def get_only_present_at_pos(fragFreqCounters, *args, **kwargs):
    """
    Find patterns in fragFreqCounters that, when present, are always
    at the same position.

    ``fragFreqCounters`` is a dictionary (usually keyed on ``fragments``)
    of whose values are dictionaries mapping positions to frequencies.

    For example::

        {
            ('a', 1, 1, 'fixed'): {1: 7, -1: 7, 3: 4},
            ('b', 1, 1, 'fixed'): {2: 6},
        }

    This indicates that the
      - pattern ``('a', 1, 1, 'fixed')`` has frequency 7 at positions 1 and -1,
        and frequency 4 at position 3;
      - pattern ``('b', 1, 1, 'fixed')`` has frequency 6 at position 2 (only)

    So this would return::

        [
            (('b', 1, 1, 'fixed'), 2)
        ]

    (sorted on ``pos``; each ``pos`` really should occur at most once.)
    """
    out = []
    print(fragFreqCounters)
    for frag, fragFreqs in fragFreqCounters.items():
        if len(fragFreqs) == 1:
            pos = fragFreqs.keys()[0]
            out.append((frag, pos))
    out.sort(key=lambda x: x[1])
    return out
