Python regex输出随机复制

Question

我有一个日志文件，格式如下。我只包含了一小部分来强调我在使用Python的regex来提取一些相关信息时遇到的问题。

Time = 1

smoothSolver:  Solving for Ux, Initial residual = 0.230812, Final residual = 0.0134171, No Iterations 2
smoothSolver:  Solving for Uy, Initial residual = 0.283614, Final residual = 0.0158797, No Iterations 3
smoothSolver:  Solving for Uz, Initial residual = 0.190444, Final residual = 0.016567, No Iterations 2
GAMG:  Solving for p, Initial residual = 0.0850116, Final residual = 0.00375608, No Iterations 3
time step continuity errors : sum local = 0.00999678, global = 0.00142109, cumulative = 0.00142109
smoothSolver:  Solving for omega, Initial residual = 0.00267604, Final residual = 0.000166675, No Iterations 3
bounding omega, min: -26.6597 max: 18468.7 average: 219.43  
smoothSolver:  Solving for k, Initial residual = 1, Final residual = 0.0862096, No Iterations 2
ExecutionTime = 4.84 s  ClockTime = 5 s


Time = 2

smoothSolver:  Solving for Ux, Initial residual = 0.0299872, Final residual = 0.00230507, No Iterations 2
smoothSolver:  Solving for Uy, Initial residual = 0.145767, Final residual = 0.00882969, No Iterations 3
smoothSolver:  Solving for Uz, Initial residual = 0.0863129, Final residual = 0.00858536, No Iterations 2
GAMG:  Solving for p, Initial residual = 0.394189, Final residual = 0.0175138, No Iterations 3
time step continuity errors : sum local = 0.00862823, global = 0.00212477, cumulative = 0.00354587
smoothSolver:  Solving for omega, Initial residual = 0.00258475, Final residual = 0.000222705, No Iterations 3
smoothSolver:  Solving for k, Initial residual = 0.112805, Final residual = 0.0054572, No Iterations 3
ExecutionTime = 5.9 s  ClockTime = 6 s

Time = 3

smoothSolver:  Solving for Ux, Initial residual = 0.128298, Final residual = 0.0070293, No Iterations 2
smoothSolver:  Solving for Uy, Initial residual = 0.138825, Final residual = 0.0116437, No Iterations 3
smoothSolver:  Solving for Uz, Initial residual = 0.0798979, Final residual = 0.00491246, No Iterations 3
GAMG:  Solving for p, Initial residual = 0.108748, Final residual = 0.00429273, No Iterations 2
time step continuity errors : sum local = 0.0073211, global = -0.00187909, cumulative = 0.00166678
smoothSolver:  Solving for omega, Initial residual = 0.00238456, Final residual = 0.000224435, No Iterations 3
smoothSolver:  Solving for k, Initial residual = 0.0529661, Final residual = 0.00280851, No Iterations 3
ExecutionTime = 6.92 s  ClockTime = 7 s

我的代码如下：

# Opening the log file for reading
with open(logFile, 'r') as logfile_read:
    for line in logfile_read:
        line = line.rstrip()

        # To extract Time or iteration
        if 'Time' in line:
            iteration_time = re.findall(r'^Time\s+=\s+(.*)', line)

        # To extract local, global and cumulative values
        if 'local' in line:
            local_global_cumu = re.findall(r'sum\s+local\s+=\s+(.*),\s+global\s+=\s+(.*),\s+cumulative\s+=\s+(.*)', line)
             if local_global_cumu:
                local_global_cumu = local_global_cumu[0]
                (cont_Local, cont_Global, cont_Cumulative) = local_global_cumu
            for t in iteration_time:
                contLocal.write("%s\t%s\n" %(t, cont_Local))
                contGlobal.write("%s\t%s\n" %(t, cont_Global))
                contCumulative.write("%s\t%s\n" %(t, cont_Cumulative))

        # To extract kinetic energy residual values
        if 'k,' in line:
            kinetic_energy = re.findall(r'k,\s+Initial\s+residual\s+=\s+(.*),\s+Final\s+residual\s+=\s+(.*),\s+No\s+Iterations\s+(.*)', line)
            if kinetic_energy:
                kinetic_energy = kinetic_energy[0]
                (k_initial, k_FinalRes, k_Iters) = kinetic_energy
            for t in iteration_time:
                k.write("%s\t%s\n" %(t, k_initial))
                kFinalRes.write("%s\t%s\n" %(t, k_FinalRes))
                kIters.write("%s\t%s\n" %(t, k_Iters))

        # To extract omega residual values
        if 'omega,' in line:
            omega_values = re.findall(r'omega,\s+Initial\s+residual\s+=\s+(.*),\s+Final\s+residual\s+=\s+(.*),\s+No\s+Iterations\s+(.*)', line)
            if omega_values:
                omega_rate = omega_values[0]
                (omega_initial, omega_FinalRes, omega_Iters) = omega_rate
            for t in iteration_time:
                print ("%s\t%s\n" %(t, omega_initial))

最后一部分(提取omega残差值)，omega_initial随机打印重复的值。print ("%s\t%s\n" %(t, omega_initial))的输出如下所示：

1   0.00267604

1   0.00267604

2   0.00258475

3   0.00238456

我不能理解为什么第一组值被写了两次，而正在读取的日志文件中只有一个这样的值。在处理完整的日志文件时，这种类型的复制对于许多值是随机发生的。

对于代码中存在的任何前面的变量，都不会观察到这种行为。

Answer 1

我能够推断出为什么我得到了重复的值。在我的日志文件中，我有两个omega实例，因此它提取的内容是我理解的两倍。诀窍是这样的。我现在用if 'Solving for omega,' in line:代替了if 'omega,' in line:，这解决了问题。

if 'Solving for omega,' in line:
        omega_values = re.findall(r'omega,\s+Initial\s+residual\s+=\s+(.*),\s+Final\s+residual\s+=\s+(.*),\s+No\s+Iterations\s+(.*)', line)
        if omega_values:
            omega_rate = omega_values[0]
            (omega_initial, omega_FinalRes, omega_Iters) = omega_rate
        for t in iteration_time:
            print ("%s\t%s\n" %(t, omega_initial))