裁剪矩阵的 nan 行和列，但保持正方形

Question

我有一个超过 1,000 行和列的方阵。在“边界”的许多字段中都有nan，例如：

grid = [[nan, nan, nan, nan, nan],
        [nan, nan, nan, nan, nan],
        [nan, nan,   1, nan, nan],
        [nan,   2,   3,   2, nan],
        [  1,   2,   2,   1, nan]]

现在我想删除我只有nan 的所有行和列。这将是 1. 和 2. 行和最后一列。但我也想收到一个方阵，所以淘汰的行数必须等于淘汰的列数。在这个例子中，我想得到这个：

grid = [[nan, nan, nan, nan],
        [nan, nan,   1, nan],
        [nan,   2,   3,   2],
        [  1,   2,   2,   1]]

我确定我可以通过循环解决这个问题：检查每一列和每一行，如果里面只有 nan，最后我使用 numpy.delete 删除我找到的行和列（但只有最小的数字，因为得到一个正方形）。 但我希望任何人都可以帮助我提供更好的解决方案或一个好的库。

Answer 1

这可行，压缩 rows\cols 的索引是关键，因此它们始终具有相同的长度，从而保持矩阵的方形。

nans_in_grid = np.isnan(grid)
nan_rows = np.all(nans_in_grid, axis=0)
nan_cols = np.all(nans_in_grid, axis=1)

indicies_to_remove = zip(np.nonzero(nan_rows)[0], np.nonzero(nan_cols)[0])
y_indice_to_remove, x_indice_to_remove = zip(*indicies_to_remove)

tmp = grid[[x for x in range(grid.shape[0]) if x not in x_indice_to_remove], :]
grid = tmp[:, [y for y in range(grid.shape[1]) if y not in y_indice_to_remove]]

继续 E 先生的解决方案，然后填充结果也可以。

def pad_to_square(a, pad_value=np.nan):
    m = a.reshape((a.shape[0], -1))
    padded = pad_value * np.ones(2 * [max(m.shape)], dtype=m.dtype)
    padded[0:m.shape[0], 0:m.shape[1]] = m
    return padded

g = np.isnan(grid) 
grid = pad_to_square(grid[:, ~np.all(g, axis=0)][~np.all(g, axis=1)])

另一个解决方案，建立在另一个答案的基础上。对于较大的矩阵，速度明显更快。

shape = grid.shape[0]

first_col  = (i for i,col in enumerate(grid.T) if np.isfinite(col).any() == True).next()
last_col  = (shape-i-1 for i,col in enumerate(grid.T[::-1]) if np.isfinite(col).any() == True).next()
first_row = (i for i,row in enumerate(grid) if np.isfinite(row).any() == True).next()
last_row  = (shape-i-1 for i,row in enumerate(grid[::-1]) if np.isfinite(row).any() == True).next()

row_len = last_row - first_row
col_len = last_col - first_col
delta_len = row_len - col_len
if delta_len == 0:
    pass
elif delta_len < 0:
    first_row = first_row - abs(delta_len)
    if first_row < 0:
        delta_len = first_row
        first_row = 0
        last_row += abs(delta_len)
elif delta_len > 0:
    first_col -= abs(delta_len)
    if first_col < 0:
        delta_len = first_col
        first_col = 0
        last_col += abs(delta_len)

grid =  grid[first_row:last_row+1, first_col:last_col+1]

Answer 2

import numpy as np
nan = np.nan
grid = [[nan, nan, nan, nan, nan],
        [nan, nan, nan, nan, nan],
        [nan, nan,   1, nan, nan],
        [nan,   2,   3,   2, nan],
        [  1,   2,   2,   1, nan]]
g = np.array(grid)
cols = np.isnan(g).all(axis=0)
rows = np.isnan(g).all(axis=1)
first_col = np.where(cols==False)[0][0]
last_col = len(cols) - np.where(cols[::-1]==False)[0][0] -1
first_row = np.where(rows==False)[0][0]
last_row = len(rows) - np.where(rows[::-1]==False)[0][0] -1
row_len = last_row - first_row
col_len = last_col - first_col
delta_len = row_len - col_len
if delta_len == 0:
    pass
elif delta_len < 0:
    first_row = first_row - abs(delta_len)
    if first_row < 0:
        delta_len = first_row
        first_row = 0
        last_row += abs(delta_len)
elif delta_len > 0:
    first_col -= abs(delta_len)
    if first_col < 0:
        delta_len = first_col
        first_col = 0
        last_col += abs(delta_len)
print g[first_row:last_row+1, first_col:last_col+1]

输出：

[[ nan  nan  nan  nan]
 [ nan  nan   1.  nan]
 [ nan   2.   3.   2.]
 [  1.   2.   2.   1.]]

Answer 3

这是一个更短的方法。它的工作原理是扫描主对角线，删除所有为 nan 的行+列，然后对辅助对角线执行相同操作：

import numpy as np
nan = np.nan
grid = [[nan, nan, nan, nan, nan],
        [nan, nan, nan, nan, nan],
        [nan, nan,   1, nan, nan],
        [nan,   2,   3,   2, nan],
        [  1,   2,   2,   1, nan]]
g = np.array(grid)
for i in [1, 2]:
    cols = np.isnan(g).all(axis=0)
    rows = np.isnan(g).all(axis=1)
    main_diagonal = np.logical_not(cols & rows)
    ind = np.nonzero(main_diagonal)[0]
    main_diagonal[ind[0]:ind[-1]+1] = True  # do not remove inner row/col
    removed_main_diag = g[main_diagonal][:, main_diagonal]
    g = removed_main_diag[:][::-1]
print g

输出：

[[ nan  nan  nan  nan]
 [ nan  nan   1.  nan]
 [ nan   2.   3.   2.]
 [  1.   2.   2.   1.]]