python - 遍历list，转换成dict

Question

我有一个list如下 {代码...} 想要得到如下的dict {代码...} 或者 {代码...} 求解 之前有sf上的同学给出了一个例子 {代码...} 但是这个上面代码里，location是覆盖的，因为dic.update(subdic)，有什么办法是append...

巴扎黑 · Answer

def single_turn2dic(lst):
    global key, value
    dic = {}
    if all([not isinstance(item, list) for item in lst]):
        if len(lst) == 2:
            key, value = lst
        elif len(lst) == 1:
            key = lst[0]
            value = ''
        elif len(lst) == 3:
            key = lst[0]
            value = lst[1]
            # dic[key] = value
        if value and dic.get(key, ''):
            dic[key] += ',' + value
        else:
            dic[key] = value

    else:
        for item in lst:
            subdic = single_turn2dic(item)
            # print subdic
            # dic.update(subdic)
            for name, name_value in subdic.iteritems():
                if name_value and dic.get(name, ''):
                    dic[name] += ',' + name_value
                else:
                    dic[name] = name_value
                    # print dic
    return dic


servers = [[['server', ''], ['port', '8800'], ['location', '/'], ['location', '/aa'], ['location', 'bb']],
           [['server', ''], ['port', '80'], ['location', '/'], ['location', '/aa'], ['location', 'bb']]]


def beautify(servers):
    # lst = []
    if not isinstance(servers[0][0], list):

        return single_turn2dic(servers)
    else:
        return [beautify(server) for server in servers]
        



print beautify(servers)

The general idea is [['server', ''], ['port', '8800'], ['location', '/'], ['location', '/aa'], ['location', 'bb']] This kind of processing is regarded as a unit (the function is modified from the code you gave), and the recursion is to the extent that it can be processed (because the number of layers may not be 2)

PHP中文网 · Answer

Code:

def turn2dic(lst):
    dic = {}
    if all([not isinstance(item, list) for item in lst]):
        key, value = lst
        dic[key] = value
    else:
        for item in lst:
            subdic = turn2dic(item)
            for key, value in subdic.items():
                dic.setdefault(key, []).append(value)
    return dic

results = [turn2dic(item) for item in lst]

print results

Result:

[{'location': ['/', '/aa', 'bb'], 'port': ['8800'], 'server': ['']}, {'location': ['/', '/aa', 'bb'], 'port': ['80'], 'server': ['']}]

I think unnecessary code can be removed. Since the problem this time is relatively simple (since it is determined that the bottom layer is a two-element list this time), there is no need to do too much judgment and processing (even this recursive solution can also be used) Change to a simple method).
I'm not sure if there are any other repetitive things in your information besides location. I think it's better to process them uniformly, so the values in my dictionary are different from the examples you gave. I use list instead of string. .

伊谢尔伦 · Answer

Break it apart to look better

from collections import defaultdict

def testtype(t):
    return lambda x: isinstance(x, t)

def test(t, lst):
    return all(map(t if callable(t) else testtype(t), lst))

def parse_dict(lst):
    dct = defaultdict(lambda: [])
    for k, v in lst:
        dct[k].append(parse(v))
    ret = {}
    for k, v in dct.items():
        if test(str, v):
            v = ','.join(v)
        elif len(v) == 1:
            v = v[0]
        ret[k] = v
        
    return ret

def parse(obj):
    if not isinstance(obj, list):
        return obj
    if test(lambda x: len(x) == 2, obj):
        return parse_dict(obj)
    return list(map(parse, obj))

src= [[['server', ''],['port','8800'],['location','/'],['location','/aa'],['location','bb']],[['server', ''],['port','80'],['location','/'],['location','/aa'],['location','bb']]]
print(parse(src))

ringa_lee · Answer

Others have already taught you how to fish, so your level will not improve if you just ask for fish

python - 遍历list，转换成dict

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