更新列表中字典中的值类型

Question

有人可以启发我吗？：

我有一个字典列表：

[{'overall': 5.0,
  'vote': 'overall',
  'reviewerID': 'AAP7PPBU72QFM'},
 {'overall': 3.0,
  'vote': '5',
  'reviewerID': 'A2E168DTVGE6SV'},
...]

如何将“整体”和“投票”项目转换为整数，同时将所有无效数字设置为 0？

更清楚地说，所需的输出是：

[{'overall': 5,
  'vote': 0,
  'reviewerID': 'AAP7PPBU72QFM'},
 {'overall': 3,
  'vote': 5,
  'reviewerID': 'A2E168DTVGE6SV'},
...]

非常感激。

Answer 1

def clean_value(x):
    try:
        return int(x)
    except ValueError:
        return 0

def clean_list_of_dicts(l):
    return [{
        k:v if k not in ('overall', 'vote') else clean_value(v) \
        for k, v in d.items()
    } for d in l]

对输入数据的测试表明此解决方案有效。

>>> clean_list_of_dicts([{'overall': 5.0,
  'vote': 'overall',
  'reviewerID': 'AAP7PPBU72QFM'},
 {'overall': 3.0,
  'vote': '5',
  'reviewerID': 'A2E168DTVGE6SV'}
])

给出输出：

[{'overall': 5,
  'vote': 0,
  'reviewerID': 'AAP7PPBU72QFM'},
 {'overall': 3,
  'vote': 5,
  'reviewerID': 'A2E168DTVGE6SV'}]

Answer 2

另一个解决方案很棒，这是可读性的另一个解决方案：

list_dicts = [{'overall': 5.0,
  'vote': 'overall',
  'reviewerID': 'AAP7PPBU72QFM'},
 {'overall': 3.0,
  'vote': '5',
  'reviewerID': 'A2E168DTVGE6SV'}]

def fix_key(d, k):
    try:
        d[k] = int(d[k])
    except:
        d[k] = 0

def fix(d):
    fix_key(d, 'vote')
    fix_key(d, 'overall')
    return d

list_dicts = [fix(d) for d in list_dicts]

# [{'overall': 5, 'vote': 0, 'reviewerID': 'AAP7PPBU72QFM'}, {'overall': 3, 'vote': 5, 'reviewerID': 'A2E168DTVGE6SV'}]
print(list_dicts)

Answer 3

我希望它对你有用，但与函数一起使用会更好

for i in data:
  i['overall'] =int(i['overall'])
  i['vote'] = int(i['vote']) if (i['vote']).isdigit() else 0