Question

这是一个嵌套的字典：

fdict = {}
fdict['apple'] = {}
fdict['banana'] = {}

fdict['apple']['green'] = 5
fdict['apple']['red'] = 0
fdict['banana']['light_yellow'] = 10
fdict['banana']['dark_yellow'] = 0
fdict['appraisal round'] = 1

{'apple': {'green': 5, 'red': 0},
 'banana': {'light_yellow': 10, 'dark_yellow': 0},
 'appraisal round': 1}

从该嵌套字典中删除值为零的键和值对的最有效方法是什么，从而得到以下字典：

{'apple': {'green': 5}, 'banana': {'light_yellow': 10}, 'appraisal round': 1}

注意：appraisal round键本身没有字典作为值。

这是到目前为止我已经实现的：

overall_dict = {}
for key in [key for key in fdict.keys() if key != 'appraisal round']:
    new_dict = {k:v for k,v in fdict[key].items() if v != 0}
    overall_dict[key] = new_dict
overall_dict['appraisal round'] = 1

但是，使用临时词典，构造全新的词典并重新添加到appraisal round中似乎不是一种干净的方法。也许有办法更有效地修改现有字典吗？

Answer 1

假设您不知道具有非字典类型的键，则解决方案可能是：

fdict = {}
fdict['apple'] = {}
fdict['banana'] = {}

fdict['apple']['green'] = 5
fdict['apple']['red'] = 0
fdict['banana']['light_yellow'] = 10
fdict['banana']['dark_yellow'] = 0
fdict['appraisal round'] = 1

def non_zero_value(item):
  k, v = item
  return v != 0

result = {}
for k, v in fdict.items():
  if isinstance(v, dict):
    result[k] = dict(filter(non_zero_value, v.items()))
  else:
    result[k] = v

print(result)
# {'apple': {'green': 5}, 'banana': {'light_yellow': 10}, 'appraisal round': 1}

Answer 2

我建议采用以下解决方案（它适用于多个嵌套级别）：

fdict = {
    'apple': {'green': 5, 'red': 0},
    'banana': {'light_yellow': 10, 'dark_yellow': 0},
    'appraisal round': 1
}

noZero = lambda d: { k1: noZero(v1) if isinstance(v1, dict) else v1 for k1, v1 in d.items() if v1 }

print(noZero(fdict))  # {'apple': {'green': 5}, 'banana': {'light_yellow': 10}, 'appraisal round': 1}

从值为零的嵌套字典中删除项目

2 个答案: