Translate every element in numpy array according to key

Question

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) 

Answer 1

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]])