我们说我有这种格式的数据(假设制表符分隔)
1 10,11,15
2 12
3 12,11
4 10,11
如何遍历列表并计算第二列中最受欢迎的对象对?假设第二列可以包含无限数量的项目。
理想的输出会返回类似
的内容pairs count
10,11 (2)
10,15 (1)
11,15 (1)
11,12 (1)
答案 0 :(得分:5)
这两个都假设您可以将输入输入到列表列表中:
如果你有Python 2.7,请与Counter
结合使用itertools
:
>>> from collections import Counter
>>> from itertools import combinations
>>> l = [[10, 11, 15], [12], [12, 11], [10, 11]]
>>> c = Counter(x for sub in l for x in combinations(sub, 2))
>>> for k, v in c.iteritems():
... print k, v
...
(10, 15) 1
(11, 15) 1
(10, 11) 2
(12, 11) 1
如果你有Python< 2.6,您可以将defaultdict
与itertools
结合使用(我敢肯定,其中一位专家会提供更清洁的解决方案。)
In [1]: from collections import defaultdict
In [2]: from itertools import combinations
In [3]: l = [[10, 11, 15], [12], [12, 11], [10, 11]]
In [4]: counts = defaultdict(int)
In [5]: for x in l:
...: for item in combinations(x, 2):
...: counts[item] += 1
...:
...:
In [6]: for k, v in counts.iteritems():
...: print k, v
...:
...:
(10, 15) 1
(11, 15) 1
(10, 11) 2
(12, 11) 1
答案 1 :(得分:0)
In [7]: with open("data1.txt") as f:
lis=[map(int,x.split(",")) for x in f]
...:
In [8]: Counter(chain(*[combinations(x,2) for x in lis]))
Out[8]: Counter({(10, 11): 2, (10, 15): 1, (11, 15): 1, (12, 11): 1})
答案 2 :(得分:0)
您可以使用combinations
和Counter
。
from itertools import combinations
import collections
newinput = []
# Removes the tabs
for line in oldinput:
newinput.append(line.partition("\t")[2])
# set up the counter
c = collections.Counter()
for line in newinput:
# Split by comma
a = line.split(',')
# make into integers from string
a = map(int, a)
# add to counter
c.update(combinations(a, 2))
然后,您最终获得了Counter
,其中包含您的所有计数:
`(10,15):1)等。