File shell脚本中的CSV文件日期处理

我有一个csv日志文件，有两列，每列都有请求（第1列）和响应（第2列）的时间戳

以下是一些示例数据：

2013-07-11 08:39:08.748,2013-07-11 08:39:08.748
2013-07-11 08:39:08.826,2013-07-11 08:39:08.826
2013-07-11 08:39:08.860,2013-07-11 08:39:08.860
2013-07-11 08:39:08.919,2013-07-11 08:39:08.919
2013-07-11 08:39:08.941,2013-07-11 08:39:08.941
2013-07-11 08:39:09.390,2013-07-11 08:39:09.390
2013-07-11 08:39:09.594,2013-07-11 08:39:09.594
2013-07-11 08:39:09.619,2013-07-11 08:39:09.619
2013-07-11 08:39:09.787,2013-07-11 08:39:09.787
2013-07-11 08:39:10.006,2013-07-11 08:39:10.006
2013-07-11 08:39:10.017,2013-07-11 08:39:10.017
2013-07-11 08:39:10.088,2013-07-11 08:39:10.088
2013-07-11 08:39:10.214,2013-07-11 08:39:10.214

我想计算完整文件的两列（响应-请求）差异的平均值。这个文件一天可以包含一百万个条目

我正在寻找一种使用shell脚本的方法。 请帮忙

谢谢Fedorqui，我试过你的剧本：

while read line; 
do 
d1=$(echo $line | cut -d, -f1); 
d2=$(echo $line | cut -d, -f2);  
ds1=$(date -d"$d1" "+%s"); 
ds2=$(date -d"$d2" "+%s"); 
echo "$ds2 - $ds1 = $(( $ds1 - $ds2))"; 
done < requestResponse.csv

---

我需要以毫秒为单位的差异。

我在基本shell脚本中以纳秒为单位完成了这项工作：

while read line
do
   d1=$(echo $line | cut -d, -f1)
   d2=$(echo $line | cut -d, -f2)
   ds1=$(date -d"$d1" "+%s.%N")
   ds2=$(date -d"$d2" "+%s.%N")
   diff=$(echo "$ds2 - $ds1" | bc)
   printf "$ds2 - $ds1 = %.3f\n" $diff
done < file

读取行时
做
d1=$（回声$线|切割-d，-f1）
d2=$（回声$行|切割-d，-f2）
ds1=$（日期-d“$d1”“+%s.%N”）
ds2=$（日期-d“$d2”“+%s.%N”）
差异=$（回声“$ds2-$ds1”| bc）
printf“$ds2-$ds1=%.3f\n”$diff
完成<文件

试验

$whilereadline；do d1=$（回声$线|切-d，-f1）；d2=$（回声$行|切割-d，-f2）；ds1=$（日期-d“$d1”“+%s.%N”）；ds2=$（日期-d“$d2”“+%s.%N”）；差异=$（回声“$ds2-$ds1”| bc）；printf“$ds2-$ds1=%.3f\n”$diff；完成<文件
1373524748.748000000 - 1373524748.748000000 = 0.000
1373524748.826000000 - 1373524748.826000000 = 0.000
1373524748.860000000 - 1373524748.860000000 = 0.000
1373524748.919000000 - 1373524748.919000000 = 0.000
1373524748.941000000 - 1373524748.941000000 = 0.000
1373524749.390000000 - 1373524749.390000000 = 0.000
1373524749.594000000 - 1373524749.594000000 = 0.000
1373524749.619000000 - 1373524749.619000000 = 0.000
1373524749.787000000 - 1373524749.787000000 = 0.000
1373524750.006000000 - 1373524750.006000000 = 0.000
1373524750.017000000 - 1373524750.017000000 = 0.000
1373524750.088000000 - 1373524750.088000000 = 0.000
1373524750.214000000 - 1373524750.214000000 = 0.000

awk可以通过mktime（）或外部

date

来完成。但我可以在不编写任何代码的情况下立即告诉您示例的输出。：），它们都是“零”。：事实上是零。这就是为什么把它贴在这里的原因（

读取行时；do d1=$（echo$line | cut-d，-f1）；d2=$（echo$line | cut-d，-f2）；ds1=$（date-d“$d1”“+%s”）；ds2=$（date-d“$d2”“+%s”）；echo“$ds2-$ds1=$（$ds1-$ds2））；完成

将为您提供“date1-date2=diff”转换为UNIX stamp后。使用此输出，您可以轻松计算平均值。@fedorqui我认为您应该将其放入答案中。此外，我猜（不确定）OP可能希望得到ms级别的结果。（仅从示例数据中猜测）@LearnMore当前数据始终为0。但是，我更新了答案，以毫秒级显示。非常适合我。非常感谢。

while read line
do
   d1=$(echo $line | cut -d, -f1)
   d2=$(echo $line | cut -d, -f2)
   ds1=$(date -d"$d1" "+%s.%N")
   ds2=$(date -d"$d2" "+%s.%N")
   diff=$(echo "$ds2 - $ds1" | bc)
   printf "$ds2 - $ds1 = %.3f\n" $diff
done < file

$ while read line; do d1=$(echo $line | cut -d, -f1); d2=$(echo $line | cut -d, -f2); ds1=$(date -d"$d1" "+%s.%N"); ds2=$(date -d"$d2" "+%s.%N"); diff=$(echo "$ds2 - $ds1" | bc); printf "$ds2 - $ds1 = %.3f\n" $diff; done < file
1373524748.748000000 - 1373524748.748000000 = 0.000
1373524748.826000000 - 1373524748.826000000 = 0.000
1373524748.860000000 - 1373524748.860000000 = 0.000
1373524748.919000000 - 1373524748.919000000 = 0.000
1373524748.941000000 - 1373524748.941000000 = 0.000
1373524749.390000000 - 1373524749.390000000 = 0.000
1373524749.594000000 - 1373524749.594000000 = 0.000
1373524749.619000000 - 1373524749.619000000 = 0.000
1373524749.787000000 - 1373524749.787000000 = 0.000
1373524750.006000000 - 1373524750.006000000 = 0.000
1373524750.017000000 - 1373524750.017000000 = 0.000
1373524750.088000000 - 1373524750.088000000 = 0.000
1373524750.214000000 - 1373524750.214000000 = 0.000