Question

假设我有一个Python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']

如何将它们转换为词典列表，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。也许每个元素总是有2个元素，或者总是有一个Name条目，或者您每次只需要看到一个键就只需要创建一个新字典。

如果每个词典中总是有 N 个元素，则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'}]

如果看到重复输入是更好的方法，则从包含空字典的列表result开始；您可以将键值对添加到result[-1]。然后一一处理您的字符串，在':'字符上分割每个字符串以创建键和值对。如果在最近的词典中已经找到该密钥，请开始一个新的空密钥：

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

通过检查是否存在key，是否交换了Name和Index条目就不再重要了。

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

我的解决方法是：

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)

具有以下输出：

[{'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'}]

如何在python

2 个答案: