如何将两个嵌套列表合并为字典？

Question

I've been trying to understand how this problem works but i don't know how to merge two nested list into a dictionary.

I need to create a function combine (). If I have a nested list called info and i have another nested list called detail. i need to combine these two nested list as a dictionary and returns the dictionary. I don't quite actually know how to start my code. As I don't actually know how to merge a nested list as dictionary.

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"])]

the output should look like this:

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

Answer 1

Using dict.setdefault

Ex:

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)

Output:

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

---

You can also use collections.defaultdict

Ex:

from collections import defaultdict

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

Output:

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

Assuming only matching names should be merged (as seen in the intended output of OP). This would work.

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)

Output:

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

Sidenote; this could be done a lot more efficiently in terms of processing time. But for a small sample size, as is asked in the original question, this works as intended.

Answer 3

You can do this like this:

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} 

This will give you a dictionary with name as the key and info as value

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

What we did was to check for names in result's key and to append the detail in it's values.

{'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']]}

I find it much easier!