将字典值转换为按另一个数组排序的数组

I have a dictionary

我有一本字典

mydict = {'jon': 12, 'alex': 17, 'jane': 13}

mydict = {'jon':12,'alex':17,'jane':13}

and I want to create a np.array which contains the values 12, 17, 13, but sorted by another array

我想创建一个包含值12,17,13的np.array,但是按另一个数组排序

sortby = np.array(['jon', 'jane', 'alex'])

sortby = np.array(['jon','jane','alex'])

which should yield the output as

应该产生的输出为

sorted_array = np.array([12, 13, 17])

sorted_array = np.array([12,13,17])

Any approaches that are more efficient than looping through the sortby array like below?

任何比循环排序数组更有效的方法如下所示?

sorted_array = []
for vals in sortby:
     sorted_array.append(mydict[vals])

return np.array(sorted_array)

1 个解决方案

#1

Use list comprehension,

使用列表理解,

In [100]: np.array([mydict[i] for i in sortby])
Out[100]: array([12, 13, 17])

Edit:

Execution timings, To make clear for mohammad and Moses Discussions

执行时间,明确*和摩西的讨论

In [119]: def test_loop():
    sorted_array = []
    for vals in sortby:
        sorted_array.append(mydict[vals])
    return np.array(sorted_array)
   .....: 

In [120]: def test_list_compres():
    return np.array([mydict[i] for i in sortby])
   .....: 

In [121]: %timeit test_list_compres
10000000 loops, best of 3: 20 ns per loop

In [122]: %timeit test_loop
10000000 loops, best of 3: 21.3 ns per loop

In [123]: %timeit test_list_compres
10000000 loops, best of 3: 20.1 ns per loop

In [124]: %timeit test_loop
10000000 loops, best of 3: 21.9 ns per loop

It's a marginal difference but it will make a significant change with huge entries.

这是一个边际差异,但它将通过巨大的条目做出重大改变。

#1