Numexpr錯誤：「a = global_dict [name]」

Question

我正在嘗試使用Numexpr快速製作Vector Norm函數來與Numpy's進行比較。我嘗試以下方法：

import numexpr as ne 
import numpy as np 

def L2_Norm(vector_in): 
    vector1 = ne.evaluate("abs(vector_in)") 
    vector2 = ne.evaluate("vector1**2") 
    vector3 = ne.evaluate("sum(vector2)") 
    vector_out = ne.evaluate("sqrt(vector3)") 
    return vector_out` 

ve = np.arange(10) 
L2_Norm(ve) 

，我得到這個：

File "C:\Folder1\Folder2\src\test.py", line 11, in L2_Norm 
    vector3 = ne.evaluate("sum(vector2)")<br> 
File "C:\Python27\lib\site-packages\numexpr\necompiler.py", line 701, in evaluate 
    a = global_dict[name]<br> 
KeyError: 'a' 

我基本上遵循同一步驟，其User Guide（這似乎是圍繞僅供參考）。我唯一的線索是這樣的：

umexpr's principal routine is this:
evaluate(ex, local_dict=None, global_dict=None, **kwargs)

where ex is a string forming an expression, like "2*a+3*b". The values for a and b will by default be taken from the calling function's frame (through the use of sys._getframe()). Alternatively, they can be specified using the local_dict or global_dict arguments, or passed as keyword arguments

......我真的不明白 - 我認爲作者保持簡單，因爲包簡單。我忽視了什麼？

Answer 1

結果「local_dict = None，global_dict = None」參數不是默認設置。您需要將它們專門添加到numexpr.evaluate函數中。