输出每次都重复第一个字典

Question

我应该从三个单独的输入接收重新格式化字典的输出。

当前输出：

{'JS': {'Paid': [200, 400, 500, 600]}, 'SK': {'Paid': [400, 1000, 1600]}}
{'JS': {'Paid': [200, 400, 500, 600]}, 'SK': {'Paid': [400, 1000, 1600]}}
{'JS': {'Paid': [200, 400, 500, 600]}, 'SK': {'Paid': [400, 1000, 1600]}}

预期输出：

print(buildCoverageDictionary(paidList1)) ->
{'JS': {'Paid': [200, 400, 500, 600]}, 'SK': {'Paid': [400, 1000, 1600]}}
print(buildCoverageDictionary(paidList2)) ->
{'JS': {'Paid': [200, 400, 500, 600]}, 'SK': {'Paid': [1010, 2000]}, 'MJ': {'Paid':
[5, 6, 7]}, 'ZF': {'Paid': [2660, 500]}}
print(buildCoverageDictionary(paidList3)) -> {'SK': {'Paid': [200, 400]}}

两个问题：

如何编辑代码以输出预期的输出？
我如何编辑代码以不需要我单独重新格式化每个字典，而是创建一个函数，无论输入是什么，都可以正确地重新格式化字典？

我的代码如下：

paidList1 = [["JS", 200, 400, 500, 600], ["SK", 400, 1000, 1600]]
paidList2 = [["JS", 200, 400, 500, 600], ["SK", 1010, 2000], ["MJ", 5, 6, 7], ["ZF", 2660, 500]]
paidList3 = [["SK", 200, 400]]


def buildCoverageDictionary(paidList):
    JS_dict = {'Paid': paidList1[0][1:5]}
    SK_dict = {'Paid': paidList1[1][1:4]}
    new_dict1 = {paidList1[0][0]: JS_dict, paidList1[1][0]: SK_dict}
    return new_dict1

    JS_dict = {'Paid': paidList2[0][1:5]}
    SK_dict = {'Paid': paidList2[1][1:3]}
    MJ_dict = {'Paid': paidList2[2][1:4]}
    ZF_dict = {'Paid': paidList2[3][1:3]}
    new_dict2 = {paidList2[0][0]: JS_dict, paidList2[1][0]: SK_dict, paidList2[2][0]: MJ_dict, paidList2[3][0]: ZF_dict}   
    return new_dict2
    
    SK_dict = {'Paid': paidList3[0][1:3]}
    ew_dict3 = {paidList3[0][0]: SK_dict}
    return new_dict3
    
print(buildCoverageDictionary(paidList1))
print(buildCoverageDictionary(paidList2))
print(buildCoverageDictionary(paidList3))

Answer 1

正如@Barmar 提到的，return 结束函数，执行后什么也没有。您可以尝试类似下面的方法，并使其更灵活地适应输入的变化，例如通过循环 paidList 而不是对索引进行硬编码，而是使用 [1:] 从索引 1 翻译到最后一个

paidList1 = [["JS", 200, 400, 500, 600], ["SK", 400, 1000, 1600]]
paidList2 = [["JS", 200, 400, 500, 600], ["SK", 1010, 2000], ["MJ", 5, 6, 7], ["ZF", 2660, 500]]
paidList3 = [["SK", 200, 400]]


def buildCoverageDictionary(paidList):
    output = {}
    for item in paidList:
        output[item[0]] = {'Paid': item[1:]}
    return output

print(buildCoverageDictionary(paidList1))
print(buildCoverageDictionary(paidList2))
print(buildCoverageDictionary(paidList3))

使用dict comprehension

def buildCoverageDictionary(paidList):
    return {item[0]: {'Paid': item[1:]} for item in paidList}

Answer 2

这应该可以回答您的第一个问题：

paidList1 = [["JS", 200, 400, 500, 600], ["SK", 400, 1000, 1600]]
paidList2 = [["JS", 200, 400, 500, 600], ["SK", 1010, 2000], ["MJ", 5, 6, 7], ["ZF", 2660, 500]]
paidList3 = [["SK", 200, 400]]


def buildCoverageDictionary():
    JS_dict = {'Paid': paidList1[0][1:5]}
    SK_dict = {'Paid': paidList1[1][1:4]}
    new_dict1 = {paidList1[0][0]: JS_dict, paidList1[1][0]: SK_dict}
    print(new_dict1)

    JS_dict = {'Paid': paidList2[0][1:5]}
    SK_dict = {'Paid': paidList2[1][1:3]}
    MJ_dict = {'Paid': paidList2[2][1:4]}
    ZF_dict = {'Paid': paidList2[3][1:3]}
    new_dict2 = {paidList2[0][0]: JS_dict, paidList2[1][0]: SK_dict, paidList2[2][0]: MJ_dict, paidList2[3][0]: ZF_dict}
    print(new_dict2)

    SK_dict = {'Paid': paidList3[0][1:3]}
    new_dict3 = {paidList3[0][0]: SK_dict}
    print(new_dict3)


buildCoverageDictionary()

我所做的更改是：

你没有使用参数 paidList，所以我去掉了它。
当您调用 return 函数时，它会 returns 您想要的东西并跳过该函数的其余部分。所以我用 print.

return

你拼错了 new_dict3(as ew_dict3)。

输出每次都重复第一个字典

Output is repeating the first dictionary every time

python

function

dictionary

format

output