python - 使用字符串含义列表从 int 'flags' 的 ndarray 创建字符串 ndarray

Question

我正在尝试从 ndarray 开始整数“标志”:

array([[1, 3, 2],
       [2, 0, 3],
       [3, 2, 0],
       [2, 0, 1]])

到ndarray字符串:

array([['Banana', 'Celery', 'Carrot'],
       ['Carrot', 'Apple', 'Celery'],
       ['Celery', 'Carrot', 'Apple'],
       ['Carrot', 'Apple', 'Banana']],
      dtype='|S6')

使用字符串列表作为“标志”到“含义”的映射:

meanings = ['Apple', 'Banana', 'Carrot', 'Celery']

我想出了以下内容:

>>> import numpy as np
>>> meanings = ['Apple', 'Banana', 'Carrot', 'Celery']
>>> flags = np.array([[1,3,2],[2,0,3],[3,2,0],[2,0,1]])
>>> flags
array([[1, 3, 2],
       [2, 0, 3],
       [3, 2, 0],
       [2, 0, 1]])
>>> mapped = np.array([meanings[f] for f in flags.flatten()]).reshape(flags.shape)
>>> mapped
array([['Banana', 'Celery', 'Carrot'],
       ['Carrot', 'Apple', 'Celery'],
       ['Celery', 'Carrot', 'Apple'],
       ['Carrot', 'Apple', 'Banana']],
      dtype='|S6')

这行得通，但我担心处理大 flatten 时相关行的效率(list comp，reshape，ndarrays) :

np.array([meanings[f] for f in flags.flatten()]).reshape(flags.shape)

有没有更好/更有效的方法来执行这样的映射？

Answer 1

花哨的索引是 numpythonic 的做法:

mapped = meanings[flags]

或通常更快的等价物:

mapped = np.take(meanings, flags)