Python词典列表问题
我对字典列表有以下输入:Python词典列表问题,python,Python,我对字典列表有以下输入: 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}] 我需要生成这个输出: op = {'a1.txt': {'shareid': 1, 'l
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}]
我需要生成这个输出:
op = {'a1.txt': {'shareid': 1, 'lid': [6, 8]},
'a2.txt': {'shareid': 2, 'lid': [7]}
}
下面是我编写的代码:
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,有人能告诉我是否有更好的方法来解决这个问题吗?虽然不是很好,但下面的解决方案很有效:
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)
输出
{'a2.txt': {'lid': [7], 'shareid': 2}, 'a1.txt': {'lid': [6, 8], 'shareid': 1}}
输出:
您可以使用
dict
的setdefault
方法将其缩短
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
输出:
{'a2.txt': {'lid': [7], 'shareid': 2}, 'a1.txt': {'lid': [6, 8], 'shareid': 1}}
我是这样做的:
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:{'a1.txt':{'lid':[6,8],'sharedid':1},'a2.txt':{'lid':[7],'sharedid':2}
{'a2.txt': {'lid': [7], 'shareid': 2}, 'a1.txt': {'lid': [6, 8], 'shareid': 1}}
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)