Python函数错误

Question

我写了一个简单的函数程序。该程序本身工作，但转换为函数时，它给出的字符数为1.我正在使用Python版本3.5。

def sink(x,y): 
    x = ['a','a','b','c'] 
    y = 'c' 

    count = {} 
    for char in x: 
    count.setdefault(char,0) 
    count[char] = count[char]+1 
    print (count.get(y,0)) 



sink (['a','a','c','d','e','e'],'e') 

Answer 1

你在一开始重写的x，y值，尝试

def sink(x, y): 
    count = {} 
    for char in x: 
     count.setdefault(char, 0) 
     count[char] = count[char] + 1 
    print(count.get(y, 0)) 


sink(['a', 'a', 'c', 'd', 'e', 'e'], 'e') 

我们也可以通过它returning值和印刷外，并使用augmented arithmetic assignment +=

def sink2(x, y): 
    count = {} 
    for char in x: 
     count.setdefault(char, 0) 
     count[char] += 1 
    return count.get(y, 0) 


print(sink(['a', 'a', 'c', 'd', 'e', 'e'], 'e')) 

提高你的函数

终于我们可以用str.join和str.count方法：

def sink3(x, y): 
    return ''.join(x).count(y) 


print(sink3(['a', 'a', 'c', 'd', 'e', 'e'], 'e')) 

Answer 2

Python有一个Counter对象，这是否和Counter返回0不已经出现了价值，所以很简单：

from collections import Counter 
def sink(x, y): 
    return Counter(x)[y] 

>>> sink(['a','a','c','d','e','e'], 'e') 
2