用python解析日志文件 问题1
我正在尝试从模拟中读取和解析日志文件。理想情况下,我想和熊猫一起这样做,但我有问题。该文件名为用python解析日志文件 问题1,python,parsing,pandas,Python,Parsing,Pandas,我正在尝试从模拟中读取和解析日志文件。理想情况下,我想和熊猫一起这样做,但我有问题。该文件名为log,下面包含一个示例。现在我试着去做 import pandas as pd import numpy as np import csv # This method works: results = [] with open('log') as inputfile: for row in csv.read
log
,下面包含一个示例。现在我试着去做
import pandas as pd
import numpy as np
import csv
# This method works:
results = []
with open('log') as inputfile:
for row in csv.reader(inputfile):
results.append(row)
# This method doesn't work:
pd.read_csv('log')
错误是
pandas.parser.CParserError: Error tokenizing data. C error: Expected 1 fields in line 77, saw 4
为什么会这样
问题2
解析日志文件最有效的方法是什么?请记住,该文件实际上非常大。基本上,我想将每个时间步(每个时间步由time=x
指定)的残差解析为一个numpy数组,用于绘图。乙二醇
smoothSolver: Solving for gamma, Initial residual = 0.813857557031, Final residual = 6.74558898691e-06, No Iterations 2
我想将0.813857557031
附加到数组res\u gamma\u init
最有效的方法是什么
/*---------------------------------------------------------------------------*\
| ========= | |
| \\ / F ield | OpenFOAM: The Open Source CFD Toolbox |
| \\ / O peration | Version: 2.3.x |
| \\ / A nd | Web: www.OpenFOAM.org |
| \\/ M anipulation | |
\*---------------------------------------------------------------------------*/
Build : 2.3.x-590d57f32fed
Exec : simpleFoam
Date : Oct 30 2015
Time : 17:24:08
Host : "jack"
PID : 4034
Case : /home/jack/OpenFOAM/jack-2.3.x/run/sim
nProcs : 1
sigFpe : Enabling floating point exception trapping (FOAM_SIGFPE).
fileModificationChecking : Monitoring run-time modified files using timeStampMaster
allowSystemOperations : Allowing user-supplied system call operations
// * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * * //
Create time
Create mesh for time = 0
Reading field p
Reading field U
Reading/calculating face flux field phi
Selecting incompressible transport model Newtonian
Selecting RAS turbulence model kOmegaSST
Using gammaReThetat-correlations by Langtry and Menter (2009)
gammaReThetatSSTCoeffs
{
ca1 2;
ce1 1;
ca2 0.06;
ce2 50;
cThetat 0.03;
sigmaf 1;
sigmaThetat 2;
s1 2;
dUds false;
alphaK1 0.85034;
alphaK2 1;
alphaOmega1 0.5;
alphaOmega2 0.85616;
gamma1 0.5532;
gamma2 0.4403;
beta1 0.075;
beta2 0.0828;
betaStar 0.09;
a1 0.31;
c1 10;
kInf 0;
omegaInf 0;
}
No finite volume options present
SIMPLE: convergence criteria
field p tolerance 1e-06
field U tolerance 1e-06
field "(k|omega|ReThetaTilda|nut|gamma)" tolerance 1e-06
Starting time loop
forceCoeffs forceCoeffs:
Not including porosity effects
Time = 1
smoothSolver: Solving for Ux, Initial residual = 0.999999999388, Final residual = 0.00692443749034, No Iterations 2
smoothSolver: Solving for Uy, Initial residual = 0.999999994742, Final residual = 0.0027182321684, No Iterations 2
DICPCG: Solving for p, Initial residual = 1, Final residual = 0.00976261323556, No Iterations 326
time step continuity errors : sum local = 8.88353600919, global = -0.154492041821, cumulative = -0.154492041821
smoothSolver: Solving for omega, Initial residual = 1, Final residual = 0.0110458877993, No Iterations 2
smoothSolver: Solving for k, Initial residual = 0.999999999804, Final residual = 0.022014322402, No Iterations 2
bounding k, min: 0 max: 0.415754458684 average: 0.410526729282
smoothSolver: Solving for ReThetatTilda, Initial residual = 1, Final residual = 0.0259070897862, No Iterations 2
smoothSolver: Solving for gamma, Initial residual = 1, Final residual = 1.51565684699e-05, No Iterations 2
ExecutionTime = 1.41 s ClockTime = 1 s
forceCoeffs forceCoeffs output:
Cm = 0
Cd = -19811.1501384
Cl = 4.01861806726e-15
Cl(f) = 2.00930903363e-15
Cl(r) = 2.00930903363e-15
Time = 2
smoothSolver: Solving for Ux, Initial residual = 0.510606164486, Final residual = 0.0151751146151, No Iterations 2
smoothSolver: Solving for Uy, Initial residual = 0.570639175137, Final residual = 0.0164524779792, No Iterations 2
DICPCG: Solving for p, Initial residual = 0.160667758287, Final residual = 0.00156013775625, No Iterations 143
time step continuity errors : sum local = 88.8245306208, global = 11.6476210117, cumulative = 11.4931289699
smoothSolver: Solving for omega, Initial residual = 0.531389189498, Final residual = 0.00899532326598, No Iterations 2
smoothSolver: Solving for k, Initial residual = 0.562103650426, Final residual = 0.0138022250836, No Iterations 2
bounding k, min: 0 max: 0.425403200237 average: 0.411310758596
smoothSolver: Solving for ReThetatTilda, Initial residual = 0.301017401945, Final residual = 0.00693946297375, No Iterations 2
smoothSolver: Solving for gamma, Initial residual = 0.813857557031, Final residual = 6.74558898691e-06, No Iterations 2
ExecutionTime = 2.13 s ClockTime = 2 s
forceCoeffs forceCoeffs output:
Cm = 0
Cd = -20283.0256988
Cl = 3.99364280461e-15
Cl(f) = 1.9968214023e-15
Cl(r) = 1.9968214023e-15
Time = 3
回答你的第二个问题。这可能会有所帮助,尽管正则表达式可能更好
line1 = 'smoothSolver: Solving for Ux, Initial residual = 0.999999999388, Final residual = 0.00692443749034, No Iterations 2'
line2 = 'smoothSolver: Solving for Uy, Initial residual = 0.999999994742, Final residual = 0.0027182321684, No Iterations 2'
test = [line1, line2]
import re
for line in test:
if line.startswith('smoothSolver'):
print re.findall(r'[residual = ][\d][.][\d]+', line)
控制台中的结果是:
[' 0.999999999388', ' 0.00692443749034']
[' 0.999999994742', ' 0.0027182321684']
也许这些答案中有一个解决了你的第一个问题:我已经试过了。没用。我想第77行不干净。您可以尝试将属性skiprows=77添加到函数调用中。我仍然很困惑,为什么我给上面的读者选择的作品。谢谢。这的确很奇怪,但我对熊猫不太熟悉。所以我希望有人能给你一个真实的答案。太好了!如果我能用另一种方法的话,我会看看使用熊猫是否有任何意义。我想这就是蟒蛇式的方法。简而言之:)你能举个例子说明文件中的科学符号是怎样的吗?我最终找到了一种使用RegExp的方法——这些东西让我头疼。谢谢你的帮助。我很乐意接受你的回答,除非有人知道如何轻松地处理熊猫。没问题。很高兴我能帮上忙,我希望有人能给你一个真实的答案