Pandas/Numpy Groupby + Aggregate (inc integer mean) + Filter

Question

I'm new to pandas/Numpy and I'm playing around to see how everything works.

I'm using this dataset for the top 1000 IMDb movie ratings: https://github.com/justmarkham/pandas-videos/blob/master/data/imdb_1000.csv

I'm trying to group by genre, filter by number of movies (> 100), and then display min/max/mean (as an integer)/median (as int)/count.

So far I have:

df.groupby("genre")['duration'].aggregate(['min', max, np.mean, np.median, 'count']).sort_values('median', ascending=False)

This shows all the genres and the duration statistics, but the mean and median are floats, and it includes those with a low count.

I want to somehow combine it with something like this:

df.groupby("genre")['duration'].filter(lambda x: x.count() > 100)

And

df.groupby("genre")['duration'].mean().astype(int)

Is this possible?

Next I'll want to graph it all, but that's for another day...

Thanks!

EDIT 1

For clarification, currently I get:

genre       min max mean        median  count
Western     85  175 136.666667  135.0   9
Adventure   89  224 134.840000  127.0   75
Biography   85  202 131.844156  127.0   77
Action      80  205 126.485294  125.0   136
Drama       64  242 126.539568  123.0   278
Crime       67  229 122.298387  118.0   124
Thriller    107 120 114.200000  116.0   5
Mystery     69  160 115.625000  115.0   16
Sci-Fi      91  132 109.000000  113.0   5
Fantasy     112 112 112.000000  112.0   1
Family      100 115 107.500000  107.5   2
Comedy      68  187 107.602564  104.0   156
Horror      70  146 102.517241  104.0   29
Animation   75  134 96.596774   94.5    62
Film-Noir   88  111 97.333333   93.0    3
History     66  66  66.000000   66.0    1

But I want:

genre       min max mean median count
Action      80  205 126  125    136
Drama       64  242 127  123    278
Crime       67  229 122  118    124
Comedy      68  187 108  104    156

Answer 1

Yes, you can simply chain the filters and groupbys:

df.groupby('genre').filter(
    lambda x: len(x) > 100
).groupby('genre')['duration'].aggregate(
    ['min','max','mean','median','count']
).sort_values('median', ascending=False)

This yields as result:

>>> df.groupby('genre').filter(lambda x: len(x) > 100).groupby('genre')['duration'].aggregate(['min','max','mean','median','count']).sort_values('median', ascending=False)
        min  max        mean  median  count
genre                                      
Action   80  205  126.485294     125    136
Drama    64  242  126.539568     123    278
Crime    67  229  122.298387     118    124
Comedy   68  187  107.602564     104    156

you can convert this to integers as well:

>>> df.groupby('genre').filter(lambda x: len(x) > 100).groupby('genre')['duration'].aggregate(['min','max','mean','median','count']).sort_values('median', ascending=False).astype(int)
        min  max  mean  median  count
genre                                
Action   80  205   126     125    136
Drama    64  242   126     123    278
Crime    67  229   122     118    124
Comedy   68  187   107     104    156