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Complete this python program:
def kendalltau(x, y, initial_lexsort=None, nan_policy='propagate'):
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#Here is the Python code you requested.

    """Calculate Kendall\s tau, a correlation measure for ordinal data.
    Kendall\s tau is a measure of the correspondence between two rankings.
    Values close to 1 indicate strong agreement, values close to -1 indicate
    strong disagreement.  This is the 1945 "tau-b" version of Kendall\s
    tau [2]_, which can account for ties and which reduces to the 1938 "tau-a"
    version [1]_ in absence of ties.
    Parameters
    x, y : array_like
    Arrays of rankings, of the same shape. If arrays are not 1-D, they will
    be flattened to 1-D.
    initial_lexsort : bool, optional
    Unused (deprecated).
    nan_policy : {\propagate\, \raise\, \omit\}, optional
    Defines how to handle when input contains nan. \propagate\ returns nan,
    \raise\ throws an error, \omit\ performs the calculations ignoring nan
    values. Default is \propagate\. Note that if the input contains nan
    \omit\ delegates to mstats_basic.kendalltau(), which has a different
    implementation.
    Returns
    correlation : float
    The tau statistic.
    pvalue : float
    The two-sided p-value for a hypothesis test whose null hypothesis is
    an absence of association, tau = 0.
    See also
    spearmanr : Calculates a Spearman rank-order correlation coefficient.
    theilslopes : Computes the Theil-Sen estimator for a set of points (x, y).
    weightedtau : Computes a weighted version of Kendall\s tau.
    Notes
    The definition of Kendall\s tau that is used is [2]_::
    tau = (P - Q) / sqrt((P + Q + T) * (P + Q + U))
    where P is the number of concordant pairs, Q the number of discordant
    pairs, T the number of ties only in `x`, and U the number of ties only in
    `y`.  If a tie occurs for the same pair in both `x` and `y`, it is not
    added to either T or U.
    References
    .. [1] Maurice G. Kendall, "A New Measure of Rank Correlation", Biometrika
    Vol. 30, No. 1/2, pp. 81-93, 1938.
    .. [2] Maurice G. Kendall, "The treatment of ties in ranking problems",
    Biometrika Vol. 33, No. 3, pp. 239-251. 1945.
    ....."""
    x = np.asarray(x).ravel()
    y = np.asarray(y).ravel()
    if (x.size != y.size):
        raise ValueError(('All    inputs    to    `kendalltau`    must    be    of    the    same    size,    found    x-size    %s    and    y-size    %s' % (x.size, y.size)))
    elif ((not x.size) or (not y.size)):
        return KendalltauResult(np.nan, np.nan)
    (cnx, npx) = _contains_nan(x, nan_policy)
    (cny, npy) = _contains_nan(y, nan_policy)
    contains_nan = (cnx or cny)
    if ((npx == 'omit') or (npy == 'omit')):
        nan_policy = 'omit'
    if (contains_nan and (nan_policy == 'propagate')):
        return KendalltauResult(np.nan, np.nan)
    elif (contains_nan and (nan_policy == 'omit')):
        x = ma.masked_invalid(x)
        y = ma.masked_invalid(y)
        return mstats_basic.kendalltau(x, y)
    if (initial_lexsort is not None):
        warnings.warn('"initial_lexsort"    is    gone!')
    def count_rank_tie(ranks):
        cnt = np.bincount(ranks).astype('int64', copy=False)
        cnt = cnt[(cnt > 1)]
        return (((cnt * (cnt - 1)) // 2).sum(), ((cnt * (cnt - 1.0)) * (cnt - 2)).sum(), ((cnt * (cnt - 1.0)) * ((2 * cnt) + 5)).sum())
    size = x.size
    perm = np.argsort(y)
    (x, y) = (x[perm], y[perm])
    y = np.r_[(True, (y[1:] != y[:(-1)]))].cumsum(dtype=np.intp)
    perm = np.argsort(x, kind='mergesort')
    (x, y) = (x[perm], y[perm])
    x = np.r_[(True, (x[1:] != x[:(-1)]))].cumsum(dtype=np.intp)
    dis = _kendall_dis(x, y)
    obs = np.r_[(True, ((x[1:] != x[:(-1)]) | (y[1:] != y[:(-1)])), True)]
    cnt = np.diff(np.where(obs)[0]).astype('int64', copy=False)
    ntie = ((cnt * (cnt - 1)) // 2).sum()
    (xtie, x0, x1) = count_rank_tie(x)
    (ytie, y0, y1) = count_rank_tie(y)
    tot = ((size * (size - 1)) // 2)
    if ((xtie == tot) or (ytie == tot)):
        return KendalltauResult(np.nan, np.nan)
    con_minus_dis = ((((tot - xtie) - ytie) + ntie) - (2 * dis))
    tau = ((con_minus_dis / np.sqrt((tot - xtie))) / np.sqrt((tot - ytie)))
    tau = min(1.0, max((-1.0), tau))
   ...