提取log文件中的一些信息

李东岳


Courant Number mean: 0.143461 max: 0.780362
deltaT = 0.000740741
Time = 0.502963

Weighted Re = 9.08136, weighted breakage frequency = 0
MULES limit is done
MULES: Solving for alpha.oil
MULES: Solving for alpha.water
oil fraction, min, max = 0.003 0.000631032 0.00359543
water fraction, min, max = 0.997 0.996405 0.999369
Phase-sum volume fraction, min, max = 1 1 1
DILUPBiCGStab:  Solving for n, Initial residual = 0.000902233, Final residual = 1.49928e-09, No Iterations 1
    Inconsistency value, average, min, max = 2.52405e-11 -0.000617714 0.00282076
    Droplet diameter max, min, mean = 0.0002, 0.0002, 0.0002
MULES limit is done
MULES: Solving for alpha.oil
MULES: Solving for alpha.water
oil fraction, min, max = 0.003 0.000631032 0.00359535
water fraction, min, max = 0.997 0.996405 0.999369
Phase-sum volume fraction, min, max = 1 1 1
DILUPBiCGStab:  Solving for n, Initial residual = 0.000901165, Final residual = 1.49419e-09, No Iterations 1
    Inconsistency value, average, min, max = 2.51988e-11 -0.000617713 0.00282694
    Droplet diameter max, min, mean = 0.0002, 0.0002, 0.0002
Constructing momentum equations
oil min/max T 300 - 300
water min/max T 300 - 300
GAMG:  Solving for p_rgh, Initial residual = 1, Final residual = 0.00559932, No Iterations 6
GAMG:  Solving for p_rgh, Initial residual = 1, Final residual = 0.00565399, No Iterations 6
GAMGPCG:  Solving for p_rgh, Initial residual = 1, Final residual = 3.20997e-08, No Iterations 15
DILUPBiCGStab:  Solving for epsilon.water, Initial residual = 0.000218679, Final residual = 6.3486e-09, No Iterations 2
DILUPBiCGStab:  Solving for k.water, Initial residual = 0.000310738, Final residual = 1.00525e-09, No Iterations 2
ExecutionTime = 91.01 s


Courant Number mean: 0.14345 max: 0.780081
deltaT = 0.000740741
Time = 0.503704

Weighted Re = 9.07954, weighted breakage frequency = 0
MULES limit is done
MULES: Solving for alpha.oil
MULES: Solving for alpha.water
oil fraction, min, max = 0.003 0.000631032 0.00359535
water fraction, min, max = 0.997 0.996405 0.999369
Phase-sum volume fraction, min, max = 1 1 1
DILUPBiCGStab:  Solving for n, Initial residual = 0.000900074, Final residual = 1.47821e-09, No Iterations 1
    Inconsistency value, average, min, max = 2.51619e-11 -0.000617711 0.00283307
    Droplet diameter max, min, mean = 0.0002, 0.0002, 0.0002
MULES limit is done
MULES: Solving for alpha.oil
MULES: Solving for alpha.water
oil fraction, min, max = 0.003 0.000631032 0.00359527
water fraction, min, max = 0.997 0.996405 0.999369
Phase-sum volume fraction, min, max = 1 1 1
DILUPBiCGStab:  Solving for n, Initial residual = 0.000899041, Final residual = 1.47284e-09, No Iterations 1
    Inconsistency value, average, min, max = 2.51262e-11 -0.00061771 0.00283924
    Droplet diameter max, min, mean = 0.0002, 0.0002, 0.0002
Constructing momentum equations
oil min/max T 300 - 300
water min/max T 300 - 300
GAMG:  Solving for p_rgh, Initial residual = 1, Final residual = 0.00559881, No Iterations 6
GAMG:  Solving for p_rgh, Initial residual = 1, Final residual = 0.00565348, No Iterations 6
GAMGPCG:  Solving for p_rgh, Initial residual = 1, Final residual = 3.20563e-08, No Iterations 15
DILUPBiCGStab:  Solving for epsilon.water, Initial residual = 0.000217858, Final residual = 6.31719e-09, No Iterations 2
DILUPBiCGStab:  Solving for k.water, Initial residual = 0.00030924, Final residual = 9.97882e-10, No Iterations 2
ExecutionTime = 91.12 s


