Python面试题分享

分享几道很有意思又简单的面试题

a = ['a,1', 'b,3,22', 'c,3,4']
b = ['a,2', 'b,1', 'd,2']
按每个字符串的第一个值，合并a和b到c
效果如下
c = ['a,1,2', 'b,3,22,1', 'c,3,4', 'd,2']

a = ['a,1', 'b,3,22', 'c,3,4']
b = ['a,2', 'b,1', 'd,2']

# 解如下：
def func(a):
    dic = dict()
    for i in a:
        k, v = i.split(',', 1)
        dic[k] = v
    return dic

a = func(a)
b = func(b)


def func_dic(a: dict, b: dict):
    for ka, va in a.items():
        if ka in b.keys():
            a[ka] = va + ',' + b[ka]
    else:
        for kb, vb in b.items():
            if kb not in a.keys():
                a[kb] = vb
    return a


a = func_dic(a, b)
c = [k + ',' + v for k, v in a.items()]
print(c)

多层嵌套列表合并 并遍历

A = [1,2,3,[4,5,6,7,[1,2,4,5,6,[12,3,4,5]]]]

A = [1,2,3,[4,5,6,7,[1,2,4,5,6,[12,3,4,5]]]]
lst_A = list()
# 生成器解法

def func(a):
    for i in a:
        if type(i) == list:
            for j in func(i):
                yield j
        else:
            yield i


for x in func(A):
    list_A.append(x)
    print(x)

# 运用到生成器 很有趣味性哦

# 看似很粗暴的递归解法
def func(A):
    for i in A:
        if type(i) == list:
            func(i)
        else:
            list_A.append(i)
            print(i)
    return list_A


func(A)