Python将元组列表转换为嵌套字典列表

我在Python编程，我有一个元组列表：

my_tuples = [ 
    ('csrf_token', u'5bc7224ef8f71c8142b4bef1597883b049b83fee'), 
    ('fields-team', u'dsada'), 
    ('fields-year_from', u'dsa'), 
    ('fields-year_to', u'dsa'), 
    ('fields-description', u'dsa') , 
    ('fields-team', u''), 
    ('fields-year_from', u''), 
    ('fields-year_to', u''), 
    ('fields-description', u'') 
]

我不得不元组的这个名单变成这样：

form_data = { 
    'csrf_token':u'5bc7224ef8f71c8142b4bef1597883b049b83fee', 
    'fields':[ 
     { 
      'team': u'dsada', 
      'year_from': u'dsa', 
      'year_to': u'dsa', 
      'description': u'dsa' 
     }, 
     { 
      'team': u'', 
      'year_from': u'', 
      'year_to': u'', 
      'description': u'' 
     } 
    ] 
}

然后，我不得不删除第二个字典，因为每个键都有一个空值。所以最终的结果将是：

form_data = { 
    'csrf_token':u'5bc7224ef8f71c8142b4bef1597883b049b83fee', 
    'fields':[ 
     { 
      'team': u'dsada', 
      'year_from': u'dsa', 
      'year_to': u'dsa', 
      'description': u'dsa' 
     } 
    ] 
}

来源

2013-04-05 Michael

将第二字典什么样子好像是 “减肥”？ – Gerrat 2013-04-05 17:57:37

确保这4个“字段”会出现吗？是否有可能出现您未显示的其他字段？ – jedwards 2013-04-05 17:59:33

是的也有可能，这些字段有不同的名字... – Michael 2013-04-05 18:02:13

如何：

def myfunc(my_tuples): 
    mydict = {} 
    d = {} 
    for kv in my_tuples: 
     if kv[0] == 'csrf_token': 
      mydict[kv[0]] = kv[1].decode('ascii') 
     elif 'fields' in kv[0]: 
      d[kv[0]] = kv[1].decode('ascii') 
    mydict['fields'] = [d] 
    return mydict 


def main(): 
    my_tuples = [ 
     ('csrf_token', u'5bc7224ef8f71c8142b4bef1597883b049b83fee'), 
     ('fields-team', u'dsada'), 
     ('fields-year_from', u'dsa'), 
     ('fields-year_to', u'dsa'), 
     ('fields-description', u'dsa'), 
     ('fields-team', u''), 
     ('fields-year_from', u''), 
     ('fields-year_to', u''), 
     ('fields-description', u'') 
    ] 

    myfunc(my_tuples) 


if __name__ == "__main__": 
    main()

来源

2013-04-05 19:08:32 ThinkCode

我的两分钱：

import collections 

def transform(loft): 
    form_data = collections.defaultdict(list) 
    internal = {} 
    numfield = 0 
    for t in my_tuples: 
     keys = t[0].split("-") 
     if len(keys) == 1: 
      form_data[keys[0]] = t[1] 
     else: 
      internal[keys[1]] = t[1] 
      if numfield == 3: 
       numfield = -1 
       form_data[keys[0]].append(internal) 
       internal = {} 
      numfield += 1 
    return form_data 

if __name__ == "__main__": 

    my_tuples = [ 
     ('csrf_token', u'5bc7224ef8f71c8142b4bef1597883b049b83fee'), 
     ('fields-team', u'dsada'), 
     ('fields-year_from', u'dsa'), 
     ('fields-year_to', u'dsa'), 
     ('fields-description', u'dsa') , 
     ('fields-team', u''), 
     ('fields-year_from', u''), 
     ('fields-year_to', u''), 
     ('fields-description', u'') 
    ] 
    trans = transform(my_tuples) 
    # cleanup 
    trans['fields'] = [f for i, f in enumerate(trans['fields']) 
     if trans['fields'][i]['team'] != '' 
     and trans['fields'][i]['year_from'] != '' 
     and trans['fields'][i]['year_to'] != '' 
     and trans['fields'][i]['description'] != ''] 
    print dict(trans)

来源

2013-04-05 20:21:16 mezzomondo

Python将元组列表转换为嵌套字典列表

回答

相关问题