使用Pythons正则表达式提取多个字符串

Question

我有一个日志文件，它有以下输出，并将其缩短为数千行：

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

我需要使用Python的正则表达式提取Time = 1、2、3和相应的累积值。更准确地说，我只需要提取与时间= 1,2和3的累积值对应的值1、2、3和0.00142109、0.00354587、0.00166678，然后写入另一个文件。

目前，这就是我所拥有的：

contCumulative_0_out = open('contCumulative_0', 'w+')

with open(logFile, 'r') as logfile_read:
for line in logfile_read:
    line = line.rstrip()
    iteration_time = re.findall(r'^Time = ([0-9]+)', line)
    print iteration_time
    contCumulative_0 = re.search(r'cumulative = ((\d|.)+)', line)
    if contCumulative_0:        
        cumvalue = contCumulative_0.groups(1)
        contCumulative_0_out.write('\n'.join(cumvalue))

变量iteration_time获取时间值，但是在下一个if循环中这是不可用的，因此我无法将时间和累积时间结合在一起，以便在输出文件中给出1 0.00142109。

Answer 1

当该行中没有“时间”或“累积”时，就没有必要覆盖该变量。你可以这样做：

...
with open(logFile, 'r') as logfile_read:
for line in logfile_read:
    line = line.rstrip()
    if 'Time' in line:
        iteration_time = re.findall(r'^Time = ([0-9]+)', line)
        print iteration_time
    if 'cumulative' in line:
        contCumulative_0 = re.search(r'cumulative = ((\d|.)+)', line)
        if contCumulative_0:
            cumvalue = contCumulative_0.groups(1)
            contCumulative_0_out.write('\n'.join(cumvalue))
...

Answer 2

您的代码在iteration_time循环的每一次迭代中都是通过for编写的。这就是问题所在。在成功地为第一个查找填充了这个变量之后，您将需要停止它的填充。

要做到这一点，在for循环中，只在不存在iteration_time或None执行正则表达式搜索时间的情况下，对iteration_time进行测试。你可以这样做：

contCumulative_0_out = open('contCumulative_0', 'w+')

with open(logFile, 'r') as logfile_read:
    iteration_time = None
    for line in logfile_read:
        line = line.rstrip()
        time_match = re.findall(r'^Time = ([0-9]+)', line)
        if time_match:
            iteration_time = time_match
            print iteration_time
        else:  # Because if there is time_match, there is no 'cumulative = ...'
            contCumulative_0 = re.search(r'cumulative = ((\d|.)+)', line)
            if contCumulative_0:        
                cumvalue = contCumulative_0.groups(1)
                # You can check and use iteration_time here
                contCumulative_0_out.write('\n'.join(cumvalue))

希望这能有所帮助。

Answer 3

您可以使用regex来完成此操作，假设您的日志格式对所有条目都是相同的。现将所发生的情况解释如下：

import re

s = """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 = 0.230812, 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 = 0.00123456
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
"""

regex = re.compile("^Time = (\d+?).*?cumulative = (\d{0,10}\.\d{0,10})",re.DOTALL|re.MULTILINE)

for x in re.findall(regex,s):
    print "{} => {}".format(x[0], x[1])

这将输出两个结果(因为我添加了两个日志条目，而不是您提供的一个日志条目)：

1 => 0.00142109
2 => 0.00123456

这是怎么回事？

正在使用的RegEx如下：

^Time = (\d+?).*?cumulative = (\d{0,10}\.\d{0,10})

这个Regex在行的开头寻找您的Time =字符串，并匹配后面的数字。然后，它对字符串cumulative =进行非贪婪的匹配，并捕获后面的数字。非贪婪很重要，否则在整个日志中只能得到一个结果，因为它将匹配Time =的第一个实例和cumulative =的最后一个实例。

然后打印每个结果。每个捕获的结果都包含时间值和累积值。如果需要，可以将代码的这一部分修改为打印为文件。

此正则表达式跨多行工作，因为它使用了两个标志：多特雷和多线。