Courant Number mean: 0.143439 max: 0.779868
deltaT = 0.000740741
Time = 0.504444

Weighted Re = 9.07807, weighted breakage frequency = 0
MULES limit is done
MULES: Solving for alpha.oil
MULES: Solving for alpha.water
oil fraction, min, max = 0.003 0.000631032 0.00359526
water fraction, min, max = 0.997 0.996405 0.999369
Phase-sum volume fraction, min, max = 1 1 1
DILUPBiCGStab:  Solving for n, Initial residual = 0.000897968, Final residual = 1.45572e-09, No Iterations 1
    Inconsistency value, average, min, max = 2.50956e-11 -0.000617709 0.00284538
    Droplet diameter max, min, mean = 0.0002, 0.0002, 0.0002
MULES limit is done
MULES: Solving for alpha.oil
MULES: Solving for alpha.water
oil fraction, min, max = 0.003 0.000631032 0.00359518
water fraction, min, max = 0.997 0.996405 0.999369
Phase-sum volume fraction, min, max = 1 1 1
DILUPBiCGStab:  Solving for n, Initial residual = 0.000896997, Final residual = 1.44647e-09, No Iterations 1
    Inconsistency value, average, min, max = 2.50645e-11 -0.000617707 0.00285154
    Droplet diameter max, min, mean = 0.0002, 0.0002, 0.0002
Constructing momentum equations
oil min/max T 300 - 300
water min/max T 300 - 300
GAMG:  Solving for p_rgh, Initial residual = 1, Final residual = 0.00559871, No Iterations 6
GAMG:  Solving for p_rgh, Initial residual = 1, Final residual = 0.00565306, No Iterations 6
GAMGPCG:  Solving for p_rgh, Initial residual = 1, Final residual = 3.20133e-08, No Iterations 15
DILUPBiCGStab:  Solving for epsilon.water, Initial residual = 0.000217072, Final residual = 6.28692e-09, No Iterations 2
DILUPBiCGStab:  Solving for k.water, Initial residual = 0.000307752, Final residual = 9.91092e-10, No Iterations 2
ExecutionTime = 91.23 s

各位大佬，这是上面的log文件，我想从中提取几个关键的值：

Time = 0.000576911里面的0.000576911
Inconsistency value, average, min, max = -1.34441e-17 -0.000583605 0.000718129里面的0.000718129

目前我使用这个语句：

awk '{if($0~"Inconsist") print}' log

可以提取出来一部分，但是达不到我的需求。我最终想要的是两列数据，类似：

0.000576911  0.000718129

上面是一行的，第一列Time，第二列是后面那个

有没有大佬会搞这个？

OItoCFD

大佬，我用的matlab，不知道大佬用不用，写了个简短的字符串处理，读取log文件，文件名aaa.log。查找Time=后的数据和Inconsistency value, average, min, max后面的数据，然后输出到data.txt，第一列时间，第二列是那个数据。用大佬提供的这三个时间步的log信息试了下，应该没啥问题。只是每个时间步Inconsistency value这个有两组，我取的第二组。

clear;
clc;

filein=['aaa.log'];
fidin=fopen(filein,'r');
ansn=zeros(500000,1);
anst=zeros(500000,1);
totn=0;
tott=0;

line=0;
flagn=0;
flagt=0;


while ~feof(fidin) % 判断是否为文件末尾  
    line=line+1;
    temp=fgetl(fidin); % 从文件读行
    if flagn ==2 
        flagn=0;
    end
    flagt=0;
    str = deblank(temp);
    s = regexp(str,  '\s+', 'split');
    if length(s) <=2 
        continue;
    end
    if (strcmp(s{2},'Inconsistency')) && (strcmp(s{3},'value,'))
        flagn=flagn+1;
    end 
    if (strcmp(s{1},'Time')) && (strcmp(s{2},'='))
        flagt=1;
    end 
    if (flagn==2)
        ss = char(s{10});
        sss=str2double(ss);
        totn=totn+1;
        ansn(totn) = sss;
    end
    if (flagt>0)
        ss = char(s{3});
        sss=str2double(ss);
        tott=tott+1;
        anst(tott) = sss;
    end 
end

fclose(fidin);
fid=fopen('data.txt','wt');
for i=1:tott
fprintf(fid, '%.6f %.8f\n',anst(i),ansn(i));
end
fclose(fid);

输出的数据：
0.502963 0.00282694
0.503704 0.00283924
0.504444 0.00285154

xpqiu

@李东岳

提供一个思路，one liner

paste <(cat log  | grep '^Time' | awk '{print $3}') <(cat log | grep 'Inconsistency' | awk '{print $9}')

paste 的作用本是将文件内容按列拼接，但是也可以将标准输入来进行拼接，输入的内容可以是来自其他命令。这里是将 Time 和 Inconsistency对应的内容分别提取出来，然后按列拼接。不过这里每个 Time 下面有两个 Inconsistency 的值，所以上面命令拼接处理的应该还不是楼主最终想要的，仅是一个思路。

Micro

很实用的提问，mark下！

李东岳

各位简直就是大佬中的大佬厉害厉害

z597288

@xpqiu 大佬您好，我最近在做重叠网格方面的问题，请问您知道如何从log提取物体centre的位移，速度这些信息吗，有这方面的脚本吗

xpqiu

@z597288 3楼用到的几个命令组合一下应该可以实现

CFD中文网

提取log文件中的一些信息