Question

问题很简单：我有两个2d np.array，我想得到一个第三个数组，它只包含与后两个不相同的行。

例如：

X = np.array([[0,1],[1,2],[4,5],[5,6],[8,9],[9,10]])
Y = np.array([[5,6],[9,10]])

Z = function(X,Y)
Z = array([[0, 1],
          [1, 2],
          [4, 5],
          [8, 9]])

我尝试了np.delete(X,Y,axis=0)，但它不起作用......

Answer 1

Z = np.vstack(row for row in X if row not in Y)

Answer 2

numpy_indexed包（免责声明：我是它的作者）将标准的numpy数组集操作扩展到这些多维用例，效率很高：

import numpy_indexed as npi
Z = npi.difference(X, Y)

Answer 3

这是基于views的方法 -

# Based on http://stackoverflow.com/a/41417343/3293881 by @Eric
def setdiff2d(a, b):
    # check that casting to void will create equal size elements
    assert a.shape[1:] == b.shape[1:]
    assert a.dtype == b.dtype

    # compute dtypes
    void_dt = np.dtype((np.void, a.dtype.itemsize * np.prod(a.shape[1:])))
    orig_dt = np.dtype((a.dtype, a.shape[1:]))

    # convert to 1d void arrays
    a = np.ascontiguousarray(a)
    b = np.ascontiguousarray(b)
    a_void = a.reshape(a.shape[0], -1).view(void_dt)
    b_void = b.reshape(b.shape[0], -1).view(void_dt)

    # Get indices in a that are also in b
    return np.setdiff1d(a_void, b_void).view(orig_dt)

示例运行 -

In [81]: X
Out[81]: 
array([[ 0,  1],
       [ 1,  2],
       [ 4,  5],
       [ 5,  6],
       [ 8,  9],
       [ 9, 10]])

In [82]: Y
Out[82]: 
array([[ 5,  6],
       [ 9, 10]])

In [83]: setdiff2d(X,Y)
Out[83]: 
array([[0, 1],
       [1, 2],
       [4, 5],
       [8, 9]])

Answer 4

navigateToView(viewName:string) {
  const { navigate } = this.props.navigation;

  navigate(viewName);
}

Numpy：如何删除2个矩阵共有的行

4 个答案: