如何将字符串列表转换为python中的字典列表

Question

假设我在Python中有一个字符串列表：

['Name: volume_test_add_volume_to_cg_2019_03_07-12_21_37', 'Index: 24', 
 'Name: volume_xx111', 'Index: 3', 'Name: volume_xx11541', 'Index: 4', 
 'Name: Volume_test_add_volume_mandatory_params_2019_03_06-16_50_10', 'Index: 6']

如何将它们转换为字典列表，以便最终结果如下：

[
 {'Name': 'volume_test_add_volume_to_cg_2019_03_07-12_21_37', 'Index': '24'}, 
 {'Name': 'volume_xx111', 'Index': '3'}, 
 {'Name': 'volume_xx11541', 'Index': '4'}, 
 {'Name': 'Volume_test_add_volume_mandatory_params_2019_03_06-16_50_10', 'Index': '6}
]

Answer 1

您需要在代码中决定如何将字符串分组为字典。也许每个都有2个元素，或者总是有一个Name条目，或者你只需​​要在每次看到一个键时创建一个新的字典。

如果每个字典总有N个元素，那么iterate in chunks of that size：

def chunks(l, n):
    """Yield successive n-sized chunks from l."""
    for i in range(0, len(l), n):
        yield l[i:i + n]

results = [
    dict(tuple(map(str.strip, entry.split(': '))) for entry in per_dict)
    for per_dict in chunks(inputlist, 2)
]

演示：

>>> from pprint import pprint
>>> inputlist = ['Name: volume_test_add_volume_to_cg_2019_03_07-12_21_37', 'Index: 24', 'Name: volume_xx111', 'Index: 3', 'Name: volume_xx11541', 'Index: 4', 'Name: Volume_test_add_volume_mandatory_params_2019_03_06-16_50_10', 'Index: 6']
>>> def chunks(l, n):
...     """Yield successive n-sized chunks from l."""
...     for i in range(0, len(l), n):
...         yield l[i:i + n]
...
>>> [
...     dict(tuple(map(str.strip, entry.split(': '))) for entry in per_dict)
...     for per_dict in chunks(inputlist, 2)
... ]
[{'Name': 'volume_test_add_volume_to_cg_2019_03_07-12_21_37', 'Index': '24'}, {'Name': 'volume_xx111', 'Index': '3'}, {'Name': 'volume_xx11541', 'Index': '4'}, {'Name': 'Volume_test_add_volume_mandatory_params_2019_03_06-16_50_10', 'Index': '6'}]
>>> pprint(_)
[{'Index': '24', 'Name': 'volume_test_add_volume_to_cg_2019_03_07-12_21_37'},
 {'Index': '3', 'Name': 'volume_xx111'},
 {'Index': '4', 'Name': 'volume_xx11541'},
 {'Index': '6',
  'Name': 'Volume_test_add_volume_mandatory_params_2019_03_06-16_50_10'}]

如果看到键重复是更好的方法，那么从包含空字典的列表result开始;你将键值对添加到result[-1]。然后逐个处理你的字符串，在':'字符上分别创建一个键和值对。如果已在最近的字典中找到密钥，请启动一个新的空字典：

results = [{}]
for entry in inputlist:
    key, value = map(str.strip, entry.split(':'))  # removing surrounding whitespace
    if key in results[-1]:
        # start a new dictionary
        results.append({})
    results[-1][key] = value

通过检查存在的key，如果Name和Index条目被交换不再重要。

演示：

>>> results = [{}]
>>> for entry in inputlist:
...     key, value = map(str.strip, entry.split(':'))  # removing surrounding whitespace
...     if key in results[-1]:
...         # start a new dictionary
...         results.append({})
...     results[-1][key] = value
...
>>> pprint(results)
[{'Index': '24', 'Name': 'volume_test_add_volume_to_cg_2019_03_07-12_21_37'},
 {'Index': '3', 'Name': 'volume_xx111'},
 {'Index': '4', 'Name': 'volume_xx11541'},
 {'Index': '6',
  'Name': 'Volume_test_add_volume_mandatory_params_2019_03_06-16_50_10'}]

Answer 2

我的解决方案如下：

lista = ['Name: volume_test_add_volume_to_cg_2019_03_07-12_21_37', 'Index: 24', 
         'Name: volume_xx111', 'Index: 3', 
         'Name: volume_xx11541', 'Index: 4', 
         'Name: Volume_test_add_volume_mandatory_params_2019_03_06-16_50_10', 'Index: 6']
result = []

for n, item in enumerate(lista):
    if n % 2 == 0:
        result.append({'Name': item[item.find(':') + 2:],
                       'Index': lista[n + 1][lista[n + 1].find(':') + 2:]})
print(result)

有了这个输出：

[{'Name': 'volume_test_add_volume_to_cg_2019_03_07-12_21_37', 'Index': '24'}, {'Name': 'volume_xx111', 'Index': '3'}, {'Name': 'volume_xx11541', 'Index': '4'}, {'Name': 'Volume_test_add_volume_mandatory_params_2019_03_06-16_50_10', 'Index': '6'}]