Question

我有一个2D数组大小（2,3）和一个列表

输入：

a=['airplane','automobile','bird','cat','deer','dog','frog','horse','ship','truck']

b=[[4 2 8][1 7 0]] #2D numpy array shape (2,3) containing indices of list a

输出：

c = [['deer','bird','ship'],['automobile','horse','airplane']]

是否有任何pythonic方式或快捷方式来实现输出而不迭代每个索引值？

Answer 1

如果您将列表设为np.array，则只需a[b]：

>>> import numpy as np
>>> keys = np.array(['airplane','automobile','bird','cat','deer','dog','frog','horse','ship','truck'])
>>> indices = np.array([[4,2,8],[1,7,0]])
>>> keys[indices]
array([['deer', 'bird', 'ship'],
       ['automobile', 'horse', 'airplane']], 
      dtype='<U10')

Answer 2

这就是工作：

a=['airplane','automobile','bird','cat','deer','dog','frog','horse','ship','truck']
b=numpy.array([[4,2,8],[1,7,0]])


c = [[a[idx] for idx in row] for row in b]

在Python中查找2D数组中存在的索引的值

2 个答案: