Question

我有以下输入的词典列表：

    links = [ {'uid': 1, 'lid': 6, 'path': 'a1.txt', 'shareid': 1},
      {'uid': 1, 'lid': 7, 'path': 'a2.txt', 'shareid': 2},
      {'uid': 1, 'lid': 8, 'path': 'a1.txt', 'shareid': 1}]

我需要生成此输出：

    op = {'a1.txt': {'shareid': 1, 'lid': [6, 8]},
          'a2.txt': {'shareid': 2, 'lid': [7]}
         }

以下是我写的代码：

def list_all_links():
       new_list = []
       result = {}

       for i in range(len(links)):
           entry = links[i]

    if not result.has_key(entry['path']):
        new_entry = {}
        lid_list = []
        new_entry['shareid'] = entry['shareid']
        if new_entry.has_key('lid'):
            lid_list = new_entry['lid']
            lid_list.append(entry['lid'])
        else:
            lid_list.append(entry['lid'])
        new_entry['lid'] = lid_list

        result[entry['path']] = new_entry

    else:
        new_entry = result[entry['path']]
        lid_list = new_entry['lid']

        if new_entry.has_key(entry['shareid']):
            new_entry['shareid'] = entry['shareid']
            lid_list = new_entry['lid']
            lid_list.append(entry['lid'])
            new_entry['lid'] = lid_list


        else:
            new_entry['shareid'] = entry['shareid']
            lid_list.append(entry['lid'])
            new_entry['lid'] = lid_list


        result[entry['path']] = new_entry

print "result = %s" %result


if __name__ == '__main__':
    list_all_links()

我能够根据需要生成相同的输出。但是，有人可以指出我是否有更好的方法来解决这个问题？

Answer 1

它不是那么漂亮，但以下解决方案有效：

links = [ {'uid': 1, 'lid': 6, 'path': 'a1.txt', 'shareid': 1},
      {'uid': 1, 'lid': 7, 'path': 'a2.txt', 'shareid': 2},
      {'uid': 1, 'lid': 8, 'path': 'a1.txt', 'shareid': 1}]

links_restructured = [(d['path'], {'shareid': d['shareid'], 'lid': [d['lid']]}) for d in links]
answer = {}
for link in links_restructured:
    if link[0] not in answer:
        answer[link[0]] = link[1]
    else:
        answer[link[0]]['lid'].extend(link[1]['lid'])
print(answer)

<强>输出

{'a2.txt': {'lid': [7], 'shareid': 2}, 'a1.txt': {'lid': [6, 8], 'shareid': 1}}

Answer 2

links = [ {'uid': 1, 'lid': 6, 'path': 'a1.txt', 'shareid': 1},
      {'uid': 1, 'lid': 7, 'path': 'a2.txt', 'shareid': 2},
      {'uid': 1, 'lid': 8, 'path': 'a1.txt', 'shareid': 1}]


def get_links(links):
    new_links = {}
    for x in links:
        path = x.get('path')
        if path in new_links.keys():
            new_links[path]['lid'].append(x['lid'])
        else:
            del x['path']
            del x['uid']
            x['lid'] = [x['lid']]
            new_links[path] = x

    return new_links

print(get_links(links))

输出：

{'a2.txt': {'lid': [7], 'shareid': 2}, 'a1.txt': {'lid': [6, 8], 'shareid': 1}}

Answer 3

您可以使用setdefault的{{1}}方法缩短

dict

<强>输出：

links = [
  {'uid': 1, 'lid': 6, 'path': 'a1.txt', 'shareid': 1},
  {'uid': 1, 'lid': 7, 'path': 'a2.txt', 'shareid': 2},
  {'uid': 1, 'lid': 8, 'path': 'a1.txt', 'shareid': 1}
]

op = dict()
for a in links:
  op.setdefault(a['path'], {}).update(shareid=a['shareid'])
  op[a['path']].setdefault('lid', []).append(a['lid'])
print op

Answer 4

以下是我的表现：

def process_links(links):
    '''
       process entries in list 'links';
       returns dictionary 'op'
    '''
    op = {}
    for dict in links:
        op_key = dict['path']
        if op_key in op:
            pass
        else:
            op[op_key] = {'shareid':None, 'lid':[]}
    return op

def fill_op(op_dict, link_list):
    for dict in link_list:
        op_key = dict['path']
        # fill shareid
        op_dict[op_key]['shareid'] = dict['shareid']
        # fill lid
        lid_list = op_dict[op_key]['lid'] 
        lid_list.append(dict['lid'])
        op_dict[op_key]['lid'] = lid_list
    return op_dict


if __name__ == "__main__":
    links = [ {'uid': 1, 'lid': 6, 'path': 'a1.txt', 'shareid': 1},
              {'uid': 1, 'lid': 7, 'path': 'a2.txt', 'shareid': 2},
              {'uid': 1, 'lid': 8, 'path': 'a1.txt', 'shareid': 1}]
    result1 = process_links(links)
    result2 = fill_op(result1, links)
    print(result2)

输出略有不同：{＆＃39; a1.txt＆＃39;：{＆＃39; lid＆＃39;：[6,8]，＆＃39; shareid＆＃39;：1}，＆＃39 ; a2.txt＆＃39;：{＆＃39; lid＆＃39;：[7]，＆＃39; shareid＆＃39;：2}}

Python词典列表问题

4 个答案:

输出：