Python - 将列表的字典展平为唯一值？

Question

我在python中有一个列表的词典:

content = {88962: [80, 130], 87484: [64], 53662: [58,80]}

我想把它变成一个唯一值的列表

[58,64,80,130]

我写了一个手动解决方案,但这是一个手动解决方案.我知道有更简洁,更优雅的方式来做这个与列表理解,map/reduce,itertools等等任何人都有线索？

content = {88962: [80, 130], 87484: [64], 53662: [58,80]}
result = set({})
for k in content.keys() :
    for i in content[k]:
        result.add(i)
# and list/sort/print just to compare the output
r2 = list( result )
r2.sort()
print r2

Answer 1

双重理解:

sorted({x for v in content.itervalues() for x in v})

Answer 2

from itertools import chain
sorted(set(chain(*content.values())))
# [58, 64, 80, 130]

# another option is `itertools.groupby`
from itertools import groupby
[k for k, g in groupby(sorted(chain(*content.values())))]

或另一种选择是.values:

from itertools import chain
sorted(set(chain.from_iterable(content.itervalues())))
# [58, 64, 80, 130]

# another option is `itertools.groupby`
[k for k, g in groupby(sorted(chain.from_iterable(content.itervalues())))]

Answer 3

使用set()和itertools.chain()：

In [83]: content = {88962: [80, 130], 87484: [64], 53662: [58,80]}

In [84]: from itertools import chain

In [94]: x=set(chain(*content.values()))

In [95]: x
Out[95]: set([58, 64, 80, 130]) # a set, the items may or may not be sorted

In [96]: sorted(x)         #convert set to a sorted list
Out[96]: [58, 64, 80, 130]