Python Sum字典元素（更好的pythonic方式）

Question

I wrote simple few lines of code to sum elements of python dictionary to make it uniqe.

test =  [{'id': 3L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test', 'sum': 5},
      {'id': 3L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test', 'sum': 1},
      {'id': 3L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test', 'sum': 3},
      {'id': 4L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test2', 'sum': 1},
      {'id': 4L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test2', 'sum': 1},
      {'id': 5L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test4', 'sum': 1},
      {'id': 6L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test5', 'sum': 1}]

newlist = []
newdict = {}

for dict_item in test:

    pixel = dict_item['id']
    country = dict_item['category']
    ac = dict_item['information']
    name = dict_item['name']
    sum = dict_item['sum']

    if newdict.has_key(pixel):

        pos = newdict[pixel]
        newlist[pos] = {'id': pixel, 'category': country, 'information': ac, 'name': name, 'sum': (newlist[pos]['sum']+sum)}
    else:
        newlist.append(dict_item)
        newdict[pixel] = len(newlist) -1

print newlist

Result of my sum:

   [{'category': 'BOOK', 'information': 'abcdefghijk', 'sum': 9, 'id': 3L, 'name': 'test'},{'category': 'BOOK', 'information': 'abcdefghijk', 'sum': 2, 'id': 4L, 'name': 'test2'},{'category': 'BOOK', 'information': 'abcdefghijk', 'sum': 2, 'id': 5L, 'name': 'test5'}]

Is there a better way to sum this dictionary to sum id to be unique?

Answer 1

No need to recreate the entire dictionary with all its keys and values. Just use dict to create a copy of the original dictionary or add to the sum of the dictionary existing in the list. Also, you can derive the newlist from the newdict .

newdict = {} # or collections.OrderedDict() if order is important
for dict_item in test:
    pixel = dict_item['id']
    if pixel in newdict:
        newdict[pixel]["sum"] += dict_item['sum']
    else:
        newdict[pixel] = dict(dict_item)
newlist = list(newdict.values())

Answer 2

I would do it like this. Basically, combine all dicts that have the same id. I'm also sticking with a dict instead of a list so that you can easily pick out the mapping from key to values.

squashed = {}
for inner in test:
    key = inner['id']
    del inner['id']
    if key in squashed:
        squashed[key]['sum'] += inner['sum']
    else:
        squashed[key] = inner

Answer 3

test =  [{'id': 3937L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test', 'sum': 5},
      {'id': 3937L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test', 'sum': 1},
      {'id': 3937L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test', 'sum': 3},
      {'id': 4215L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test2', 'sum': 1},
      {'id': 4215L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test2', 'sum': 1},
      {'id': 4217L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test4', 'sum': 1},
      {'id': 4218L, 'category': 'BOOK', 'information': 'abcdefghijk', 'name': 'test5', 'sum': 1}]

dct = {}

for d in test:
    if d["id"] in dct:
        dct[d["id"]]["sum"] += d["sum"]
    else:
        dct[d["id"]] = d.copy()

from pprint import pprint as pp

pp(dct.values())

Output matches running your own code:

[{'category': 'BOOK',
  'id': 4217L,
  'information': 'abcdefghijk',
  'name': 'test4',
  'sum': 1},
 {'category': 'BOOK',
  'id': 3937L,
  'information': 'abcdefghijk',
  'name': 'test',
  'sum': 9},
 {'category': 'BOOK',
  'id': 4218L,
  'information': 'abcdefghijk',
  'name': 'test5',
  'sum': 1},
 {'category': 'BOOK',
  'id': 4215L,
  'information': 'abcdefghijk',
  'name': 'test2',
  'sum': 2}]

You can simply call dict.copy() to create new dicts, you don't need to create a new dict manually each time.