[英]pythonic way of parsing a string into a dictionary with dict comprehension
[英]Pythonic way to achieve a list/dict comprehension
比方说,我有一个元组列表:
lst = [('NP', (2, 4)), ('VP', (1, 4)), ('VP', (0, 4)),
('S-SBJ', (0, 4)), ('ADJP-PRD', (5, 7)), ('VP', (4, 7))]
获得以下内容的最标准/pythonic 方式是什么:
out = {(2, 4): 1, (1, 4): 1, (0, 4): 2, (5, 7): 1, (4, 7): 1}
我试着做:
from collections import defaultdict
d = defaultdict(list)
for item in lst:
d[item[1]].append(item[0])
dct = {}
for k, v in d.items():
dct[k] = len(v)
print(dct)
# {(2, 4): 1, (1, 4): 1, (0, 4): 2, (5, 7): 1, (4, 7): 1}
我会使用一个Counter
:
>>> from collections import Counter
>>> lst = [('NP', (2, 4)), ('VP', (1, 4)), ('VP', (0, 4)),
... ('S-SBJ', (0, 4)), ('ADJP-PRD', (5, 7)), ('VP', (4, 7))]
>>> c = Counter(item[1] for item in lst)
>>> c
Counter({(0, 4): 2, (2, 4): 1, (1, 4): 1, (5, 7): 1, (4, 7): 1})
您可以使用itertools.groupby
:
from itertools import groupby as gb
lst = [('NP', (2, 4)), ('VP', (1, 4)), ('VP', (0, 4)), ('S-SBJ', (0, 4)), ('ADJP-PRD', (5, 7)), ('VP', (4, 7))]
r = {j:i for _, b in gb(lst, key=lambda x:x[0]) for i, j in enumerate(b, 1)}
Output:
{('NP', (2, 4)): 1, ('VP', (1, 4)): 1, ('VP', (0, 4)): 2, ('S-SBJ', (0, 4)): 1, ('ADJP-PRD', (5, 7)): 1, ('VP', (4, 7)): 1}
如果您正在寻找纯粹的理解:
>>> lst = [('NP', (2, 4)), ('VP', (1, 4)), ('VP', (0, 4)),
... ('S-SBJ', (0, 4)), ('ADJP-PRD', (5, 7)), ('VP', (4, 7))]
>>> {tup[1]: sum([1 for itup in lst if itup[1] == tup[1]]) for tup in lst}
{(2, 4): 1, (1, 4): 1, (0, 4): 2, (5, 7): 1, (4, 7): 1}
>>>
但是上面的理解是相当简洁的。 可读的方式是:
>>> d = {};
>>> for x, y in lst:
... d[y] = 1 if y not in d else d[y] + 1
...
>>> d
{(2, 4): 1, (1, 4): 1, (0, 4): 2, (5, 7): 1, (4, 7): 1}
>>>
然而,OP 的defaultdict
解决方案也是可读的; 但我会使用0
作为默认值,然后在每次匹配时递增1
。 这样,您就不需要额外的循环来计算len
。
我看到的最简单的方法是:
from collections import Counter
d = dict(Counter([i[1] for i in lst]))
它准确地给出了您要求的 output:
{(2, 4): 1, (1, 4): 1, (0, 4): 2, (5, 7): 1, (4, 7): 1}
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