Question

这是对here提出的问题（在下面引用）的扩展

我有一个矩阵（准确地说是2d numpy ndarray）：
A = np.array([[4, 0, 0],
              [1, 2, 3],
              [0, 0, 5]])
根据滚动，我想独立滚动A的每一行另一个数组中的值：
r = np.array([2, 0, -1])
也就是说，我要这样做：
print np.array([np.roll(row, x) for row,x in zip(A, r)])

[[0 0 4]
 [1 2 3]
 [0 5 0]]
有没有办法有效地做到这一点？也许使用花式索引招吗？

可接受的解决方案是：

rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]

# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:,np.newaxis]

result = A[rows, column_indices]

我基本上想做同样的事情，除了当索引被“粘贴”到行的末尾时，我希望用NaN填充行的另一侧，而不是将值移到定期排在行的“前”。

也许以某种方式使用np.pad？但是我无法弄清楚如何用不同的数量填充不同的行。

Answer 1

受Roll rows of a matrix independently's solution的启发，这是一个基于np.lib.stride_tricks.as_strided的矢量化矢量-

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样品运行-

from skimage.util.shape import view_as_windows as viewW

def strided_indexing_roll(a, r):
    # Concatenate with sliced to cover all rolls
    p = np.full((a.shape[0],a.shape[1]-1),np.nan)
    a_ext = np.concatenate((p,a,p),axis=1)

    # Get sliding windows; use advanced-indexing to select appropriate ones
    n = a.shape[1]
    return viewW(a_ext,(1,n))[np.arange(len(r)), -r + (n-1),0]

Answer 2

我能够将其与线性索引结合使用……它获得了正确的结果，但是在大型阵列上的执行速度却很慢。

A = np.array([[4, 0, 0],
              [1, 2, 3],
              [0, 0, 5]]).astype(float)

r = np.array([2, 0, -1])

rows, column_indices = np.ogrid[:A.shape[0], :A.shape[1]]

# Use always a negative shift, so that column_indices are valid.
# (could also use module operation)
r_old = r.copy()
r[r < 0] += A.shape[1]
column_indices = column_indices - r[:,np.newaxis]

result = A[rows, column_indices]

# replace with NaNs
row_length = result.shape[-1]

pad_inds = []
for ind,i in np.enumerate(r_old):
    if i > 0:
        inds2pad = [np.ravel_multi_index((ind,) + (j,),result.shape) for j in range(i)]
        pad_inds.extend(inds2pad)
    if i < 0:
        inds2pad = [np.ravel_multi_index((ind,) + (j,),result.shape) for j in range(row_length+i,row_length)]
        pad_inds.extend(inds2pad)
result.ravel()[pad_inds] = nan

给出预期的结果：

print result

[[ nan  nan   4.]
 [  1.   2.   3.]
 [  0.   5.  nan]]

Answer 3

基于@Seberg 和@yann-dubois 在非 nan 情况下的回答，我编写了一个方法：

比当前答案更快
适用于任何形状的 ndarrays（使用 axis 参数指定行轴）
允许将 fill 设置为 np.nan、任何其他“填充值”或 False 以允许定期滚动数组边缘。

基准测试

cols, rows = 1024, 2048
arr = np.stack(rows*(np.arange(cols,dtype=float),))
shifts = np.random.randint(-cols, cols, rows)

np.testing.assert_array_almost_equal(row_roll(arr, shifts), strided_indexing_roll(arr, shifts))
# True

%timeit row_roll(arr, shifts)
# 25.9 ms ± 161 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

%timeit strided_indexing_roll(arr, shifts)
# 29.7 ms ± 446 µs per loop (mean ± std. dev. of 7 runs, 10 loops each)

def row_roll(arr, shifts, axis=1, fill=np.nan):
    """Apply an independent roll for each dimensions of a single axis.

    Parameters
    ----------
    arr : np.ndarray
        Array of any shape.

    shifts : np.ndarray, dtype int. Shape: `(arr.shape[:axis],)`.
        Amount to roll each row by. Positive shifts row right.

    axis : int
        Axis along which elements are shifted. 
        
    fill: bool or float
        If True, value to be filled at missing values. Otherwise just rolls across edges.
    """
    if np.issubdtype(arr.dtype, int) and isinstance(fill, float):
        arr = arr.astype(float)

    shifts2 = shifts.copy()
    arr = np.swapaxes(arr,axis,-1)
    all_idcs = np.ogrid[[slice(0,n) for n in arr.shape]]
    # Convert to a positive shift
    shifts2[shifts2 < 0] += arr.shape[-1] 
    all_idcs[-1] = all_idcs[-1] - shifts2[:, np.newaxis]

    result = arr[tuple(all_idcs)]

    if fill is not False:
        # Create mask of row positions above negative shifts
        # or below positive shifts. Then set them to np.nan.
        *_, nrows, ncols  = arr.shape

        mask_neg = shifts < 0
        mask_pos = shifts >= 0
        
        shifts_pos = shifts.copy()
        shifts_pos[mask_neg] = 0
        shifts_neg = shifts.copy()
        shifts_neg[mask_pos] = ncols+1 # need to be bigger than the biggest positive shift
        shifts_neg[mask_neg] = shifts[mask_neg] % ncols

        indices = np.stack(nrows*(np.arange(ncols),))
        nanmask = (indices < shifts_pos[:, None]) | (indices >= shifts_neg[:, None])
        result[nanmask] = fill

    arr = np.swapaxes(result,-1,axis)

    return arr

独立移动numpy数组的行

3 个答案:

基准测试