实现类似这样的东西，它遍历每一列

awk '{for (i=1;i<=NF;i++) sum[i]+=$i;}; END{for (i in sum) print "for column "i" is " sum[i];}' FileA

作为和附加的问题，如果我能实现这样的目标，有什么好处呢？例如：

>20MB/30 is 684
>200MB/30 is x
>2048MB/30 is y
>5120MB/30 is z
>10240MB/30 is aa

我用

NR==1

尝试过这个方法，但没有成功

EDIT1可能在这里发现了什么

awk -F'|' 'NR>1{for (i=9;i<NF;i++) n=$i; gsub(/"/,"",n); sum[i]+=n} END {print sum[i]}' test6
24

---

EDIT3 对于以下答案，只需进行一些格式化：

awk -F'|' 'NR>1{for (i=9;i<=NF;i++) {gsub(/"/,"",$i); sum[i]+=$i}}; NR==1{for (i=9;i<=NF;i++) {col[i]=$i}};  END{for (i in sum) print "for column "col[i]"  the sum is " sum[i];}' test6
for column ">200MB/30"  the sum is 457
for column ">2048MB/30"  the sum is 86
for column ">5120MB/30"  the sum is 24
for column ">10240MB/30"  the sum is 6
for column ">20MB/30"  the sum is 684

awk-F'|''NR>1{对于（i=9；i200MB/30），总和为457
对于“>2048MB/30”列，总和为86
对于“>5120MB/30”列，总和为24
对于“>10240MB/30”列，总和为6
对于“>20MB/30”列，总和为684

试试：

gsub

修改

$i

并返回它所做的替换数，通常为2。上面的语句将所做的替换数相加。将其替换为：

{gsub(/"/,"",$i); sum[i]+=$i}

---

这将修改

$i

，然后将其添加到

sum[i]

中，而不是使用

|

作为分隔符，您可以使用

“

来代替（并去掉

gsub

）：

awk-F'\''NR>1{for（i=9；i
awk -F'|' 'NR>1{for (i=9;i<=NF;i++) {gsub(/"/,"",$i); sum[i]+=$i}}; NR==1{for (i=9;i<=NF;i++) {col[i]=$i}};  END{for (i in sum) print "for column "col[i]"  the sum is " sum[i];}' test6
for column ">200MB/30"  the sum is 457
for column ">2048MB/30"  the sum is 86
for column ">5120MB/30"  the sum is 24
for column ">10240MB/30"  the sum is 6
for column ">20MB/30"  the sum is 684

$ awk -F'|' '{for (i=9;i<=NF;i++) {gsub(/"/,"",$i); sum[i]+=$i}}; END{for (i in sum) print "for column "i" is " sum[i];}' test6
for column 9 is 5
for column 10 is 4
for column 11 is 1
for column 12 is 0
for column 13 is 0

sum[i]+=gsub(/"/,"",$i)

{gsub(/"/,"",$i); sum[i]+=$i}

 awk -F'\"' 'NR>1{for(i=9;i<NF/2;i++)sum[i]+=$(i*2)}END{for(i in sum) print "for column "i" is "sum[i]}' test