python只在需要的时候应用装饰排序，python应用装饰排序, def lazy

python只在需要的时候应用装饰排序，python应用装饰排序, def lazy

    def lazyDSU_sort(seq, keys, warn=False):        """Return sorted seq, breaking ties by applying keys only when needed.        If ``warn`` is True then an error will be raised if there were no        keys remaining to break ties.        Examples        ========        Here, sequences are sorted by length, then sum:        >>> seq, keys = [[[1, 2, 1], [0, 3, 1], [1, 1, 3], [2], [1]], [        ...    lambda x: len(x),        ...    lambda x: sum(x)]]        ...        >>> lazyDSU_sort(seq, keys)        [[1], [2], [1, 2, 1], [0, 3, 1], [1, 1, 3]]        If ``warn`` is True, an error will be raised if there were not        enough keys to break ties:        >>> lazyDSU_sort(seq, keys, warn=True)        Traceback (most recent call last):        ...        ValueError: not enough keys to break ties        Notes        =====        The decorated sort is one of the fastest ways to sort a sequence for        which special item comparison is desired: the sequence is decorated,        sorted on the basis of the decoration (e.g. making all letters lower        case) and then undecorated. If one wants to break ties for items that        have the same decorated value, a second key can be used. But if the        second key is expensive to compute then it is inefficient to decorate        all items with both keys: only those items having identical first key        values need to be decorated. This function applies keys successively        only when needed to break ties.        """        from collections import defaultdict        d = defaultdict(list)        keys = list(keys)        f = keys.pop(0)        for a in seq:            d[f(a)].append(a)        seq = []        for k in sorted(d.keys()):            if len(d[k]) > 1:                if keys:                    d[k] = lazyDSU_sort(d[k], keys, warn)                elif warn:                  raise ValueError('not enough keys to break ties')                seq.extend(d[k])            else:                seq.append(d[k][0])        return seq