如何对按多个字段分组的数量求和？

Question

如何对按 [pid][sid][eid][dbid] 字段分组的 'qty' 字段求和：

d = [
   {'pid': 146, 'sid': 6, 'eid': 6212, 'dbid': 1, 'qty': 10},
   {'pid': 146, 'sid': 6, 'eid': 6212, 'dbid': 1, 'qty': 20},
   {'pid': 232, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 1},
   {'pid': 232, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 1},
   {'pid': 146, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 100},
   {'pid': 146, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 100},
   ]

# non-working code
sum_of_qty = []
for row in d:
    sum_of_qty[[row['pid']][row['sid']][row['eid']][row['dbid']]] += row['qty']

这是我想要的最终结果：

sum_of_qty = [
    {'pid': 146, 'sid': 6, 'eid': 6121, 'dbid': 1, 'qty': 30},
    {'pid': 232, 'sid': 56, 'eid': 6121, 'dbid': 1, 'qty': 2},
    {'pid': 146, 'sid': 56, 'eid': 6121, 'dbid': 1, 'qty': 200}
    ]

Answer 1

您可以使用 itertools.groupby:

import itertools
d = [
  {'pid': 146, 'sid': 6, 'eid': 6212, 'dbid': 1, 'qty': 10},
  {'pid': 146, 'sid': 6, 'eid': 6212, 'dbid': 1, 'qty': 20},
  {'pid': 232, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 1},
  {'pid': 232, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 1},
  {'pid': 146, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 100},
  {'pid': 146, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 100},
]
new_d = [list(b) for _, b in itertools.groupby(d, key=lambda x:[x['pid'], x['sid']])]
final_result = [{**i[0], **{'qty':sum(b['qty'] for b in i)}} for i in new_d]

输出：

[{'pid': 146, 'sid': 6, 'eid': 6212, 'dbid': 1, 'qty': 30}, {'pid': 232, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 2}, {'pid': 146, 'sid': 56, 'eid': 6212, 'dbid': 1, 'qty': 200}]

Answer 2

试试这个：

sum_of_qty = []
temp = {} # temporary datastructure

for element in d:
    tup = (element['pid'],element['sid'],element['eid'],element['dbid']) # tuple
    if tup not in temp:
       temp[tup] = element['qty']
    else:
       temp[tup] += element['qty']

for key in temp:
   dic = {'pid':key[0],'sid': key[1], 'eid': key[2], 'dbid': key[3], 'qty': temp[key]}
   sum_of_qty.append(dic)

Answer 3

基于 Ajax1234 答案，但更具可读性：

import itertools
from collections import ChainMap

sum_of_qty = [
    list(b) for _, b in itertools.groupby(d, key=lambda x: [x[k] for k in x if k != 'qty'])
]

for i, item in enumerate(sum_of_qty):
    sum_of_qty[i] = dict(ChainMap(*item))
    sum_of_qty[i]['qty'] = sum(map(lambda x: x['qty'], item))

如何对按多个字段分组的数量求和？

How to sum the quantity grouped by more than one field?

python

python-3.x

python-3.4