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Solve the following problem using Python, implementing the functions described below, one line at a time
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Description:
 def diff(x, lag=1, differences=1):
 """Difference an array. A python implementation of the R ``diff`` function [1]. This computes lag differences from an array given a ``lag`` and ``differencing`` term. If ``x`` is a vector of length :math:`n`, ``lag=1`` and ``differences=1``, then the computed result is equal to the successive differences ``x[lag:n] - x[:n-lag]``. Examples -------- Where ``lag=1`` and ``differences=1``: array([ -6., -2., 7., 25.], dtype=float32) Where ``lag=1`` and ``differences=2``: array([ 4., 9., 18.], dtype=float32) Where ``lag=3`` and ``differences=1``: array([ -1., 30.], dtype=float32) Where ``lag=6`` (larger than the array is) and ``differences=1``: array([], dtype=float32) For a 2d array with ``lag=1`` and ``differences=1``: array([[ 1., 1., 1.], [ 1., 1., 1.]], dtype=float32) Parameters x : array-like, shape=(n_samples, [n_features]) The array to difference. lag : int, optional (default=1) An integer > 0 indicating which lag to use. differences : int, optional (default=1) An integer > 0 indicating the order of the difference. Returns ------- res : np.ndarray, shape=(n_samples, [n_features]) The result of the differenced arrays. References .. [1] https://stat.ethz.ch/R-manual/R-devel/library/base/html/diff.html """

    if any(v < 1 for v in (lag, differences)):
        raise ValueError('lag and differences must be positive (> 0) integers')

    x = check_array(x, ensure_2d=False, dtype=np.float32)  # type: np.ndarray
    fun = _diff_vector if x.ndim == 1 else _diff_matrix
    res = x

    # "recurse" over range of differences
    for i in range(differences):
        res = fun(res, lag)
        # if it ever comes back empty, just return it as is
        if not res.shape[0]:
            return res

    return res