Question

我在代码部分的下面有这段代码。

我必须返回一个列表，其中第二个参数中的所有函数已经以元组的格式类型应用于第一个参数的元素例如这两个列表

objs = [(1,2),[1,3,4,5,6,7],[0]]
funcs = [len,sum]
I must return #--> [(2, 3, <class 'tuple'>), (6, 26, <class 'list'>), (1, 0, <class 'list'>)]

谢谢您的帮助

objs = [(1,2),[1,3,4,5,6,7],[0]]
funcs = [len,sum,type]


def calc_len(objs, funcs):
    t = []
    z = []
    for i in objs:
        print(i) # To check output
        z.append(i) # Create a list Z of "i"
        #print("Z : ", z)
        # Then I apply each function from the second list
        for j in funcs:
            if j==len:
                print("Lenght:",j(i))
            elif j==sum:
                print("Sum :" ,j(i))
            else:
                print("Type :",j(i))
            t.append(j(i))
 # How Turn that list into a tuple and add to list to return
        print (t)



print(calc_len(objs, funcs))
Should return #--> [(2, 3), (6, 26), (1, 0)]

结果下方

[358]: print(calc_len(objs, funcs))
(1, 2)
Z :  [(1, 2)]
Lenght: 2
Sum : 3
Type : <class 'tuple'>
[2, 3, <class 'tuple'>]
[1, 3, 4, 5, 6, 7]
Z :  [(1, 2), [1, 3, 4, 5, 6, 7]]
Lenght: 6
Sum : 26
Type : <class 'list'>
[2, 3, <class 'tuple'>, 6, 26, <class 'list'>]
[0]
Z :  [(1, 2), [1, 3, 4, 5, 6, 7], [0]]
Lenght: 1
Sum : 0
Type : <class 'list'>
[2, 3, <class 'tuple'>, 6, 26, <class 'list'>, 1, 0, <class 'list'>]

Answer 1

您可以使用列表推导将每个函数应用于参数，并使用结果构建元组。

result = [ tuple(f(x) for f in funcs) for x in objs ]

Answer 2

objs = [(1,2),[1,3,4,5,6,7],[0]]
result =  [(len(item),sum(item),type(item)) for item in objs]
print (result)

输出：

[(2, 3, <class 'tuple'>), (6, 26, <class 'list'>), (1, 0, <class 'list'>)]

。

objs = [(1,2),[1,3,4,5,6,7],[0]]
funcs = ['len','sum','type']

def calc_len(objs, funcs):
    result = []
    for item in objs:
        r = []
        for func in funcs:
            if func == 'len':
                r.append(len(item))
            if func == 'sum':
                r.append(sum(item))
            if func == 'type':
                r.append(type(item))
        result.append(tuple(r))
    return result

print(calc_len(objs, funcs))

输出：

[(2, 3, <class 'tuple'>), (6, 26, <class 'list'>), (1, 0, <class 'list'>)]

加入2列表并转换为元组

2 个答案: