線性方程的Numpy語法

Question

我是新來的numpy，但不是python。有一個關於numpy辦法做到這一點的問題，可以考慮：

編輯：修正功能**

def _my_function(weights, features, bias): 
    # the pure python way 
    value = 0. 
    for i in range(len(weights)): 
     value += (weights[i]*features[i]) 

    return value+bias 

什麼是numpy的方式做到這一點？

Answer 1

如果weigths和features大小相同的numpy的陣列，可以實現elementwise乘法：

values = np.sum(weights * features + bias) 

如果要避免增加偏壓到產品陣列的每個元素，可以做

values = np.sum(weights * features) + bias * weights.size 

你甚至可以使用Python的內置sum功能：

values = sum(weights * features + bias) 

或

values = sum(weights * features) + bias * weights.size 

之所以這樣工作原理是，表達式weights * features和weights * features + bias都是臨時numpy的陣列，其是可迭代沿着其第一尺寸並且因此可以被傳遞給sum。

時序

我跑的IPython以下定時測試：

%timeit weights.dot(features) 
The slowest run took 145.86 times longer than the fastest. This could mean that an intermediate result is being cached. 
1000000 loops, best of 3: 649 ns per loop 

%timeit np.sum(weights * features) 
The slowest run took 17.09 times longer than the fastest. This could mean that an intermediate result is being cached. 
100000 loops, best of 3: 2.83 µs per loop 

使用ndarray.dot是五倍量級比在產品使用np.sum更快。但是，警告表明，在您第一次運行代碼時，這可能不正確。 0.649μs* 145.86 =94.66μs，而2.83μs* 17.09 =48.36μs。

Answer 2

方法1：使用dot-product與np.dot -

weights.dot(features) + bias*len(weights) 

方法2：引進來np.einsum執行sum-reduction -

np.einsum('i,i->',weights,features) + bias*len(weights) 

我想辦法＃1人做得更好。