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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 trim(self, neighborhood, method="ztest", factor=3, replace="nan", verbose=True):
 """Remove outliers from the dataset. Identifies outliers by comparing each point to its neighbors using a statistical test. Parameters neighborhood : list of integers Size of the neighborhood in each dimension. Length of the list must be equal to the dimensionality of the channel. method : {'ztest'} (optional) Statistical test used to detect outliers. Default is ztest. ztest Compare point deviation from neighborhood mean to neighborhood standard deviation. factor : number (optional) Tolerance factor. Default is 3. replace : {'nan', 'mean', 'exclusive_mean', number} (optional) Behavior of outlier replacement. Default is nan. nan Outliers are replaced by numpy nans. mean Outliers are replaced by the mean of its neighborhood, including itself. exclusive_mean Outilers are replaced by the mean of its neighborhood, not including itself. number Array becomes given number. Returns ------- list of tuples Indicies of trimmed outliers. See Also -------- clip Remove pixels outside of a certain range. """

        warnings.warn("trim", category=wt_exceptions.EntireDatasetInMemoryWarning)
        outliers = []
        means = []
        ex_means = []
        # find outliers
        for idx in np.ndindex(self.shape):
            slices = []
            for i, di, size in zip(idx, neighborhood, self.shape):
                start = max(0, i - di)
                stop = min(size, i + di + 1)
                slices.append(slice(start, stop, 1))
            neighbors = self[slices]
            mean = np.nanmean(neighbors)
            sum_ = np.nansum(neighbors)
            limit = np.nanstd(neighbors) * factor
            if np.abs(self[idx] - mean) > limit:
                outliers.append(idx)
                means.append(mean)
                # Note, "- 1" is to exclude the point itself, which is not nan, in order
                # to enter this if block, as `np.abs(nan - mean)` is nan, which would
                # evaluate to False
                ex_means.append((sum_ - self[idx]) / (np.sum(~np.isnan(neighbors)) - 1))

        # replace outliers
        i = tuple(zip(*outliers))

        if len(i) == 0:
            if verbose:
                print("No outliers found")
            return []

        replace = {"nan": np.nan, "mean": means, "exclusive_mean": ex_means}.get(replace, replace)

        # This may someday be available in h5py directly, but seems that day is not yet.
        # This is annoying because it is the only reason we hold the whole set in memory.
        # KFS 2019-03-21
        arr = self[:]
        arr[i] = replace
        self[:] = arr

        # finish
        if verbose:
            print("%i outliers removed" % len(outliers))
        return outliers