Question

我想根据字典键从当前字典列表中创建一个子列表。

我的数据：

[{'0': 2}, {'0': 1}, {'1': 2}, {'2': 2}, {'2': 2}]

我想要实现的数据：

[ [{'0': 2}, {'0': 1}], [{'1': 2}], [{'2': 2}, {'2': 2}] ]

如您所见，内部数组包含一个具有相同键值的字典。

我的代码当前代码是这样的：

dicts = [{'0': 2}, {'0': 1}, {'1': 2}, {'2': 2}, {'2': 2}]

ex_list = []
sublist = []
for group in dicts:
  if group.keys() in sublist:
    sublist.append(group)
  else:
    sublist.append(group)
    if group.keys() != sublist[-1]:
      sublist = []
      sublist.append(group)
ex_list.append(sublist)

非常感谢任何帮助。

Answer 1

请参阅内嵌注释以获取解释。

from collections import defaultdict

dicts = [{'0': 2}, {'0': 1}, {'1': 2}, {'2': 2}, {'2': 2}]

# keep track of mapping between key and values.
result = defaultdict(list)

for d in dicts:
    # d.items() returns an iterable of key/value pairs.
    # assuming each dictionary only has one key/value pair,
    # using next(iter()), we get the first pair, and pattern-match on key and val.
    key, val = next(iter(d.items())):

    # with defaultdict, if key is not present in the result dictionary,
    # the list will be created automatically.
    result[key].append(val)

# results = {0: [2,1], 1: [2], 2: [2,2]}
# for each key, values pair in results, create a list of {key: value}
# dictionaries for each value in values.

print([[{key: value} for value in values] for key, values in result.items()])

Answer 2

如果你想接近你的程序，你应该跟踪当前和最后一个键，我已经重写了你的一些代码，它完成了工作。

dicts = [{'0': 2}, {'0': 1}, {'1': 2}, {'2': 2}, {'2': 2}]

ex_list = []
sublist = []
lastkey = list(dicts[0].keys())[0]

for group in dicts:
  key = list(group.keys())[0]
  if key == lastkey:
    sublist.append(group)
  else: # If key has change
    ex_list.append(sublist)
    sublist = []
    lastkey = key
    sublist.append(group)
ex_list.append(sublist) #Don't forget to include last sublist as the loop doesn't include it since no change in key

print(ex_list)

从字典列表创建子列表

2 个答案: