我有一个Python元组列表,如下所示:
listoftups = [('A', 'B'), ('C','D'), ('E','F'), ('G','H'), ('A','B'), ('C','D')]
我希望能够计算这个元组列表中的重复项数量,并希望输出结果如下:
A -> B 2
C -> D 2
E -> F 1
G -> H 1
我该如何用Python实现这个功能?我正在考虑使用计数器,但不确定是否可行。谢谢。
listoftups = [('A', 'B'), ('C','D'), ('E','F'), ('G','H'), ('A','B'), ('C','D')]
from collections import Counter
for k, v in Counter(listoftups).most_common():
print "{} -> {} {}".format(k[0], k[1], v)
输出
A -> B 2
C -> D 2
G -> H 1
E -> F 1
sorted(Counter([('A', 'B'), ('C','D'), ('E','F'), ('G','H'), ('A','B'), ('C','D')]).items())
- iruvarcount
listoftups = [('A', 'B'), ('C','D'), ('E','F'), ('G','H'), ('A','B'), ('C','D')]
tup = listoftups.count(('A', 'B')) # returns 2
result = dict()
for tup in set(listoftups):
result[tup] = listoftups.count(tup)
result = {tup:listoftups.count(tup) for tup in set(listoftups)}
在你手上有一个字典:
result = { ('A', 'B'): 2, ('C', 'D'): 2, ('E','F'): 1, ('G', 'H'): 1}
你可以以与fourtheye相同的方式打印它,或者:
for k, v in result.items():
print k[0] + "->" + k[1] + " ", v
import collections
result = collections.defaultdict(int)
def f(tup):
result[tup] += 1
map(lambda t: f(t), listoftups)
defaultdict(<type 'int'>, {('G', 'H'): 1, ('A', 'B'): 2, ('C', 'D'): 2, ('E', 'F'): 1})
listoftups = [('A', 'B'), ('C','D'), ('E','F'), ('G','H'), ('A','B'), ('C','D')]
listoftups = list(set(listoftups))
for el in listoftups:
print "{} -> {} {}".format(el[0], el[1], listoftups.count(el))
tmp = []
for el in listoftups:
if el not in tmp:
tmp.append(el)
from collections import Counter
tuples = [('A', 'B'), ('C', 'D'), ('E', 'F'), ('G', 'H'), ('A', 'B'), ('C', 'D')]
counted = Counter(tuples).most_common()
s_t = sorted(counted, key=lambda x: x[0][0])
for key, value in s_t:
print key, value
>>> from collections import Counter
>>> tuples = [('A', 'B'), ('C', 'D'), ('E', 'F'), ('G', 'H'), ('A', 'B'), ('C', 'D'), ('C', 'D'), ('C', 'D')]
>>> counted = Counter(tuples).most_common()
>>> counted
Out[8]: [(('C', 'D'), 4), (('A', 'B'), 2), (('G', 'H'), 1), (('E', 'F'), 1)]
>>> sorted_tuples = sorted(counted, key=lambda x: x[0][0])
>>> sorted_tuples
Out[10]: [(('A', 'B'), 2), (('C', 'D'), 4), (('E', 'F'), 1), (('G', 'H'), 1)]
G->H
特征,考虑到它不是重复项? - iruvar