pivot & unstack with nan in the index

closes https://github.com/pydata/pandas/issues/7466
on branch:

    >>> df
         a   b   c
    0   R1  C1  10
    1   R2  C2  15
    2  NaN  C3  17
    3   R4  C4  20
    >>> df.pivot('a', 'b', 'c').fillna('.')
    b    C1  C2  C3  C4
    a
    NaN   .   .  17   .
    R1   10   .   .   .
    R2    .  15   .   .
    R4    .   .   .  20
    >>> df.pivot('b', 'a', 'c').fillna('.')
    a  NaN  R1  R2  R4
    b
    C1   .  10   .   .
    C2   .   .  15   .
    C3  17   .   .   .
    C4   .   .   .  20

the `pivot` function requires the `unstack` function to work with `nan`s in the index:

    >>> df
                    4th     5th
    1st 2nd 3rd
    d   y   67   d.y.67  67.y.d
            39   d.y.39  39.y.d
        w   53   d.w.53  53.w.d
    NaN w   72    .w.72  72.w. 
            57    .w.57  57.w. 
        NaN 80    . .80  80. . 
            31    . .31  31. . 
            18    . .18  18. . 
    a   z   11   a.z.11  11.z.a
            30   a.z.30  30.z.a
    c   z   59   c.z.59  59.z.c
            50   c.z.50  50.z.c
        NaN 62   c. .62  62. .c
    e   NaN 59   e. .59  59. .e
            76   e. .76  76. .e
    b   x   52   b.x.52  52.x.b
            14   b.x.14  14.x.b
            53   b.x.53  53.x.b
        NaN 60   b. .60  60. .b
            51   b. .51  51. .b
the `4th` and `5th` column are so that it is easy to verify the frame:

    >>> df.unstack(level=1).fillna('-')
                4th                                     5th
    2nd         NaN       w       x       y       z     NaN       w       x       y       z
    1st 3rd
    NaN 18    . .18       -       -       -       -  18. .        -       -       -       -
        31    . .31       -       -       -       -  31. .        -       -       -       -
        57        -   .w.57       -       -       -       -  57.w.        -       -       -
        72        -   .w.72       -       -       -       -  72.w.        -       -       -
        80    . .80       -       -       -       -  80. .        -       -       -       -
    a   11        -       -       -       -  a.z.11       -       -       -       -  11.z.a
        30        -       -       -       -  a.z.30       -       -       -       -  30.z.a
    b   14        -       -  b.x.14       -       -       -       -  14.x.b       -       -
        51   b. .51       -       -       -       -  51. .b       -       -       -       -
        52        -       -  b.x.52       -       -       -       -  52.x.b       -       -
        53        -       -  b.x.53       -       -       -       -  53.x.b       -       -
        60   b. .60       -       -       -       -  60. .b       -       -       -       -
    c   50        -       -       -       -  c.z.50       -       -       -       -  50.z.c
        59        -       -       -       -  c.z.59       -       -       -       -  59.z.c
        62   c. .62       -       -       -       -  62. .c       -       -       -       -
    d   39        -       -       -  d.y.39       -       -       -       -  39.y.d       -
        53        -  d.w.53       -       -       -       -  53.w.d       -       -       -
        67        -       -       -  d.y.67       -       -       -       -  67.y.d       -
    e   59   e. .59       -       -       -       -  59. .e       -       -       -       -
        76   e. .76       -       -       -       -  76. .e       -       -       -       -

`nan` values are kept out of levels and are handled by labels:

    >>> df.unstack(level=1).index
    MultiIndex(levels=[['a', 'b', 'c', 'd', 'e'], [11, 14, 18, 30, 31, 39, 50, 51, 52, 53, 57, 59, 60, 62, 67, 72, 76, 80]],
               labels=[[-1, -1, -1, -1, -1, 0, 0, 1, 1, 1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4], [2, 4, 10, 15, 17, 0, 3, 1, 7, 8, 9, 12, 6, 11, 13, 5, 9, 14, 11, 16]],
               names=['1st', '3rd'])
    >>> df.unstack(level=1).columns
    MultiIndex(levels=[['4th', '5th'], ['w', 'x', 'y', 'z']],
               labels=[[0, 0, 0, 0, 0, 1, 1, 1, 1, 1], [-1, 0, 1, 2, 3, -1, 0, 1, 2, 3]],
               names=[None, '2nd'])


> the problem is the resulting index are full of nan levels

@jreback  no! there will be no nan in the levels and there is no nan level. see the last lines in the very first comment, or check `df.unstack().index.levels`. nan's are handled by labels. the point of running benchmarks is this way of handling nan's does not impact performance.