3

我试图弄清楚如何在句子中获得辅音的表示。我刚刚使用的代码似乎没有完成这项工作：

vowels = ("aeiou")
count = 0
for x in text:
    if not x in vowels:
        count += 1

在示例“hello world”作为输入中，我收到 8 个辅音。提前谢谢了。

4

10 回答 10

9

你最好直接检查辅音，而不是“不是元音，不是空格，不是标点符号和......”

consonants = "bcdfghjklmnpqrstvwxyz"
count = 0
for x in text:
    if x in consonants:
        count += 1
于 2012-04-23T23:44:48.240 回答
3

您正在计算所有字符，包括空格。此外，您还需要包含标点符号、空格和任何其他非辅音字符。

于 2012-04-23T23:40:04.423 回答
3

更直接

consonants = set("bcdfghjklmnpqrstvwxyz")
count = sum(1 for c in text if c in consonants)

为辅音使用一组应该会使查找更快一点

于 2012-04-23T23:48:47.100 回答
1 
import string

all_letters = string.ascii_letters

consonants = set(all_letters).difference(set(('a','e','i','o','u','A','E','I','O','U')))

my_sentence = 'Here is my Sentence'

sum_of_cons = sum(ele in consonants for ele in my_sentence)

结果

>>> sum_of_cons
10
于 2012-04-23T23:48:39.843 回答
1

如果速度确实起作用，编译的正则表达式似乎是获得计数的最快方法。

计时结果 

Found 8292 Consonants in 0.002978 seconds using compiled regex
Found 8292 Consonants in 0.009412 seconds using sets
Found 8292 Consonants in 0.024511 seconds by looking at each character

测试代码 

import re
import time
import os
string_length = 100000
random_string = os.urandom(string_length)

con_re = re.compile("[bcdfghjklmnpqrstvwxyz]")
start = time.clock()
re_results = con_re.findall(random_string)
print "Found %d Consonants in %f seconds using compiled regex" % (len(re_results), time.clock() - start)

consonants = set("bcdfghjklmnpqrstvwxyz")
start = time.clock()
count = sum(1 for c in random_string if c in consonants)
print "Found %d Consonants in %f seconds using sets" % (count, time.clock() - start)

cnt = 0
consonants = "bcdfghjklmnpqrstvwxyz"
start = time.clock()
for x in range(string_length):
    if random_string[x] in consonants:
        cnt += 1
print "Found %d Consonants in %f seconds by looking at each character" % (cnt, time.clock() - start)
于 2012-04-24T02:44:32.970 回答
0

不要忘记考虑空格。

根据您的代码，8 看起来像是正确答案。

空格字符不在您的元音列表中。

于 2012-04-23T23:36:27.210 回答
0

也许您可以使用以下代码：

consonants = list("bcdfghjklmnpqrstvwxyz")
word=" hello world "
number_of_consonants = sum(word.count(c) for c in consonants)
于 2012-04-23T23:46:49.143 回答
0

使用非元音条件、字典理解和Counter.

import collections as ct

text = "Hello world!"
vowels = ("aeiou")
letters = ct.Counter(text.lower())

对于前面显示的类似技术，不需要一长串辅音。isalpha您仍然可以使用and搜索不是元音的字母 not in vowels。 isalpha消除空格和标点符号：

# Option 1: Non-vowels

sum(1 for letter in text.lower() if letter.isalpha() and not letter in vowels)
 # 7

对于更精细的细节，我们可以使用一个Counter对象来计算文本中的每个唯一字母，并使用字典理解来保留所有辅音：

# Option 2: Dictionary comprehension

comp = {k:v for k, v in letters.items() if letter.isalpha() and not letter in vowels}    
comp 
# {'d': 1, 'h': 1, 'l': 3, 'r': 1, 'w': 1}

sum(comp.values())
# 7

此外，Counter对象具有有用的属性，例如.most_common()和.elements()。当然，您也可以轻松地汇总所有辅音。Counter在这里，我们将单独为辅音制作一个特殊的：

# Option 3: Consonants counter

consonants = letters.copy()
for k in letters:
    if k in vowels or not k.isalpha():
        consonants.pop(k)

consonants
# Counter({'d': 1, 'h': 1, 'l': 3, 'r': 1, 'w': 1})

consonants.most_common(3)
#[('l', 3), ('d', 1), ('h', 1)] 

list(consonants.elements())
# ['d', 'h', 'l', 'l', 'l', 'r', 'w']

sum(consonants.values())
# 7
于 2016-12-25T01:06:10.487 回答
0 
def count_consonants(sample_string):
    spaces_removed = sample_string.replace(" ", "")
    count = 0
    vowels = ['a', 'e', 'i', 'o', 'u']
    for char in spaces_removed.lower():
        if char not in vowels:
            count = count + 1
    return count
于 2019-03-02T06:44:58.073 回答
-1 
def count_consonants(some_string):
    some_string=some_string.lower();
    consonants = list("bcdfghjklmnpqrstvwxyz")
    number_of_consonants = sum(some_string.count(c) for c in consonants)
    return number_of_consonants

test_str="Hercules was a hero"
result_str=count_consonants(test_str)
print(result_str)
于 2016-02-28T13:42:02.313 回答