Question

有没有办法找出抛出（和捕获）的异常数量，以及异常处理花了多少时间？

我正在使用pyparsing（稍加修改），它严重依赖于异常，我想找出（或至少估计）是否值得努力将其重写为无异常工作。

异常的提升和捕获广泛分散在模块上，所以我正在寻找一种不需要修改每个try-except-block的方法

Answer 1

每当您想要分析代码时，使用上下文管理器可能是个好主意

import time
from contextlib import contextmanager

# global
NUMBER_OF_EXCEPTIONS = 0

# define the timer
class Timer(object):
    def __init__(self):
        self.t1 = time.time()
    def stop(self):
        self.t2 = time.time()
        self.elapsed = self.t2 - self.t1

# define the profiler
@contextmanager
def profiler():
    t = Timer()
    yield t
    t.stop()
    print("elapsed: ", t.elapsed)

# use the profiler!
with profiler():
    try:
        1/0
    except ZeroDivisionError:
        # handle exception
        NUMBER_OF_EXCEPTIONS += 1
        time.sleep(1.1)

# and use it again!
with profiler():
    try:
        1/2
    except ZeroDivisionError:
        # handle exception
        NUMBER_OF_EXCEPTIONS += 1
        time.sleep(1.1)

print("Total handled exceptions: ", NUMBER_OF_EXCEPTIONS)

输出应该是这样的：

elapsed:  1.10120511055
elapsed:  4.05311584473e-06
Total handled exceptions:  1

现在，如果你想变得非常花哨，你也可以编写上下文管理器来处理异常（source）。这种技术可以产生非常干净的代码：

NUMBER_OF_EXCEPTIONS = 0

@contextmanager
def handle_zero_division_error():
    try:
        yield
    except ZeroDivisionError as err:
        global NUMBER_OF_EXCEPTIONS
        NUMBER_OF_EXCEPTIONS += 1
        time.sleep(1.1)

with profiler():
    with handle_zero_division_error():
        1/0

with profiler():
    with handle_zero_division_error():
        1/0

# you can even write it like this
with profiler(), handle_zero_division_error():
    1/2

print("Total handled exceptions: ", NUMBER_OF_EXCEPTIONS)

输出：

elapsed:  1.10123705864
elapsed:  1.10085892677
elapsed:  1.90734863281e-05
Total handled exceptions:  2

如何在Python中分析异常处理？

1 个答案: