是否有比此列表理解更简单的Numpy阵列广播表达式？

发布于 2025-01-27 02:56:11 字数 477 浏览 4 评论 0原文

在此代码段中：

triangles = np.float32([[[0, -2], [-2, 3], [1, 1]], [[0, -1], [-1, 3], [1, 1]]])
centers = np.average(triangles, axis=1)
samples = np.float32([t-centers[i] for i, t in enumerate(triangles)])

我想将样本作为阵列广播扣除，即类似于triangles-Centers的东西，由于以下原因是不起作用的：

ValueError: operands could not be broadcast together with shapes (2,3,2) (2,2)

是否有：定义样本的简单方法比列表理解？

原文

In this code snippet:

triangles = np.float32([[[0, -2], [-2, 3], [1, 1]], [[0, -1], [-1, 3], [1, 1]]])
centers = np.average(triangles, axis=1)
samples = np.float32([t-centers[i] for i, t in enumerate(triangles)])

I would like to express samples as an array broadcast subtraction, i.e. something similar to triangles-centers, which doesn't work due to:

ValueError: operands could not be broadcast together with shapes (2,3,2) (2,2)

Is there a simpler way to define samples than a list comprehension?

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ペ泪落弦音 2025-02-03 02:56:11

使用 numpy.mean.mean 与keepdims = true在轴上保持一个长度-1维度，您要遵守：

samples = triangles - np.mean(triangles, axis=1, keepdims=True)

Use numpy.mean with keepdims=True to keep a length-1 dimension in the axis you're taking the mean over:

samples = triangles - np.mean(triangles, axis=1, keepdims=True)

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ゝ偶尔ゞ 2025-02-03 02:56:11

我相信这做了您想要的。只需重塑中心以匹配行大小即可。

samples = triangles - centers.reshape(triangles.shape[0],-1,triangles.shape[2])

I believe this does what you want. Just reshape the centers to match the row size.

samples = triangles - centers.reshape(triangles.shape[0],-1,triangles.shape[2])

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烟雨扶苏 2025-02-03 02:56:11

受@Tim Roberts上面答案的启发：

triangles = np.float32([[[0, -2], [-2, 3], [1, 1]], [[0, -1], [-1, 3], [1, 1]]])
centers = np.average(triangles, axis=1)
samples = triangles-centers[:, None, :]

Inspired by an answer above from @Tim Roberts:

triangles = np.float32([[[0, -2], [-2, 3], [1, 1]], [[0, -1], [-1, 3], [1, 1]]])
centers = np.average(triangles, axis=1)
samples = triangles-centers[:, None, :]

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