从日志文件中提取信息

Question

我有一个日志文件，它具有以下结构。我想在每一行中的"Initial residual =“文本之后获取信息，并将其导出到.txt文件中。

Time = 1996

smoothSolver:  Solving for Ux, Initial residual = 0.000674461, Final residual = 9.44251e-07, No Iterations 5
smoothSolver:  Solving for Uy, Initial residual = 0.00478922, Final residual = 7.06574e-07, No Iterations 7
GAMG:  Solving for p, Initial residual = 0.00899, Final residual = 9.67531e-07, No Iterations 10
time step continuity errors : sum local = 2.63988e-05, global = 4.03835e-06, cumulative = -1.29495e-05
smoothSolver:  Solving for epsilon, Initial residual = 0.00209075, Final residual = 3.67614e-07, No Iterations 6
smoothSolver:  Solving for k, Initial residual = 0.0017321, Final residual = 9.33393e-07, No Iterations 6
ExecutionTime = 27.9 s  ClockTime = 28 s

Time = 1997

smoothSolver:  Solving for Ux, Initial residual = 0.000659293, Final residual = 9.46747e-07, No Iterations 5
smoothSolver:  Solving for Uy, Initial residual = 0.00479541, Final residual = 7.03185e-07, No Iterations 7
GAMG:  Solving for p, Initial residual = 0.00975341, Final residual = 9.64034e-07, No Iterations 10
time step continuity errors : sum local = 2.6336e-05, global = 3.87822e-06, cumulative = -9.07129e-06
smoothSolver:  Solving for epsilon, Initial residual = 0.00213882, Final residual = 3.64484e-07, No Iterations 6
smoothSolver:  Solving for k, Initial residual = 0.00172913, Final residual = 9.36427e-07, No Iterations 6
ExecutionTime = 27.91 s  ClockTime = 28 s

.txt文件的结构如下

1996 0.000674461 0.00478922 0.00899 0.00209075 0.0017321
1997 0.000659293 0.00479541 0.00975 0.00213882 0.0017291   
.
.

使用grep，我可以找到包含某个字符串的行。然后，我可以使用cut来提取某个列位置的数字，并将其写入文本文件中。例如，我使用以下命令从包含指定字符串的行中提取信息：

cat log | grep 'Solving for Ux' | cut -d' ' -f9 | tr -d ',' > resy.txt
cat log | grep 'Solving for Uy' | cut -d' ' -f9 | tr -d ',' > resy.txt

但我想检查是否有其他方法可以提取与两个字符串对应的所有信息。谢谢。

Answer 1

这是一种丑陋的方式，但可以为您工作：

#!/bin/bash
INPUT="/path/to/log.txt"
OUTPUT="/path/to/output.txt"
LINE=""
while IFS= read -r line
do
  echo "$line" | grep ^Time > /dev/null
  if [ $? -eq 0 ] ; then
    if [ -n "$LINE" ] ; then
      echo $LINE
    fi
    LINE=""
    LINE="${LINE}${line#*= }"
  else
    echo -e "$line" | grep "Initial residual" > /dev/null
    if [ $? -eq 0 ] ; then
      value=$(echo ${line} | sed -e 's/.*Initial residual = \(.*\), Final.*/\1/')
      LINE="${LINE} $value"
    fi
  fi
done < "$INPUT"
echo $LINE

输出：

1996 0.000674461 0.00478922 0.00899 0.00209075 0.0017321
1997 0.000659293 0.00479541 0.00975341 0.00213882 0.00172913

Answer 2

使用awk

awk -F 'Initial residual = ' -v out=out.txt '
/^Time =/ {
sub("Time = ","")
i++
if (i>1){nl="\n"}
printf nl$0" " > out
}

/Initial residual/ {
sub(",.*","",$2)
printf $2" " > out
}

END {print "" > out}' log