Question

例如

String = "The boy who cried wolf made many many silly mistakes."
def f(n):

。

f(3)

应输出：

因为只有2个字母，“a”，“m”，“y”和“e”出现三次以上。

到目前为止，我已经尝试过以及到目前为止的目标：

 String = "The boy who cried wolf made many many silly mistakes."
 a={}
 def f(n):
  for s in String:
    if s in a:
      a[s] +=1
    else:
      a[s] = 1

我不确定下一步该做什么

Answer 1

您可以使用集合模块中的Counter Class。

>>> from collections import Counter
>>> string = "The boy who cried wolf made many many silly mistakes."
>>> Counter(string)
Counter({' ': 9, 'e': 4, 'a': 4, 'y': 4, 'm': 4, 's': 3, 'i': 3, 'o': 3, 'l': 3, 'n': 2, 'd': 2, 'h': 2, 'w': 2, 'T': 1, 'b': 1, '.': 1, 'k': 1, 'r': 1, 't': 1, 'c': 1, 'f': 1})
>>> m = 0
>>> c = Counter(string)
>>> for i in c:
        if i.isalpha() and c[i] > 3:
            m += 1


>>> m
4

将其定义为单独的功能。

from collections import Counter
def f(n):
    c = Counter(n)
    m = 0
    for i in c:
        if i.isalpha() and c[i] > 3:
            m += 1
    return m
stri = "The boy who cried wolf made many many silly mistakes."    
print f(stri)

Answer 2

使用collections.Counter和sum

from collections import Counter    
s = "The boy who cried wolf made many many silly mistakes."
sum(1 for k,  v in Counter(s).items() if v > 3 and k.isalpha())

<强>输出

或者您可以定义一个函数。

def f(s, n):
    return sum(1 for k,  v in Counter(s).items() if v > n and k.isalpha())

和

f(s, 3) #return 4

Answer 3

使用collections.Counter肯定是一种很好的方法;但是，如果您不想导入任何内容，请考虑以下内容。

string = "The boy who cried wolf made many many silly mistakes."
def f(n):
    alpha_chars = filter(lambda char: char.isalpha(), string)
    greater_chars = 0
    for char in list(set(alpha_chars)):
        if string.count(char) > n:
            greater_chars += 1
    return greater_chars

f(3)将输出4。

但是，as explained here，最好使用字符串作为参数。

my_string = "The boy who cried wolf made many many silly mistakes."
def f(string, n):
    ...

然后，您可以使用f(my_string, 3)进行调用，这将输出4。

查找在python上出现超过特定次数的字母数？

3 个答案: