[英]Sort lists in a Pandas Dataframe column
I have a Dataframe column which is a collection of lists 我有一个Dataframe列,它是一个列表集合
a
['a', 'b']
['b', 'a']
['a', 'c']
['c', 'a']
I would like to use this list to group by its unique values (['a', 'b'] & ['a', 'c']). 我想使用此列表按其唯一值(['a','b']和['a','c'])进行分组。 However, this generates an error 但是,这会产生错误
TypeError: unhashable type: 'list'
Is there any way around this. 有没有办法解决。 Ideally I would like to sort the values in place and create an additional column of a concatenated string. 理想情况下,我想对值进行排序,并创建一个连接字符串的附加列。
You can also sort values by column. 您还可以按列对值进行排序。
Example: 例:
x = [['a', 'b'], ['b', 'a'], ['a', 'c'], ['c', 'a']]
df = pandas.DataFrame({'a': Series(x)})
df.a.sort_values()
a
0 [a, b]
2 [a, c]
1 [b, a]
3 [c, a]
However, for what I understand, you want to sort [b, a]
to [a, b]
, and [c, a]
to [a, c]
and then set
values in order to get only [a, b][a, c]
. 但是,根据我的理解,你想要将[b, a]
为[a, b]
和[c, a]
到[a, c]
,然后set
值以便只获得[a, b][a, c]
。
i'd recommend use lambda
我建议使用lambda
Try: 尝试:
result = df.a.sort_values().apply(lambda x: sorted(x))
result = DataFrame(result).reset_index(drop=True)
It returns: 它返回:
0 [a, b]
1 [a, c]
2 [a, b]
3 [a, c]
Then get unique values: 然后获得唯一值:
newdf = pandas.DataFrame({'a': Series(list(set(result['a'].apply(tuple))))})
newdf.sort_values(by='a')
a
0 (a, b)
1 (a, c)
list are unhashable. 列表是不可用的。 however, tuples are hashable 但是,元组是可以清洗的
use 使用
df.groupby([df.a.apply(tuple)])
setup 建立
df = pd.DataFrame(dict(a=[list('ab'), list('ba'), list('ac'), list('ca')]))
results 结果
df.groupby([df.a.apply(tuple)]).size()
a
(a, b) 1
(a, c) 1
(b, a) 1
(c, a) 1
dtype: int64
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