判断列表中某一元素有多少个

pythonic

# 这个是一个很程序化的写法，但是没有利用好python的特性
def number_of_occurrences(s, xs):
    num = 0
    for i in sorted(xs):
        if i == s:
            num += 1
    return num

# 这种写法就很pythonic
def number_of_occurrences2(s, xs):
    return xs.count(s)

print number_of_occurrences(2, [1, 2, 2, 3, 4, 5])
print number_of_occurrences2(2, [1, 2, 2, 3, 4, 5])
# result 2