使用 Numpy 对除第一个维度之外的所有维度进行平面索引
Flat indexing of all but first dimension with Numpy
有没有什么方法可以通过 NumPy 对剩余的维度使用平面索引?我正在尝试将以下 MATLAB 函数转换为 Python
function [indices, weights] = locate(values, gridpoints)
indices = ones(size(values));
weights = zeros([2, size(values)]);
for ix = 1:numel(values)
if values(ix) <= gridpoints(1)
indices(ix) = 1;
weights(:, ix) = [1; 0];
elseif values(ix) >= gridpoints(end)
indices(ix) = length(gridpoints) - 1;
weights(:, ix) = [0; 1];
else
indices(ix) = find(gridpoints <= values(ix), 1, 'last');
weights(:, ix) = ...
[gridpoints(indices(ix) + 1) - values(ix); ...
values(ix) - gridpoints(indices(ix))] ...
/ (gridpoints(indices(ix) + 1) - gridpoints(indices(ix)));
end
end
end
但我无法理解与 MATLAB 的 weights(:, ix)
等效的 NumPy 是什么——也就是说,仅在剩余维度中进行线性索引。
我希望语法可以直接翻译,但是假设 values
是一个 3×4 的数组,那么 weights
就变成了 2×3×4大批。在 MATLAB 中,weights(:, ix)
是一个 2×1 数组,而在 Python 中,weights[:, ix]
是一个 2×3 数组。
我认为我已经处理了下面函数中的所有其他内容。
import numpy as np
def locate(values, gridpoints):
indices = np.zeros(np.shape(values), dtype=int)
weights = np.zeros((2,) + np.shape(values))
for ix in range(values.size):
if values.flat[ix] <= gridpoints[0]:
indices.flat[ix] = 0
# weights[:, ix] = [1, 0]
elif values.flat[ix] >= gridpoints[-1]:
indices.flat[ix] = gridpoints.size - 2
# weights[:, ix] = [0, 1]
else:
indices.flat[ix] = (
np.argwhere(gridpoints <= values.flat[ix]).flatten()[-1]
)
# weights[:, ix] = (
# np.array([gridpoints[indices.flat[ix] + 1] - values.flat[ix],
# values.flat[ix] - gridpoints[indices.flat[ix]]])
# / (gridpoints[indices.flat[ix] + 1] - gridpoints[indices.flat[ix]])
# )
return indices, weights
你有什么建议吗?也许我只是在想这个问题。我也尝试尽可能简单地编写代码,因为我打算稍后使用 Numba 来加速它。
根据 hpaulj's , there doesn't seem to be a direct NumPy equivalent. In lack thereof, the best I can think of is to reshape the weights
array as in the code below and the suggestion from NumPy for Matlab Users.
import numpy as np
def locate(values, gridpoints):
indices = np.zeros(values.shape, dtype=int)
weights = np.zeros((2, values.size)) # Temporarily make weights 2-by-N
for ix in range(values.size):
if values.flat[ix] <= gridpoints[0]:
indices.flat[ix] = 0
weights[:, ix] = [1, 0]
elif values.flat[ix] >= gridpoints[-1]:
indices.flat[ix] = gridpoints.size - 2
weights[:, ix] = [0, 1]
else:
indices.flat[ix] = (
np.argwhere(gridpoints <= values.flat[ix]).flatten()[-1]
)
weights[:, ix] = (
np.array([gridpoints[indices.flat[ix] + 1] - values.flat[ix],
values.flat[ix] - gridpoints[indices.flat[ix]]])
/ (gridpoints[indices.flat[ix] + 1] - gridpoints[indices.flat[ix]])
)
# Give weights correct dimensions
weights.shape = (2,) + values.shape
return indices, weights
有没有什么方法可以通过 NumPy 对剩余的维度使用平面索引?我正在尝试将以下 MATLAB 函数转换为 Python
function [indices, weights] = locate(values, gridpoints)
indices = ones(size(values));
weights = zeros([2, size(values)]);
for ix = 1:numel(values)
if values(ix) <= gridpoints(1)
indices(ix) = 1;
weights(:, ix) = [1; 0];
elseif values(ix) >= gridpoints(end)
indices(ix) = length(gridpoints) - 1;
weights(:, ix) = [0; 1];
else
indices(ix) = find(gridpoints <= values(ix), 1, 'last');
weights(:, ix) = ...
[gridpoints(indices(ix) + 1) - values(ix); ...
values(ix) - gridpoints(indices(ix))] ...
/ (gridpoints(indices(ix) + 1) - gridpoints(indices(ix)));
end
end
end
但我无法理解与 MATLAB 的 weights(:, ix)
等效的 NumPy 是什么——也就是说,仅在剩余维度中进行线性索引。
我希望语法可以直接翻译,但是假设 values
是一个 3×4 的数组,那么 weights
就变成了 2×3×4大批。在 MATLAB 中,weights(:, ix)
是一个 2×1 数组,而在 Python 中,weights[:, ix]
是一个 2×3 数组。
我认为我已经处理了下面函数中的所有其他内容。
import numpy as np
def locate(values, gridpoints):
indices = np.zeros(np.shape(values), dtype=int)
weights = np.zeros((2,) + np.shape(values))
for ix in range(values.size):
if values.flat[ix] <= gridpoints[0]:
indices.flat[ix] = 0
# weights[:, ix] = [1, 0]
elif values.flat[ix] >= gridpoints[-1]:
indices.flat[ix] = gridpoints.size - 2
# weights[:, ix] = [0, 1]
else:
indices.flat[ix] = (
np.argwhere(gridpoints <= values.flat[ix]).flatten()[-1]
)
# weights[:, ix] = (
# np.array([gridpoints[indices.flat[ix] + 1] - values.flat[ix],
# values.flat[ix] - gridpoints[indices.flat[ix]]])
# / (gridpoints[indices.flat[ix] + 1] - gridpoints[indices.flat[ix]])
# )
return indices, weights
你有什么建议吗?也许我只是在想这个问题。我也尝试尽可能简单地编写代码,因为我打算稍后使用 Numba 来加速它。
根据 hpaulj's weights
array as in the code below and the suggestion from NumPy for Matlab Users.
import numpy as np
def locate(values, gridpoints):
indices = np.zeros(values.shape, dtype=int)
weights = np.zeros((2, values.size)) # Temporarily make weights 2-by-N
for ix in range(values.size):
if values.flat[ix] <= gridpoints[0]:
indices.flat[ix] = 0
weights[:, ix] = [1, 0]
elif values.flat[ix] >= gridpoints[-1]:
indices.flat[ix] = gridpoints.size - 2
weights[:, ix] = [0, 1]
else:
indices.flat[ix] = (
np.argwhere(gridpoints <= values.flat[ix]).flatten()[-1]
)
weights[:, ix] = (
np.array([gridpoints[indices.flat[ix] + 1] - values.flat[ix],
values.flat[ix] - gridpoints[indices.flat[ix]]])
/ (gridpoints[indices.flat[ix] + 1] - gridpoints[indices.flat[ix]])
)
# Give weights correct dimensions
weights.shape = (2,) + values.shape
return indices, weights