有效地重新排列2D NumPy数组 [英] Rearrange 2D NumPy Array Efficiently

问题描述

假设我有一个2D NumPy数组：

Let's say I have a 2D NumPy array:

x = np.random.rand(100, 100000)

然后我检索按列排序的索引（即，每个列都独立于其他列和索引进行排序返回）：

And I retrieve the column-wise sorted indices (i.e., each column is sorted independently from the others and the indices are returned):

idx = np.argsort(x, axis=0) 

然后，对于每一列，我都需要索引= [10、20、30、40、50]中的值成为前5行中的第一行（该列的），然后是其余的排序值（不是索引！）。

Then, for each column, I need the values from indices = [10, 20, 30, 40, 50] to be first the first 5 rows (of that column) and then followed by the rest of the sorted values (not the indices!).

一种幼稚的方法可能是：

A naive approach might be:

indices = np.array([10, 20, 30, 40, 50])
out = np.empty(x.shape, dtype=int64)

for col in range(x.shape[1]):
    # For each column, fill the first few rows with `indices`
    out[:indices.shape[0], col] = x[indices, col]  # Note that we want the values, not the indices

    # Then fill the rest of the rows in this column with the remaining sorted values excluding `indices`
    n = indices.shape[0]
    for row in range(indices.shape[0], x.shape[0]):
        if idx[row, col] not in indices:
            out[n, col] = x[row, col]  # Again, note that we want the value, not the index
            n += 1

推荐答案

方法＃1

这里是一个基于 上一个帖子 的那个不需要 idx -

xc = x.copy()
xc[indices] = (xc.min()-np.arange(len(indices),0,-1))[:,None]
out = np.take_along_axis(x,xc.argsort(0),axis=0)

方法＃2

另一个与 np.isin 使用 idx -

mask = np.isin(idx, indices)
p2 = np.take_along_axis(x,idx.T[~mask.T].reshape(x.shape[1],-1).T,axis=0)
out = np.vstack((x[indices],p2))

方法＃2-替代
如果您连续地编辑到 out 来更改除 indices 之外的所有内容，分配可能适合您-

Approach #2- Alternative If you are continously editing into out to change everything except those indices, an array-assignment might be the one for you -

n = len(indices)
out[:n] = x[indices]

mask = np.isin(idx, indices)
lower = np.take_along_axis(x,idx.T[~mask.T].reshape(x.shape[1],-1).T,axis=0)
out[n:] = lower

这篇关于有效地重新排列2D NumPy数组的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！