有效地重新排列 2D NumPy 数组

Question

假设我有一个 2D NumPy 数组：

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

然后我检索按列排序的索引（即，每列独立于其他列排序并返回索引）：

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

然后，对于每一列，我需要 index = [10, 20, 30, 40, 50] 中的值首先是（该列的）前 5 行，然后是其余的排序值（不是索引！）。

一种天真的方法可能是：

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

Answer 1

方法#1

这是一个基于 previous post 的，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

另一个使用np.isin 掩码的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 - 替代方案 如果您不断编辑 out 以更改除 indices 之外的所有内容，则数组赋值可能适合您 -

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

Answer 2

这应该可以帮助您消除最内层循环和if 条件。首先，您可以传入x[:, col] 作为输入参数x。

def custom_ordering(x, idx, indices):
    # First get only the desired indices at the top
    out = x[indices, :]

    # delete `indices` from `idx` so `idx` doesn't have the values in `indices`
    idx2 = np.delete(idx, indices)

    # select `idx2` rows and concatenate
    out = np.concatenate((out, x[idx2, :]), axis=0)

    return out

Answer 3

这是我对问题的解决方案：

rem_indices = [_ for _ in range(x.shape[0]) if _ not in indices]    # get all remaining indices
xs = np.take_along_axis(x, idx, axis = 0)                                        # the sorted array
out = np.empty(x.shape)

out[:indices.size, :] = xs[indices, :]                                                  # insert specific values at the beginning
out[indices.size:, :] = xs[rem_indices, :]                                         # insert the remaining values after the previous

如果我理解你的问题，请告诉我。

Answer 4

我使用较小的数组和较少的索引来执行此操作，以便我可以轻松地检查结果，但它应该转化为您的用例。我认为这个解决方案非常有效，因为一切都已到位。

import numpy as np
x = np.random.randint(10, size=(12,3)) 
indices = np.array([5,7,9])

# Swap top 3 rows with the rows 5,7,9 and vice versa
x[:len(indices)], x[indices] = x[indices], x[:len(indices)].copy()
# Sort the wanted portion of array
x[len(indices):].sort(axis=0) 

这是输出：

>>> import numpy as np
>>> x = np.random.randint(10, size=(10,3))
>>> indices = np.array([5,7,9])
>>> x
array([[7, 1, 8],
       [7, 4, 6],
       [6, 5, 2],
       [6, 8, 4],
       [2, 0, 2],
       [3, 0, 4],  # 5th row
       [4, 7, 4],
       [3, 1, 1],  # 7th row
       [3, 5, 3],
       [0, 5, 9]]) # 9th row

>>> # We want top of array to be
>>> x[indices]
array([[3, 0, 4],
       [3, 1, 1],
       [0, 5, 9]])

>>> # Swap top 3 rows with the rows 5,7,9 and vice versa
>>> x[:len(indices)], x[indices] = x[indices], x[:len(indices)].copy()
>>> # Assert that rows have been swapped correctly
>>> x
array([[3, 0, 4],  #
       [3, 1, 1],  # Top of array looks like above
       [0, 5, 9],  #
       [6, 8, 4],
       [2, 0, 2],
       [7, 1, 8],  # Previous top row
       [4, 7, 4],
       [7, 4, 6],  # Previous second row
       [3, 5, 3],
       [6, 5, 2]]) # Previous third row

>>> # Sort the wanted portion of array
>>> x[len(indices):].sort(axis=0)
>>> x
array([[3, 0, 4], #
       [3, 1, 1], # Top is the same, below is sorted
       [0, 5, 9], #
       [2, 0, 2],
       [3, 1, 2],
       [4, 4, 3],
       [6, 5, 4],
       [6, 5, 4],
       [7, 7, 6],
       [7, 8, 8]])

编辑： 如果indices 中的任何元素小于len(indices)，则此版本应处理

import numpy as np
x = np.random.randint(10, size=(12,3)) 
indices = np.array([1,2,4])

tmp = x[indices]

# Here I just assume that there aren't any values less or equal to -1. If you use 
# float, you can use -np.inf, but there is no such equivalent for ints (which I 
# use in my example).
x[indices] = -1

# The -1 will create dummy rows that will get sorted to be on top of the array,
# which can switch with tmp later
x.sort(axis=0) 
x[indices] = tmp