util.py 文件源码

def squish_versions(releases):
    """
    Given a list of `dict`s containing (at least) ``"name"`` and ``"version"``
    fields, return for each name the `dict` with the highest version.

    It is assumed that `dict`s with the same name are always adjacent.
    """
    for _, versions in groupby(releases, itemgetter("name")):
        yield max(versions, key=lambda v: parse(v["version"]))