我可以找出一个 numpy 向量是否显示为另一个向量的一部分吗? [英] Can I find out if one numpy vector appears as a slice of another?

问题描述

我想知道我的 numpy 向量 needle 是否作为切片或连续子向量出现在另一个向量 haystack 中.

I want to find out if my numpy vector, needle, appears inside another vector, haystack, as a slice, or contiguous sub-vector.

我想要一个函数 find(needle, haystack) 当且仅当存在可能的整数索引 p 和 q 时返回 trueneedle 等于 haystack[p:q]，其中equals"表示元素在所有位置都相等.

I want a function find(needle, haystack) that returns true if and only if there are possible integer indexes p and q such that needle equals haystack[p:q], where "equals" means elements are equal at all positions.

示例:

find([2,3,4], [1,2,3,4,5]) == True
find([2,4], [1,2,3,4,5]) == False  # not contiguous inside haystack
find([2,3,4], [0,1,2,3]) == False  # incomplete

这里我使用列表来简化说明，但实际上它们将是 numpy 向量(一维数组).

Here I am using lists to simplify the illustration, but really they would be numpy vectors (1-dimensional arrays).

对于 Python 中的字符串，等效操作很简单:它是 in: "bcd" in "abcde" == True.

For strings in Python, the equivalent operation is trivial: it's in: "bcd" in "abcde" == True.

关于维度的附录.

亲爱的读者，您可能会被类似的问题所吸引，例如 测试 Numpy 数组是否包含给定的行，或 检查一个 NumPy 数组是否包含另一个数组.但是我们可以认为这种相似性对维度的考虑没有帮助.

Dear reader, you might be tempted by similar looking questions, such as testing whether a Numpy array contains a given row, or Checking if a NumPy array contains another array. But we can dismiss this similarity as not being helpful by a consideration of dimensions.

向量是一维数组.在 numpy 术语中，长度为 N 的向量将具有 .shape == (N,);其形状的长度为 1.

A vector is a one-dimensional array. In numpy terms a vector of length N will have .shape == (N,); its shape has length 1.

其他引用的问题是，通常寻求为二维矩阵中的行找到精确匹配.

The other referenced questions are, generally seeking to find an exact match for a row in a matrix that is 2-dimensional.

我试图像窗户一样沿着一维干草堆的同一轴滑动我的一维针，直到整个针em> 匹配通过窗口可见的 haystack 部分.

I am seeking to slide my 1-dimensional needle along the same axis of my 1-dimensional haystack like a window, until the entire needle matches the portion of the haystack that is visible through the window.

推荐答案

如果您可以创建两个数组的副本，那么您可以使用 Python in 运算符来处理字节对象:

If you are fine with creating copies of the two arrays, you could fall back on Python in operator for byte objects:

def find(a, b):
  return a.tobytes() in b.tobytes()

print(
    find(np.array([2,3,4]), np.array([1,2,3,4,5])),
    find(np.array([2,4]),   np.array([1,2,3,4,5])),
    find(np.array([2,3,4]), np.array([0,1,2,3])),
    find(np.array([2,3,4]), np.array([0,1,2,3,4,5,2,3,4])),
)

# True False False True

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