这是基于

Here's a vectorized one based on np.searchsorted to trace back the locations for each of those keys in the array and then replacing and please excuse the almost sexist function name here (couldn't help it though) -

def replace_with_dict(ar, dic):
    # Extract out keys and values
    k = np.array(list(dic.keys()))
    v = np.array(list(dic.values()))

    # Get argsort indices
    sidx = k.argsort()

    # Drop the magic bomb with searchsorted to get the corresponding
    # places for a in keys (using sorter since a is not necessarily sorted).
    # Then trace it back to original order with indexing into sidx
    # Finally index into values for desired output.
    return v[sidx[np.searchsorted(k,ar,sorter=sidx)]]

样品运行-

In [82]: dic ={334:0, 4:22, 8:31, 12:16, 16:17, 24:27, 28:18, 32:21, 36:1}
    ...: 
    ...: np.random.seed(0)
    ...: a = np.random.choice(dic.keys(), 20)
    ...: 

In [83]: a
Out[83]: 
array([ 28,  16,  32,  32, 334,  32,  28,   4,   8, 334,  12,  36,  36,
        24,  12, 334, 334,  36,  24,  28])

In [84]: replace_with_dict(a, dic)
Out[84]: 
array([18, 17, 21, 21,  0, 21, 18, 22, 31,  0, 16,  1,  1, 27, 16,  0,  0,
        1, 27, 18])

改进

对于大型数组，一种更快的方法是对值和键数组进行排序，然后在不使用sorter的情况下使用searchsorted，就像这样-

A faster one for big arrays would be sort the values and keys arrays and then use searchsorted without sorter, like so -

def replace_with_dict2(ar, dic):
    # Extract out keys and values
    k = np.array(list(dic.keys()))
    v = np.array(list(dic.values()))

    # Get argsort indices
    sidx = k.argsort()

    ks = k[sidx]
    vs = v[sidx]
    return vs[np.searchsorted(ks,ar)]

运行时测试-

In [91]: dic ={334:0, 4:22, 8:31, 12:16, 16:17, 24:27, 28:18, 32:21, 36:1}
    ...: 
    ...: np.random.seed(0)
    ...: a = np.random.choice(dic.keys(), 20000)

In [92]: out1 = replace_with_dict(a, dic)
    ...: out2 = replace_with_dict2(a, dic)
    ...: print np.allclose(out1, out2)
True

In [93]: %timeit replace_with_dict(a, dic)
1000 loops, best of 3: 453 µs per loop

In [95]: %timeit replace_with_dict2(a, dic)
1000 loops, best of 3: 341 µs per loop

所有数组元素都不在字典中的一般情况

如果不能保证输入数组中的所有元素都在字典中，那么我们需要做更多的工作，如下所列-

If all elements in the input array are not guaranteed to be in the dictionary, we need a bit more work as listed below -

def replace_with_dict2_generic(ar, dic, assume_all_present=True):
    # Extract out keys and values
    k = np.array(list(dic.keys()))
    v = np.array(list(dic.values()))

    # Get argsort indices
    sidx = k.argsort()

    ks = k[sidx]
    vs = v[sidx]
    idx = np.searchsorted(ks,ar)

    if assume_all_present==0:
        idx[idx==len(vs)] = 0
        mask = ks[idx] == ar
        return np.where(mask, vs[idx], ar)
    else:
        return vs[idx]

样品运行-

In [163]: dic ={334:0, 4:22, 8:31, 12:16, 16:17, 24:27, 28:18, 32:21, 36:1}
     ...: 
     ...: np.random.seed(0)
     ...: a = np.random.choice(dic.keys(), (20))
     ...: a[-1] = 400

In [165]: a
Out[165]: 
array([ 28,  16,  32,  32, 334,  32,  28,   4,   8, 334,  12,  36,  36,
        24,  12, 334, 334,  36,  24, 400])

In [166]: replace_with_dict2_generic(a, dic, assume_all_present=False)
Out[166]: 
array([ 18,  17,  21,  21,   0,  21,  18,  22,  31,   0,  16,   1,   1,
        27,  16,   0,   0,   1,  27, 400])