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如果我想查看一个方法的作用，例如模块 random 中的方法 gauss，我将如何使用 Python 解释器来做呢？例如，在我在 Python 解释器的控制台中键入 import random 之后，我可以做些什么来找出模块 random 中方法 gauss 的实际代码，而无需查看实际文件。提前致谢！

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3 回答 3

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尝试：

import inspect
inspect.getsource(random.gauss)
于 2012-07-15T05:23:45.297 回答
3

如果您使用的是IPython，您可以这样做：

>>> random.gauss??
于 2012-07-15T05:22:43.300 回答
2

虽然它在默认的 python shell 中不可用，但这是ipython.

In [4]: %psource random.gauss
    def gauss(self, mu, sigma):
        """Gaussian distribution.

        mu is the mean, and sigma is the standard deviation.  This is
        slightly faster than the normalvariate() function.

        Not thread-safe without a lock around calls.

        """

        # When x and y are two variables from [0, 1), uniformly
        # distributed, then
        #
        #    cos(2*pi*x)*sqrt(-2*log(1-y))
        #    sin(2*pi*x)*sqrt(-2*log(1-y))
        #
        # are two *independent* variables with normal distribution
        # (mu = 0, sigma = 1).
        # (Lambert Meertens)
        # (corrected version; bug discovered by Mike Miller, fixed by LM)

        # Multithreading note: When two threads call this function
        # simultaneously, it is possible that they will receive the
        # same return value.  The window is very small though.  To
        # avoid this, you have to use a lock around all calls.  (I
        # didn't want to slow this down in the serial case by using a
        # lock here.)

        random = self.random
        z = self.gauss_next
        self.gauss_next = None
        if z is None:
            x2pi = random() * TWOPI
            g2rad = _sqrt(-2.0 * _log(1.0 - random()))
            z = _cos(x2pi) * g2rad
            self.gauss_next = _sin(x2pi) * g2rad

        return mu + z*sigma

这与random.gauss??@icktoofay 建议的打字相同

于 2012-07-15T05:24:16.880 回答