怎样找出一个序列中出现次数最多的元素呢？
1：超极简单的方法
from collections import Counter
words = [
'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
'my', 'eyes', "you're", 'under'
]
print(Counter(words))
#OUT
# Counter({'eyes': 8, 'the': 5, 'look': 4, 'into': 3, 'my': 3, 'around': 2, "you're": 1, "don't": 1, 'under': 1, 'not': 1})
print (Counter(words).most_common(4))
#OUT
# [('eyes', 8), ('the', 5), ('look', 4), ('into', 3)]
2：稍微复杂的
words = [
'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
'my', 'eyes', "you're", 'under'
]
dict_num = {}
for item in words:
if item not in dict_num.keys():
dict_num[item] = words.count(item)
print (dict_num)
# 未做排序
# {"you're": 1, 'eyes': 8, 'look': 4, "don't": 1, 'into': 3, 'under': 1, 'not': 1, 'the': 5, 'my': 3, 'around': 2}
# 排序
import operator
sorted(dict_num.items(),key=operator.itemgetter(1))
#OUT
#[('not', 1),("don't", 1),("you're", 1),('under', 1),('around', 2),('into', 3),('my', 3),('look', 4),('the', 5),('eyes', 8)]
3：字典推导
words = [
'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
'my', 'eyes', "you're", 'under'
]
dict_num = dict_num = {i:words.count(i) for i in set(words)}
4：字典setdefault
words = [
'look', 'into', 'my', 'eyes', 'look', 'into', 'my', 'eyes',
'the', 'eyes', 'the', 'eyes', 'the', 'eyes', 'not', 'around', 'the',
'eyes', "don't", 'look', 'around', 'the', 'eyes', 'look', 'into',
'my', 'eyes', "you're", 'under'
]
d = dict()
for item in words:
d[item] = d.setdefault(item,0) + 1
