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)

大致思路就是 [['server', ''], ['port', '8800'], ['location', '/'], ['location', '/aa'], ['location', 'bb']] 这种的视为一个单元处理(函数修改自你给的代码),递归到可以处理程度(因为层数可能不为2)

PHP中文网 · Answer

代碼:

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

結果:

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

我覺得不必要的代碼可以拿掉，這次的問題既然比較單純(既然這次都確定底層是雙元素的 list)，那就不用做太多判斷和處理(甚至這個 recursive 的解法也可以改成簡單的做法)。
我不確定你的資料裡面，除了 location 以外還有沒有別的會重複的東西，我覺得統一處理比較好，所以我字典的值跟你給出的範例不太一樣，用 list 而不是 string。

伊谢尔伦 · Answer

分拆开来好看些

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

别人已经授你渔了，一味的要鱼水平不会有提高的

python - 遍历list，转换成dict

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