我正在尝试并行化这个numba jitted函数。我最初认为这是微不足道的，因为这是一个令人尴尬的并行问题，但它产生了意想不到的结果（顺序和并行“实现”的不同输出）。你知道那里会发生什么吗？是否有办法让这一切顺利进行？
参见以下可重现示例：

import numba
import numpy as np
from sklearn.datasets import make_regression

@numba.jit(nopython=True)
def nanlstsq(X, y):
    """Return the least-squares solution to a linear matrix equation

    Analog to ``numpy.linalg.lstsq`` for dependant variable containing ``Nan``

    Args:
        X ((M, N) np.ndarray): Matrix of independant variables
        y ({(M,), (M, K)} np.ndarray): Matrix of dependant variables

    Returns:
        np.ndarray: Least-squares solution, ignoring ``Nan``
    """
    beta = np.zeros((X.shape[1], y.shape[1]), dtype=np.float64)
    for idx in range(y.shape[1]):
        # subset y and X
        isna = np.isnan(y[:,idx])
        X_sub = X[~isna]
        y_sub = y[~isna,idx]
        # Compute beta on data subset
        XTX = np.linalg.inv(np.dot(X_sub.T, X_sub))
        XTY = np.dot(X_sub.T, y_sub)
        beta[:,idx] = np.dot(XTX, XTY)
    return beta

@numba.jit(nopython=True, parallel=True)
def nanlstsq_parallel(X, y):
    beta = np.zeros((X.shape[1], y.shape[1]), dtype=np.float64)
    for idx in numba.prange(y.shape[1]):
        # subset y and X
        isna = np.isnan(y[:,idx])
        X_sub = X[~isna]
        y_sub = y[~isna,idx]
        # Compute beta on data subset
        XTX = np.linalg.inv(np.dot(X_sub.T, X_sub))
        XTY = np.dot(X_sub.T, y_sub)
        beta[:,idx] = np.dot(XTX, XTY)
    return beta



# Generate random data
n_targets = 10000
n_features = 3
X, y = make_regression(n_samples=200, n_features=n_features,
                       n_targets=n_targets)
# Add random nan to y array
y.ravel()[np.random.choice(y.size, 5*n_targets, replace=False)] = np.nan
# Run the regression
beta = nanlstsq(X, y)
beta_parallel = nanlstsq_parallel(X, y)

np.testing.assert_allclose(beta, beta_parallel)

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