Question

python的新手，我有以下字典，并得到一个字典表格，这将不包含重复但如果找到重复，那么数据必须附加到第一个键和值。例如，在41中有重复并且有pending = 1并且已交付14.我想从此创建这样的列表，它只包含一行41和contians计数挂起和交付以及这两种状态的加号。

    temp = [
(41, 1, 2015-1-22 12:37:58.631670, 'Pending'), 
(37, 1, 2015-1-21 13:56:3.632057, 'Delivered'), 
(41, 14, 2015-1-22 12:37:58.631670, 'Delivered'), 
(36, 1, 2015-1-21 13:22:52.705818, 'Delivered'), 
(40, 2, 2015-1-22 12:37:58.631670, 'Delivered'), 
(38, 1, 2015-1-21 14:4:10.206100,, 'Delivered')
]

第一列是id，第二列是状态计数，状态是待处理，已交付，失败。

如果想要像这样制作字典

dict = {id : { id : id, Pending : pending_count, Failed : failed_count, Delivered : delivered_count, total : pending+failed+delivered, date-time : date-time}}

喜欢

dict = { 
id : { 'id' : 41, 'Pending' : 1, 'Failed' : 0, 'Delivered' : 14, 'total' : 15, 'date time' : 2015-1-22 12:37:58.631670},
id : { 'id' : 37, 'Pending' : 0, 'Failed' : 0, 'Delivered' : 1, 'total' : 1, 'date-time' : 2015-1-21 13:56:3.632057}
}

Answer 1

由于输入列表具有恒定的结构。列表元素为tuple 因此，对于输出字典，每个id内的第一个项tuple为key，输出字典的值也是字典。

迭代temp列表中的每个项目。
根据状态获取所有计数并分配尊重计数。使用if语句。
如果输出字典中存在键，则更新现有值，即更新所有计数值以及总计和日期值。
如果不存在，则添加输出字典。
使用try除了处理异常，即当输出字典中没有键时。

代码：

import pprint

temp = [
(41, 1, "2015-1-22 12:37:58.631670", 'Pending'), 
(37, 1, "2015-1-21 13:56:3.632057", 'Delivered'), 
(41, 14, "2015-1-22 12:37:58.631670", 'Delivered'), 
(36, 1, "2015-1-21 13:22:52.705818", 'Delivered'), 
(40, 2, "2015-1-22 12:37:58.631670", 'Delivered'), 
(38, 1, "2015-1-21 14:4:10.206100", 'Delivered')
]

output = {}
for i in temp:
    id = i[0]
    count = i[1]
    date_v = i[2]
    status = i[3]
    p_count = 0
    d_count = 0
    f_count = 0
    if status=="Pending":
        p_count = count
    elif status=="Delivered":
        d_count = count
    elif status=="Failed":
        f_count= count

    try:
        output[i[0]]["Pending"] = output[i[0]]["Pending"]+p_count
        output[i[0]]["Failed"] = output[i[0]]["Failed"]+f_count
        output[i[0]]["Delivered"] = output[i[0]]["Delivered"]+d_count
        output[i[0]]["total"] = output[i[0]]["Pending"]+count
        output[i[0]]["date time"] = date_v
    except KeyError, e:
        total = count
        output[i[0]] = {'id':id, 'Pending':p_count, 'Failed':f_count,\
                        'Delivered':d_count, 'total':total, 'date time':date_v}


pprint.pprint(output)

输出：

{36: {'Delivered': 1,
      'Failed': 0,
      'Pending': 0,
      'date time': '2015-1-21 13:22:52.705818',
      'id': 36,
      'total': 1},
 37: {'Delivered': 1,
      'Failed': 0,
      'Pending': 0,
      'date time': '2015-1-21 13:56:3.632057',
      'id': 37,
      'total': 1},
 38: {'Delivered': 1,
      'Failed': 0,
      'Pending': 0,
      'date time': '2015-1-21 14:4:10.206100',
      'id': 38,
      'total': 1},
 40: {'Delivered': 2,
      'Failed': 0,
      'Pending': 0,
      'date time': '2015-1-22 12:37:58.631670',
      'id': 40,
      'total': 2},
 41: {'Delivered': 14,
      'Failed': 0,
      'Pending': 1,
      'date time': '2015-1-22 12:37:58.631670',
      'id': 41,
      'total': 15}}

Answer 2

我认为你想要的是：

from collections import defaultdict

d = defaultdict(dict)

for row in temp:
  result = d[row[0]]
  result[row[-1]] = result.setdefault(row[-1], 0) + row[1]
  result['total'] = result.setdefault('total', 0) + row[1]
  result['{}-date'.format(row[-1])] = row[2]

对于d[41]，这会给你：

{'Delivered': 14,
 'total': 15,
 'Pending-date': '2015-1-22 12:37:58.631670',
 'Pending': 1,
 'Delivered-date': '2015-1-22 12:37:58.631670'}

过滤掉删除重复项的字典

2 个答案: