我有一个以下格式的文件:
Total:89.3
User: user1
Count:3
Sum:80
departmentId: dept1
Amount by departmentId: 20
departmentId: dept1
Amount by departmentId: 35
departmentId: dept2
Amount by departmentId: 25
User: user2
Count:3
Sum:7.199999999999999
departmentId: dept1
Amount by departmentId: 2.4
departmentId: dept2
Amount by departmentId: 2.4
departmentId: dept3
Amount by departmentId: 2.4
User: user3
Count:1
Sum:0.2
departmentId: dept2
Amount by departmentId: 0.2
User: user4
Count:2
Sum:2
departmentId: dept3
Amount by departmentId: 1
departmentId: dept3
Amount by departmentId: 1
文件列表基本上是部门的用户费用。如果同一用户多次到达某个部门,则需要将其合并为一行。输出文件需要采用以下格式。对于user1,他有dept1的2个会费和dept2的1个会费。因此在输出文件中,dept1的2个需要合并为1个计数需要为no。每个部门的唯一用户。
Format:
count_of_uique_user_dept_rows total_sum -- note** header row-->total sum and total no. of unique user dues
userId+deptId sum for that dept
Example:
7 89.3
user1dept1 55
user1dept2 25
user2dept1 2.4
user2dept2 2.4
user2dept3 2.4
user3dept2 0.2
user4dept3 2
我到目前为止,
# This awk script is used to convert the input of library credit/debit's to the required Student Accounts Load format
BEGIN { FS=": *" }
{
gsub(/^ +| +$/,"")
f[$1] = $2
}
/Amount/ {
dept = f["departmentId"]
total = f["Total"]
sum[dept] += $2
amount += $2
}
$1 == "User" {
if (NR>1) {
format()
}
user = $2
}
END { format() }
function format() {
if ( length(sum) > 0 ) {
for (dept in sum) {
printf "%-9s%-12s%10.2f\n", substr(user,1,9), substr(dept,1,12), sum[dept]
}
delete sum
amount = 0
}
}
上面的脚本为我们提供了数据行。我无法弄清楚如何获得7 89.3的标题行请帮忙。
我决定不再读取文件两次,只是在打印前将输出保存在数组中。以下是如何做到这一点:
第1步:修复当你假设sum
是标量时你会从某些awk中获得的语法错误,因为你在通过在BEGIN部分添加length(sum)
数组操作将它用作数组之前调用delete sum
(你可以删除它测试长度(总和),因为它没有对你的代码做任何有用的事情,但我想解释这个问题以及如何解决它一般)。
BEGIN { FS=": *"; delete sum }
第2步:更改format()
函数以加载稍后要输出的值数组,而不是立即输出这些值:
function format() {
if ( length(sum) > 0 ) {
for (dept in sum) {
vals[++numVals] = sprintf("%-9s%-12s%10.2f", substr(user,1,9), substr(dept,1,12), sum[dept])
}
delete sum
amount = 0
}
}
第3步:在END部分添加一个循环来实际执行打印:
END {
format()
for (valNr=1; valNr<=numVals; valNr++) {
print vals[valNr]
}
}
此时,您获得的输出将与现有脚本完全相同,但它使我们能够添加您需要的新功能:
第4步:将每个用户+ dept组合保存为数组usrdpt[]
的索引:
/Amount/ {
dept = f["departmentId"]
total = f["Total"]
sum[dept] += $2
usrdpt[user,dept]
amount += $2
}
步骤5:在打印先前的值之前,在END部分中打印新usrdpt[]
数组的唯一索引的数量:
END {
format()
print length(usrdpt)
for (valNr=1; valNr<=numVals; valNr++) {
print vals[valNr]
}
}
结果是:
$ cat tst.awk
BEGIN { FS=": *"; delete sum }
{
gsub(/^ +| +$/,"")
f[$1] = $2
}
/Amount/ {
dept = f["departmentId"]
total = f["Total"]
sum[dept] += $2
usrdpt[user,dept]
amount += $2
}
$1 == "User" {
if (NR>1) {
format()
}
user = $2
}
END {
format()
print length(usrdpt)
for (valNr=1; valNr<=numVals; valNr++) {
print vals[valNr]
}
}
function format() {
if ( length(sum) > 0 ) {
for (dept in sum) {
vals[++numVals] = sprintf("%-9s%-12s%10.2f", substr(user,1,9), substr(dept,1,12), sum[dept])
}
delete sum
amount = 0
}
}
.
$ awk -f tst.awk file
7
user1 dept1 55.00
user1 dept2 25.00
user2 dept1 2.40
user2 dept2 2.40
user2 dept3 2.40
user3 dept2 0.20
user4 dept3 2.00
我假设您可以弄清楚如何保存并稍后打印Total
值。
使用GNU awk和2d数组:
$ awk '
$1=="User:" { # store user
u=$NF
}
$1=="departmentId:" { # store dept
d=$NF
}
$1=="Amount" {
if(a[u][d]=="") # count uniq user/depts
c++
s+=$NF # total sum
a[u][d]+=$NF # user/dept sum
}
END {
printf "%s, %.2f\n",c,s # output count and total
for(u in a)
for(d in a[u])
printf "%s %s %.2f\n",u,d,a[u][d] # output user/dept sums
}' file
输出:
7 89.40
user1 dept1 55.00
user1 dept2 25.00
user2 dept1 2.40
user2 dept2 2.40
user2 dept3 2.40
user3 dept2 0.20
user4 dept3 2.00