根据键翻译numpy数组中的每个元素 [英] Translate every element in numpy array according to key

问题描述

我正在尝试根据给定的键翻译 numpy.array 的每个元素:

例如:

a = np.array([[1,2,3],[3,2,4]])my_dict = {1:23, 2:34, 3:36, 4:45}

我想得到:

array([[ 23., 34., 36.],[ 36., 34., 45.]])

我可以看到如何用循环来做到这一点:

def loop_translate(a, my_dict):new_a = np.empty(a.shape)对于 i,row in enumerate(a):new_a[i,:] = map(my_dict.get, row)返回 new_a

有没有更有效和/或纯粹的麻木方式?

我计时了，DSM 提出的 np.vectorize 方法对于更大的数组要快得多:

在 [13]: def loop_translate(a, my_dict):....: new_a = np.empty(a.shape)....: 对于 i,row in enumerate(a):....: new_a[i,:] = map(my_dict.get, row)....: 返回 new_a....:在 [14]: def vec_translate(a, my_dict):....:返回 np.vectorize(my_dict.__getitem__)(a)....:在 [15] 中:a = np.random.randint(1,5, (4,5))在 [16] 中:一个出[16]:数组([[2, 4, 3, 1, 1],[2, 4, 3, 2, 4],[4, 2, 1, 3, 1],[2, 4, 3, 4, 1]])在 [17]: %timeit loop_translate(a, my_dict)10000 个循环，最好的 3 个:每个循环 77.9 us在 [18]: %timeit vec_translate(a, my_dict)10000 个循环，最好的 3 个:每个循环 70.5 us在 [19] 中:a = np.random.randint(1, 5, (500,500))在 [20]: %timeit loop_translate(a, my_dict)1 个循环，最好的 3 个:每个循环 298 毫秒在 [21]: %timeit vec_translate(a, my_dict)10 个循环，最好的 3 个:每个循环 37.6 毫秒在 [22]: %timeit loop_translate(a, my_dict)

解决方案

我不知道效率，但你可以在 .get 上使用 np.vectorize字典的方法:

<预><代码>>>>a = np.array([[1,2,3],[3,2,4]])>>>my_dict = {1:23, 2:34, 3:36, 4:45}>>>np.vectorize(my_dict.get)(a)数组([[23, 34, 36],[36, 34, 45]])

I am trying to translate every element of a numpy.array according to a given key:

For example:

a = np.array([[1,2,3],
              [3,2,4]])

my_dict = {1:23, 2:34, 3:36, 4:45}

I want to get:

array([[ 23.,  34.,  36.],
       [ 36.,  34.,  45.]])

I can see how to do it with a loop:

def loop_translate(a, my_dict):
    new_a = np.empty(a.shape)
    for i,row in enumerate(a):
        new_a[i,:] = map(my_dict.get, row)
    return new_a

Is there a more efficient and/or pure numpy way?

Edit:

I timed it, and np.vectorize method proposed by DSM is considerably faster for larger arrays:

In [13]: def loop_translate(a, my_dict):
   ....:     new_a = np.empty(a.shape)
   ....:     for i,row in enumerate(a):
   ....:         new_a[i,:] = map(my_dict.get, row)
   ....:     return new_a
   ....: 

In [14]: def vec_translate(a, my_dict):    
   ....:     return np.vectorize(my_dict.__getitem__)(a)
   ....: 

In [15]: a = np.random.randint(1,5, (4,5))

In [16]: a
Out[16]: 
array([[2, 4, 3, 1, 1],
       [2, 4, 3, 2, 4],
       [4, 2, 1, 3, 1],
       [2, 4, 3, 4, 1]])

In [17]: %timeit loop_translate(a, my_dict)
10000 loops, best of 3: 77.9 us per loop

In [18]: %timeit vec_translate(a, my_dict)
10000 loops, best of 3: 70.5 us per loop

In [19]: a = np.random.randint(1, 5, (500,500))

In [20]: %timeit loop_translate(a, my_dict)
1 loops, best of 3: 298 ms per loop

In [21]: %timeit vec_translate(a, my_dict)
10 loops, best of 3: 37.6 ms per loop

In [22]:  %timeit loop_translate(a, my_dict)

解决方案

I don't know about efficient, but you could use np.vectorize on the .get method of dictionaries:

>>> a = np.array([[1,2,3],
              [3,2,4]])
>>> my_dict = {1:23, 2:34, 3:36, 4:45}
>>> np.vectorize(my_dict.get)(a)
array([[23, 34, 36],
       [36, 34, 45]])

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