如何计算字符串中没有空格的字母数？

Question

这是我的解决方案导致错误.返回0

PS:我仍然喜欢修复我的代码:)

from collections import Counter
import string


def count_letters(word):
    global count
    wordsList = string.split(word)
    count = Counter()
    for words in wordsList:
        for letters in set(words):
            return count[letters]

word = "The grey old fox is an idiot"
print count_letters(word)

Answer 1

def count_letters(word):
    return len(word) - word.count(' ')

或者,如果您有多个要忽略的字母,则可以过滤字符串:

def count_letters(word):
    BAD_LETTERS = " "
    return len([letter for letter in word if letter not in BAD_LETTERS])

Answer 2

使用sum函数简单解决:

sum(c != ' ' for c in word)

它是一种内存有效的解决方案,因为它使用生成器而不是创建临时列表,然后计算它的总和.

值得一提的是c != ' '返回True or False值,它是类型的值bool,但是bool是子类型int,所以你可以总结bool值(True对应1并False对应0)

您可以使用以下mro方法检查固有情况:

>>> bool.mro() # Method Resolution Order
[<type 'bool'>, <type 'int'>, <type 'object'>]

在这里,您可以看到它bool的子类型int是子类型object.

Answer 3

MattBryant的答案很好,但是如果你想要排除更多类型的字母而不仅仅是空格,它就会变得笨重.以下是您当前使用的代码的变体Counter:

from collections import Counter
import string

def count_letters(word, valid_letters=string.ascii_letters):
    count = Counter(word) # this counts all the letters, including invalid ones
    return sum(count[letter] for letter in valid_letters) # add up valid letters

示例输出:

>>> count_letters("The grey old fox is an idiot.") # the period will be ignored
22