Question

使用np.delete时，当使用越界索引时会引发indexError。当一个越界索引在使用的np.array中并且该数组用作np.delete中的参数时，为什么这不会引发indexError？

np.delete(np.array([0, 2, 4, 5, 6, 7, 8, 9]), 9)

这给出了索引错误，因为它应该（索引9超出界限）

而

np.delete(np.arange(0,5), np.array([9]))

和

np.delete(np.arange(0,5), (9,))

得到：

array([0, 1, 2, 3, 4])

Answer 1

这是一个已知的＆＃34;功能＆＃34;并将在以后的版本中弃用。

From the source of numpy：

# Test if there are out of bound indices, this is deprecated
inside_bounds = (obj < N) & (obj >= -N)
if not inside_bounds.all():
    # 2013-09-24, 1.9
    warnings.warn(
        "in the future out of bounds indices will raise an error "
        "instead of being ignored by `numpy.delete`.",
        DeprecationWarning)
    obj = obj[inside_bounds]

在python中启用DeprecationWarning实际上会显示此警告。 Ref

In [1]: import warnings

In [2]: warnings.simplefilter('always', DeprecationWarning)

In [3]: warnings.warn('test', DeprecationWarning)
C:\Users\u31492\AppData\Local\Continuum\Anaconda\Scripts\ipython-script.py:1: De
precationWarning: test
  if __name__ == '__main__':

In [4]: import numpy as np

In [5]: np.delete(np.arange(0,5), np.array([9]))
C:\Users\u31492\AppData\Local\Continuum\Anaconda\lib\site-packages\numpy\lib\fun
ction_base.py:3869: DeprecationWarning: in the future out of bounds indices will
 raise an error instead of being ignored by `numpy.delete`.
  DeprecationWarning)
Out[5]: array([0, 1, 2, 3, 4])

当越界索引在np数组中时，为什么python numpy.delete不会引发indexError

1 个答案: