Add df.split with predicate function

[ghost]

xref: http://stackoverflow.com/questions/13353233/best-way-to-split-a-dataframe-given-an-edge/15449992#15449992

Here's one for you, jeff.

[wesm]

Maybe this merits a new API? Essentially "split with predicate function". so we'd do:

df.split(lambda x: x == 'B', axis=0)

[jreback]

I think more of a scalar or list of values for that axis,
to allow multiple pieces
but then how do u control where the split value goes

maybe interval=open/closed? and don't return empty groups (eg if I select the last column)

if cols are list('abcdefg')
df.split(['a','c'])

groups of
a
bc
defg

[ghost]

Another addition would be to introspect the lambda for it's argcount
and provide a moving window of values:

df.groupby(lambda prev,curr: curr != prev).

and add a win_offset arg to specify nvals before, nvals ahead.

[ghost]

implementing that would also answer #414

[dalejung]

I wonder if a general edge binner would be useful here.

def edge_groupby(df, edges):
    edges[0] = True
    edges.iloc[-1] = True

    trues = edges[edges].index.values
    trues[-1] = trues[-1] + 1 # make sure we include last value

    bins = lib.generate_bins_dt64(edges.index, trues, closed='left')
    binlabels = [0] + list(bins[:-1]) # label=left
    grouper = BinGrouper(bins, binlabels)
    return df.groupby(grouper)

grouped = edge_groupby(df, df.a == 'B')

That would take in a bool series where the True values are the edges.

[ghost]

Maybe a sliding window with a reduce style operation?

df.groupby_reduce(lambda acc,prev,curr: acc + (prev and prev == 'B'))

or

df.groupby_reduce(lambda acc,*vs: acc + (vs[0] and vs[0] == 'B'),2,'right')

with acc=0 on init.

[ghost]

closing this in favor of cleaner implementation for #4059