Question

我一直在试图了解此问题的工作原理，但是我不知道如何将两个嵌套列表合并到字典中。

我需要创建一个功能Combine（）。如果我有一个名为info的嵌套列表，而我还有另一个称为detail的嵌套列表。我需要结合这两个嵌套列表作为字典，并返回字典。我实际上并不知道如何启动我的代码。因为我实际上并不知道如何将嵌套列表合并为字典。

info = [["Kean", 36, "Comp Sci", "Dept 2"], ["Ethan", 24, "Engineer", "Dept 5"], ["Kin", 23, "Med Tech", "Dept 1"]]
detail = [("Kean", ['good', "very good", "pass"]), ("Ethan", ["fail", "good", "fail"])]

输出应如下所示：

{"Kean": [["Kean", 36, "Comp Sci", "Dept 2"], ['good', "very good", "pass"]], 'Ethan': [["Ethan", 24, "Engineer", "Dept 5"], ["fail", "good", "fail"]]}

Answer 1

使用dict.setdefault

例如：

result = {}
info = [["Kean", 36, "Comp Sci", "Dept 2"], ["Ethan", 24, "Engineer", "Dept 5"], ["Kin", 23, "Med Tech", "Dept 1"]]
detail = [("Kean", ['good', "very good", "pass"]), ("Ethan", ["fail", "good", "fail"])]
for key, *value in info + detail:
    result.setdefault(key, []).append(value)
print(result)

输出：

{'Ethan': [[24, 'Engineer', 'Dept 5'], [['fail', 'good', 'fail']]],
 'Kean': [[36, 'Comp Sci', 'Dept 2'], [['good', 'very good', 'pass']]],
 'Kin': [[23, 'Med Tech', 'Dept 1']]}

您也可以使用collections.defaultdict

例如：

from collections import defaultdict

result = defaultdict(list)
for key, *value in info + detail:
    result[key].append(value)

输出：

defaultdict(<class 'list'>,
            {'Ethan': [[24, 'Engineer', 'Dept 5'], [['fail', 'good', 'fail']]],
             'Kean': [[36, 'Comp Sci', 'Dept 2'],
                      [['good', 'very good', 'pass']]],
             'Kin': [[23, 'Med Tech', 'Dept 1']]})

Answer 2

假定只应合并匹配的名称（如OP的预期输出所示）。这会起作用。

info = [["Kean", 36, "Comp Sci", "Dept 2"], ["Ethan", 24, "Engineer", "Dept 5"], ["Kin", 23, "Med Tech", "Dept 1"]]
detail = [("Kean", ['good', "very good", "pass"]), ("Ethan", ["fail", "good", "fail"])]

outdict = {}

for i in info:    
    nameInfo = i[0]
    for j in detail:
        nameDetail = j[0]
        if nameDetail == nameInfo:
            outdict[nameDetail] = [i, j[1]]
            
print(outdict)

输出：

{'Kean': [['Kean', 36, 'Comp Sci', 'Dept 2'], ['good', 'very good', 'pass']], 'Ethan': [['Ethan', 24, 'Engineer', 'Dept 5'], ['fail', 'good', 'fail']]}

侧注；就处理时间而言，这可以更有效地完成。但是，如原始问题所要求的那样，对于较小的样本量，这可以按预期进行。

Answer 3

您可以这样做：

info = [["Kean", 36, "Comp Sci", "Dept 2"], ["Ethan", 24, "Engineer", "Dept 5"], ["Kin", 23, "Med Tech", "Dept 1"]]
detail = [("Kean", ['good', "very good", "pass"]), ("Ethan", ["fail", "good", "fail"])]

result={element[0]:[element] for element in info}

这将为您提供以名称为键，信息为值的字典

for name in detail:
    if name[0] in result.keys():
        result[name[0]].append(name[1])
print(result)

我们要做的是检查结果键中的名称，并将详细信息附加在其值中。

{'Kean': [['Kean', 36, 'Comp Sci', 'Dept 2'], ['good', 'very good', 'pass']], 'Ethan': [['Ethan', 24, 'Engineer', 'Dept 5'], ['fail', 'good', 'fail']], 'Kin': [['Kin', 23, 'Med Tech', 'Dept 1']]}

我发现它容易得多！

如何合并两个嵌套列表作为字典？

3 个答案: