В отсутствие хорошего ответа, это то, что я собрал вместе.Это требует кодирования фильтров в дерево, чтобы его можно было обойти, и соответствующие последовательные фильтры обнуляются при каждом проходе.Я все еще заинтересован в лучших решениях.
Пример вызова, основанный на приведенном выше вопросе, где df
- кадр данных панд:
crossfilter(df, ('eq', 'Make', 'Ford'), ['Make', 'Year', 'Color'])
Код:
# filter operators of the form (operator, filter1, filter2)
group_ops = {
'and': operator.and_,
'or': operator.or_,
}
# hokie way of forcing all-pass or all-fail filters
nan = float('nan')
# recursive function that turns a tree of python-dict encoded filters into
# bitwise operators for pandas
def build_filter(df, payload, nullify_series=None, nullify_value=True):
if not payload:
# no filters, but we have to return something
# so grab the first series and filter out all NaN values
return operator.ne(df.ix[:,0], nan)
op = payload[0]
if op in value_ops:
# format: (operator, series, val)
series = payload[1]
value = payload[2]
if series == nullify_series:
# nullify filter
if nullify_value:
# push toward True (e.g. nested in an 'and' operator)
return operator.ne(df[series], nan)
else:
# push toward False (e.g. nested in an 'or' operator)
return operator.eq(df[series], nan)
return value_ops[op](df[series], value)
elif op == 'not':
# format: ('not', nested_filter)
value = payload[1]
return operator.inv(build_filter(df, value, nullify_series, False))
else:
# format: (operator, nested_filter_1, nested_filter_2)
group1 = payload[1]
group2 = payload[2]
return group_ops[op](build_filter(df, group1, nullify_series, True),
build_filter(df, group2, nullify_series, True))
# returns value counts for all series in `gather`, applying filters in `filters` in all other series
def crossfilter(df, filters, gather):
df_scoped = df[gather]
results = { series: df_filtered[series].value_counts().to_dict()
for series in gather
for df_filtered in [ df_scoped[build_filter(df_scoped, filters, series)]
if filters else df_scoped ]}
return results