计算字符串中char出现次数的最佳方法

Question

您好我试图在一行中编写这些python行,但由于代码正在进行字典修改而出现一些错误.

for i in range(len(string)):
    if string[i] in dict:
        dict[string[i]] += 1

我相信的一般语法

abc = [i for i in len(x) if x[i] in array]

考虑到我在字典中添加1值,有人可以告诉我这是如何工作的

谢谢

Answer 1

您正在尝试做的是dict,生成器表达式和str.count():

abc = dict((c, string.count(c)) for c in string)

替代使用set(string) (来自下面的评论soulcheck):

abc = dict((c, string.count(c)) for c in set(string))

定时

看到下面的评论,我在这个和其他答案中进行了一些测试.(使用python-3.2)

测试功能:

@time_me
def test_dict(string, iterations):
    """dict((c, string.count(c)) for c in string)"""
    for i in range(iterations):
        dict((c, string.count(c)) for c in string)

@time_me
def test_set(string, iterations):
    """dict((c, string.count(c)) for c in set(string))"""
    for i in range(iterations):
        dict((c, string.count(c)) for c in set(string))

@time_me
def test_counter(string, iterations):
    """Counter(string)"""
    for i in range(iterations):
        Counter(string)

@time_me
def test_for(string, iterations, d):
    """for loop from cha0site"""
    for i in range(iterations):
        for c in string:
            if c in d:
                d[c] += 1

@time_me
def test_default_dict(string, iterations):
    """defaultdict from joaquin"""
    for i in range(iterations):
        mydict = defaultdict(int)
        for mychar in string:
            mydict[mychar] += 1

测试执行:

d_ini = dict((c, 0) for c in string.ascii_letters)
words = ['hand', 'marvelous', 'supercalifragilisticexpialidocious']

for word in words:
    print('-- {} --'.format(word))
    test_dict(word, 100000)
    test_set(word, 100000)
    test_counter(word, 100000)
    test_for(word, 100000, d_ini)
    test_default_dict(word, 100000)
    print()

print('-- {} --'.format('Pride and Prejudcie - Chapter 3 '))

test_dict(ch, 1000)
test_set(ch, 1000)
test_counter(ch, 1000)
test_for(ch, 1000, d_ini)
test_default_dict(ch, 1000)

检测结果:

-- hand --
389.091 ms -  dict((c, string.count(c)) for c in string)
438.000 ms -  dict((c, string.count(c)) for c in set(string))
867.069 ms -  Counter(string)
100.204 ms -  for loop from cha0site
241.070 ms -  defaultdict from joaquin

-- marvelous --
654.826 ms -  dict((c, string.count(c)) for c in string)
729.153 ms -  dict((c, string.count(c)) for c in set(string))
1253.767 ms -  Counter(string)
201.406 ms -  for loop from cha0site
460.014 ms -  defaultdict from joaquin

-- supercalifragilisticexpialidocious --
1900.594 ms -  dict((c, string.count(c)) for c in string)
1104.942 ms -  dict((c, string.count(c)) for c in set(string))
2513.745 ms -  Counter(string)
703.506 ms -  for loop from cha0site
935.503 ms -  defaultdict from joaquin

# !!!: Do not compare this last result with the others because is timed
#      with 1000 iterations instead of 100000
-- Pride and Prejudcie - Chapter 3  --
155315.108 ms -  dict((c, string.count(c)) for c in string)
982.582 ms -  dict((c, string.count(c)) for c in set(string))
4371.579 ms -  Counter(string)
1609.623 ms -  for loop from cha0site
1300.643 ms -  defaultdict from joaquin

Answer 2

Python 2.7+的替代方案:

from collections import Counter

abc = Counter('asdfdffa')
print abc
print abc['a']

输出:

Counter({'f': 3, 'a': 2, 'd': 2, 's': 1})
2

Answer 3

这是集合模块的工作:

选项1.- 集合.defaultdict:

>>> from collections import defaultdict
>>> mydict = defaultdict(int)

然后你的循环变成:

>>> for mychar in mystring: mydict[mychar] += 1

选项2.- collections.Counter (来自Felix评论):

对于这种特定情况更好的替代方案,并且来自同一collections模块:

>>> from collections import Counter

那么你只需要(!!!):

>>> mydict = Counter(mystring)

计数器仅适用于python 2.7.所以对于python <2.7,你应该使用defaultdict