有一个numpy内置函数可以帮助实现这一点。

import numpy as np
>>> a = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])
>>> b = np.array([3,4,7])
>>> c = np.setdiff1d(a,b)
>>> c
array([1, 2, 5, 6, 8, 9])

删除特定索引(我从矩阵中删除了16和21)

import numpy as np
mat = np.arange(12,26)
a = [4,9]
del_map = np.delete(mat, a)
del_map.reshape(3,4)

输出:

array([[12, 13, 14, 15],
      [17, 18, 19, 20],
      [22, 23, 24, 25]])

列表理解也是一种有趣的方法。

a = np.array([1, 2, 3, 4, 5, 6, 7, 8, 9])
index = np.array([2, 3, 6]) #index is changed to an array.  
out = [val for i, val in enumerate(a) if all(i != index)]
>>> [1, 2, 5, 6, 8, 9]

过滤不需要的部分:

import numpy as np
a = np.array([1,2,3,4,5,6,7,8,9])
a = a[(a!=3)&(a!=4)&(a!=7)]

如果你有一个要删除的索引列表:

to_be_removed_inds = [2,3,6]
a = np.array([1,2,3,4,5,6,7,8,9])
a = a[[x for x in range(len(a)) if x not in to_be_removed]]

如果不知道索引，就不能使用logical_and

x = 10*np.random.randn(1,100)
low = 5
high = 27
x[0,np.logical_and(x[0,:]>low,x[0,:]<high)]

按值删除:

modified_array = np.delete(original_array, np.where(original_array == value_to_delete))