Translate every element in numpy array according to key

I think it'd be better to iterate over the dictionary, and set values in all the rows and columns "at once":

>>> a = np.array([[1,2,3],[3,2,1]])
>>> a
array([[1, 2, 3],
       [3, 2, 1]])
>>> d = {1 : 11, 2 : 22, 3 : 33}
>>> for k,v in d.iteritems():
...     a[a == k] = v
... 
>>> a
array([[11, 22, 33],
       [33, 22, 11]])

Edit:

While it may not be as sexy as DSM's (really good) answer using numpy.vectorize, my tests of all the proposed methods show that this approach (using @jamylak's suggestion) is actually a bit faster:

from __future__ import division
import numpy as np
a = np.random.randint(1, 5, (500,500))
d = {1 : 11, 2 : 22, 3 : 33, 4 : 44}

def unique_translate(a,d):
    u,inv = np.unique(a,return_inverse = True)
    return np.array([d[x] for x in u])[inv].reshape(a.shape)

def vec_translate(a, d):    
    return np.vectorize(d.__getitem__)(a)

def loop_translate(a,d):
    n = np.ndarray(a.shape)
    for k in d:
        n[a == k] = d[k]
    return n

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


if __name__ == '__main__':
    import timeit
    n_exec = 100
    print 'orig'
    print timeit.timeit("orig_translate(a,d)", 
                        setup="from __main__ import np,a,d,orig_translate",
                        number = n_exec) / n_exec
    print 'unique'
    print timeit.timeit("unique_translate(a,d)", 
                        setup="from __main__ import np,a,d,unique_translate",
                        number = n_exec) / n_exec
    print 'vec'
    print timeit.timeit("vec_translate(a,d)",
                        setup="from __main__ import np,a,d,vec_translate",
                        number = n_exec) / n_exec
    print 'loop'
    print timeit.timeit("loop_translate(a,d)",
                        setup="from __main__ import np,a,d,loop_translate",
                        number = n_exec) / n_exec

Outputs:

orig
0.222067718506
unique
0.0472617006302
vec
0.0357889199257
loop
0.0285375618935

The numpy_indexed package (disclaimer: I am its author) provides an elegant and efficient vectorized solution to this type of problem:

import numpy_indexed as npi
remapped_a = npi.remap(a, list(my_dict.keys()), list(my_dict.values()))

The method implemented is similar to the approach mentioned by John Vinyard, but even more general. For instance, the items of the array do not need to be ints, but can be any type, even nd-subarrays themselves.

If you set the optional 'missing' kwarg to 'raise' (default is 'ignore'), performance will be slightly better, and you will get a KeyError if not all elements of 'a' are present in the keys.

Here's another approach, using numpy.unique:

>>> a = np.array([[1,2,3],[3,2,1]])
>>> a
array([[1, 2, 3],
       [3, 2, 1]])
>>> d = {1 : 11, 2 : 22, 3 : 33}
>>> u,inv = np.unique(a,return_inverse = True)
>>> np.array([d[x] for x in u])[inv].reshape(a.shape)
array([[11, 22, 33],
       [33, 22, 11]])

This approach is much faster than np.vectorize approach when the number of unique elements in array is small. Explanaion: Python is slow, in this approach the in-python loop is used to convert unique elements, afterwards we rely on extremely optimized numpy indexing operation (done in C) to do the mapping. Hence, if the number of unique elements is comparable to the overall size of the array then there will be no speedup. On the other hand, if there is just a few unique elements, then you can observe a speedup of up to x100.

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]])