2

I want to translate a dataframe to frozensets and keep the dataframe columns within the frozenset.

Example

x=pd.DataFrame(data=dict(sample=["A","B","C"],lane=[1,1,2]))
>>> x
   lane sample
0     1      A
1     1      B
2     2      C

And I would like something as :

x2= {frozenset({("sample", "A"), ("lane", 1)}),
    frozenset({("sample", "B"), ("lane", 1)}),
    frozenset({("sample", "C"), ("lane", 2)})}

>>> x2
{frozenset({('sample', 'B'), ('lane', 1)}), frozenset({('sample', 'A'), ('lane', 1)}), frozenset({('lane', 2), ('sample', 'C')})}

I tried x.apply(frozenset,1) but it gives me that :

0    (1, A)
1    (1, B)
2    (C, 2)
dtype: object

Any help will be useful. Thank you

4

1 回答 1

3

您可以使用以下方法将数据框转换为records您需要的格式pd.DataFrame.to_dict：

x.to_dict('records')

# [{'sample': 'A', 'lane': 1}, 
#  {'sample': 'B', 'lane': 1}, 
#  {'sample': 'C', 'lane': 2}]

由于这会产生一个列表，因此您可以像这样映射frozenset到该列表：

# using abbreviation 'r' instead of 'records'
map(lambda y: frozenset(y.iteritems()), x.to_dict('r'))

# [frozenset([('sample', 'A'), ('lane', 1)]), 
#  frozenset([('sample', 'B'), ('lane', 1)]), 
#  frozenset([('sample', 'C'), ('lane', 2)])]

或者，使用集合理解，如果您的输出应该是一组冻结集：

{frozenset(y.iteritems()) for y in x.to_dict('records')}

# set([frozenset([('sample', 'C'), ('lane', 2)]),  
#      frozenset([('sample', 'B'), ('lane', 1)]), 
#      frozenset([('sample', 'A'), ('lane', 1)])])
于 2018-02-13T14:41:42.660 回答