<SYSTEM_TASK:>
Solve the following problem using Python, implementing the functions described below, one line at a time
<END_TASK>
<USER_TASK:>
Description:
 def lowess(x, w, x0, kernel=epanechnikov, l=1, robust=False):
 """ Locally linear regression with the LOWESS algorithm. Parameters x: float n-d array Values of x for which f(x) is known (e.g. measured). The shape of this is (n, j), where n is the number the dimensions of the and j is the number of distinct coordinates sampled. w: float array The known values of f(x) at these points. This has shape (j,) x0: float or float array. Values of x for which we estimate the value of f(x). This is either a single scalar value (only possible for the 1d case, in which case f(x0) is estimated for just that one value of x), or an array of shape (n, k). kernel: callable A kernel function. {'epanechnikov', 'tri_cube'} l: float or float array with shape = x.shape The metric window size for the kernel robust: bool Whether to apply the robustification procedure from [Cleveland79], page 831 Returns ------- The function estimated at x0. Notes ----- The solution to this problem is given by equation 6.8 in Friedman, Hastie and Tibshirani (2008). The Elements of Statistical Learning (Chapter 6). Example ------- # For the 1D case: """


    if robust:
        # We use the procedure described in 
        # Start by calling this function with robust set to false and the x0
        # input being equal to the x input:
        w_est = lowess(x, w, x, kernel=epanechnikov, l=1, robust=False)
        resid = w_est - w
        median_resid = stats.nanmedian(np.abs(resid))
        robustness_weights = bi_square(resid/(6*median_resid))
        # Calculate the bi-cube function on the
        # Sub-select
        
    # For the case where x0 is provided as a scalar: 
    if not np.iterable(x0):
       x0 = np.asarray([x0])
    ans = np.zeros(x0.shape[-1]) 
    # We only need one design matrix:
    B = np.vstack([np.ones(x.shape[-1]), x]).T
    for idx, this_x0 in enumerate(x0.T):
        # This is necessary in the 1d case (?):
        if not np.iterable(this_x0):
            this_x0 = np.asarray([this_x0])
        # Different weighting kernel for each x0:
        W = np.diag(do_kernel(this_x0, x, l=l, kernel=kernel))

        # XXX It should be possible to calculate W outside the loop, if x0 and
        # x are both sampled in some regular fashion (that is, if W is the same
        # matrix in each iteration). That should save time.

        if robust:
            # We apply the robustness weights to the weighted least-squares
            # procedure:
            robustness_weights[np.isnan(robustness_weights)] = 0
            W = np.dot(W, np.diag(robustness_weights))

        try: 
            # Equation 6.8 in FHT:
            BtWB = np.dot(np.dot(B.T, W), B)
            BtW = np.dot(B.T, W)
            # Get the params:
            beta = np.dot(np.dot(la.inv(BtWB), BtW), w.T)
            # Estimate the answer based on the parameters:
            ans[idx] += beta[0] + np.dot(beta[1:], this_x0)
        # If we are trying to sample far away from where the function is
        # defined, we will be trying to invert a singular matrix. In that case,
        # the regression should not work for you and you should get a nan:
        except la.LinAlgError:
            ans[idx] += np.nan

            

    return ans.T