awk '/^[+r]/{$1~/r/?r[$15]=$2:r[$15]?d[$15]=r[$15]-$2:1} END {for(p in d){sum+=r[p];num++}print sum/num}' trace.file
它会跳过所有不以“+”或“r”开头的行。如果该行以“r”开头，则会向 r 数组添加时间。否则，如果在 r 数组中找到该元素，它会计算延迟并将其添加到 d 数组中。最后，它循环 d 数组中的元素，将总延迟和元素数量相加，并计算平均值。在你的例子中，平均值是 0。
主块末尾的 :1 就在那里，所以我可以使用三元表达式而不是明显更冗长的 if 语句。
编辑：用于处理添加条件的新表达式：
awk '/^[+r]/{$1~/r/?$3>r[$15]?r[$15]=$3:1:!a[$15]||$3<a[$15]?a[$15]=$3:1} END {for(i in r){sum+=r[i]-a[i];num++}print "Average delay", sum/num}'
或作为 awk 文件
/^[+r]/ {
if ($1 ~ /r/) {
if ($3 > received[$15])
received[$15] = $3;
} else {
if (!added[$15] || $3 < added[$15])
added[$15] = $3;
}
} END {
for (packet in received) {
sum += received[packet] - added[packet];
num++
}
print "Average delay", sum/num
}
根据您的算法，似乎 1.745 将是开始时间，而您写的是 2.134。
There's not much to work with here, as Iain said in the comments to your question, but if I understand what you want to do correctly, something like this should work:
awk '/^[+r]/{$1~/r/?r[$15]=$2:r[$15]?d[$15]=r[$15]-$2:1} END {for(p in d){sum+=r[p];num++}print sum/num}' trace.file
It skips all lines not starting with '+' or 'r'. If the line starts with 'r' it adds time to the r array. Otherwise, it calculates the delay and adds it to the d array if the element is found in the r array. Finally it loops over the elements in the d array, adds up the total delay and number of elements and calculates the average from this. In your case the average is 0.
The :1 at the end of the main block is just in there so I can get away with a ternary expression instead of the significantly more verbose if statement.
EDIT: New expression to work with the added conditions:
awk '/^[+r]/{$1~/r/?$3>r[$15]?r[$15]=$3:1:!a[$15]||$3<a[$15]?a[$15]=$3:1} END {for(i in r){sum+=r[i]-a[i];num++}print "Average delay", sum/num}'
or as an awk-file
/^[+r]/ {
if ($1 ~ /r/) {
if ($3 > received[$15])
received[$15] = $3;
} else {
if (!added[$15] || $3 < added[$15])
added[$15] = $3;
}
} END {
for (packet in received) {
sum += received[packet] - added[packet];
num++
}
print "Average delay", sum/num
}
According to your algorithm it seems like 1.745 would be the start time, while you write that 2.134 is.
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正如伊恩在对你的问题的评论中所说，这里没有太多可以处理的内容，但是如果我理解你想要正确执行的操作，那么这样的事情应该可以工作：
它会跳过所有不以“+”或“r”开头的行。如果该行以“r”开头，则会向 r 数组添加时间。否则，如果在 r 数组中找到该元素，它会计算延迟并将其添加到 d 数组中。最后，它循环 d 数组中的元素，将总延迟和元素数量相加，并计算平均值。在你的例子中，平均值是 0。
主块末尾的
:1就在那里，所以我可以使用三元表达式而不是明显更冗长的 if 语句。编辑：用于处理添加条件的新表达式：
或作为 awk 文件
根据您的算法，似乎 1.745 将是开始时间，而您写的是 2.134。
There's not much to work with here, as Iain said in the comments to your question, but if I understand what you want to do correctly, something like this should work:
It skips all lines not starting with '+' or 'r'. If the line starts with 'r' it adds time to the r array. Otherwise, it calculates the delay and adds it to the d array if the element is found in the r array. Finally it loops over the elements in the d array, adds up the total delay and number of elements and calculates the average from this. In your case the average is 0.
The
:1at the end of the main block is just in there so I can get away with a ternary expression instead of the significantly more verbose if statement.EDIT: New expression to work with the added conditions:
or as an awk-file
According to your algorithm it seems like 1.745 would be the start time, while you write that 2.134 is.