閱讀庫方法定義

如果我想要查看一個方法的作用，例如隨機模塊中的gauss方法，我將如何使用Python解釋器進行操作？例如，我輸入隨機導入Python解釋器的控制檯後，我可以做什麼來隨機地在模塊中找到方法gauss的實際代碼，而無需查看實際文件。提前致謝！閱讀庫方法定義

來源

2012-07-15 Q Liu

嘗試：

import inspect 
inspect.getsource(random.gauss)

來源

2012-07-15 05:23:45 SuperSaiyan

的建議爲何不乾脆：'inspect.getsource（random.gauss）' – jamylak 2012-07-15 05:29:55

@jamylak：哦，是的！謝謝：） – SuperSaiyan 2012-07-15 05:31:42

如果您使用IPython，你可以這樣做：

>>> random.gauss??

來源

2012-07-15 05:22:43 icktoofay

雖然不能從默認的Python殼，這是衆多功能之一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

這是一樣的打字random.gauss??通過@icktoofay

來源

2012-07-15 05:24:16

閱讀庫方法定義

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