在python中的dicts列表中对特定键的值求和答案

【问题标题】：Summing values for a particular key in a list of dicts in python在python中的dicts列表中对特定键的值求和
【发布时间】：2016-03-01 17:10:30
【问题描述】：

这是一个棘手的小问题，我被困住了。我有一个字典列表，如下所示：

[{'medication_name': 'Actemra IV',
  'total_prescriptions': 4},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 3},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 1},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 6},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 8},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 1},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 3}]

我想要做的是总结每种药物的各种字典的总处方，并将最终总和作为每个字典的条目附加如下：

[{'medication_name': 'Actemra IV',
  'total_prescriptions': 4,
  'final_count': 14},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 3,
  'final_count': 14},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 1,
  'final_count': 14},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 6,
  'final_count': 14},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 8,
  'final_count': 12},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 1,
  'final_count': 12},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 3,
  'final_count': 12}
 ]

最有效的方法是什么？

【问题讨论】：

你为什么选择了另一个答案，和我的一样，只是慢了点？
哦，我看到他对其进行了编辑以纠正其不必要的复杂性。

标签： list python-2.7 dictionary

【解决方案1】：

从集合导入计数器

counts = Counter()
for dct in lst:
    counts[dct['medication_name']] += dct['total_prescriptions']
for dct in lst:
    dct['final_count'] = counts[dct['medication_name']]

from pprint import pprint    as pp

输出：

[{'final_count': 14,
  'medication_name': 'Actemra IV',
  'total_prescriptions': 4},
 {'final_count': 14,
  'medication_name': 'Actemra IV',
  'total_prescriptions': 3},
 {'final_count': 14,
  'medication_name': 'Actemra IV',
  'total_prescriptions': 1},
 {'final_count': 14,
  'medication_name': 'Actemra IV',
  'total_prescriptions': 6},
 {'final_count': 12,
  'medication_name': 'Actemra SC',
  'total_prescriptions': 8},
 {'final_count': 12,
  'medication_name': 'Actemra SC',
  'total_prescriptions': 1},
 {'final_count': 12,
  'medication_name': 'Actemra SC',
  'total_prescriptions': 3}]

【讨论】：

【解决方案2】：

from collections import defaultdict
tot = defaultdict(int)
for d in your_dict_list:
    tot[d['medication_name']] += d['total_prescriptions']

for d in your_dict_list:
    d['final_count'] = tot[d['medication_name']]

类似的东西应该是相当有效的。

【讨论】：

【解决方案3】：

from collections import defaultdict


dlist = [{'medication_name': 'Actemra IV',
  'total_prescriptions': 4},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 3},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 1},
 {'medication_name': 'Actemra IV',
  'total_prescriptions': 6},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 8},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 1},
 {'medication_name': 'Actemra SC',
  'total_prescriptions': 3}]


indexes = defaultdict(list)

for i in xrange(0, len(dlist)):
    indexes[dlist[i]['medication_name']].append(i)

for med_k, med_list in indexes.iteritems():
    tot = sum([dlist[i]['total_prescriptions'] for i in med_list])
    for i in med_list:
        dlist[i]['final_count'] = tot

这是相当有效的，因为它只循环列表中的每个字典一次，然后每个不同的医学名称循环一次。

【讨论】：

【解决方案4】：

使用 groupby 和 itemgetter：

from itertools import groupby
from operator import itemgetter
L = [ (i['medication_name'],i['total_prescriptions']) for i in dlist]
sum_dict = dict([(x, sum(map(itemgetter(1), y))) for x, y in groupby(L, itemgetter(0))])
for i, v in enumerate(dlist): dlist[i]['final_count'] = sum_dict[v['medication_name']]
print dlist

输出：

[{'medication_name': 'Actemra IV', 'final_count': 14, 'total_prescriptions': 4}, 
{'medication_name': 'Actemra IV', 'final_count': 14, 'total_prescriptions': 3}, 
{'medication_name': 'Actemra IV', 'final_count': 14, 'total_prescriptions': 1}, 
{'medication_name': 'Actemra IV', 'final_count': 14, 'total_prescriptions': 6}, 
{'medication_name': 'Actemra SC', 'final_count': 12, 'total_prescriptions': 8}, 
{'medication_name': 'Actemra SC', 'final_count': 12, 'total_prescriptions': 1}, 
{'medication_name': 'Actemra SC', 'final_count': 12, 'total_prescriptions': 3}]

【讨论】：