Question

前言：为了帮助解释为什么我这样做，我将解释最终目标。基本上我有一个以非常具体的语法定义的帐户列表。以下是一些例子：

Assets:Bank:Car
Assets:Bank:House
Assets:Savings:Emergency
Assets:Savings:Goals:Roof
Assets:Reserved

如上所示，帐户可以包含任意数量的父母和子女。最终目标是将上述帐户解析为Python中的树结构，用于在Sublime文本编辑器中提供帐户自动完成（即，如果我键入 Assets：，然后查询为auto - 完成后，我会看到一个列表：银行，储蓄，保留）

结果：使用前言中的帐户列表，Python中的所需结果如下所示：

[  
   {  
      "Assets":[  
         {  
            "Bank":[  
               "Car",
               "House"
            ]
         },
         {  
            "Savings":[  
               "Emergency",
               {  
                  "Goals":[  
                     "Roof"
                  ]
               }
            ]
         },
         "Reserved"
      ]
   }
]

半解决方案：我能够使用递归将两个基本帐户添加到一起。这适用于添加以下两项：Assets:Bank:Car和Assets:Bank:House。但是，一旦它们开始出现差异，它就会开始崩溃并且递归变得混乱，所以我不确定它是否是最佳方式。

import re

def parse_account(account_str):
    subs = account_str.split(":")

    def separate(subs):
        if len(subs) == 1:
            return subs
        elif len(subs):
            return [{subs[0]: separate(subs[1:])}]

    return separate(subs)

def merge_dicts(a, b):
    # a will be a list with dictionaries and text values and then nested lists/dictionaries/text values
    # b will always be a list with ONE dictionary or text value

    key = b[0].keys()[0] # this is the dictionary key of the only dictionary in the b list

    for item in a: # item is a dictionary or a text value
        if isinstance(item, dict): # if item is a dictionary
            if key in item:
                # Is the value a list with a dict or a list with a text value
                if isinstance(b[0][key][0], str):
                    # Extend the current list with the new value
                    item[key].extend(b[0][key])
                else:
                    # Recurse to the next child
                    merge_dicts(item[key], b[0][key])
            else:


    return a

# Accounts have an "open [name]" syntax for defining them
text = "open Assets:Bank:Car\nopen Assets:Bank:House\nopen Assets:Savings:Emergency\nopen Assets:Savings:Goals:Roof\nopen Assets:Reserved"
EXP = re.compile("open (.*)")
accounts = EXP.findall(text) # This grabs all accounts

# Create a list of all the parsed accounts
dicts = []
for account in accounts:
    dicts.append(parse_account(account))

# Attempt to merge two accounts together
final = merge_dicts(dicts[0], dicts[1])
print final

# In the future we would call: reduce(merge_dicts, dicts) to merge all accounts

我可能会以完全错误的方式解决这个问题，我会对不同意见感兴趣。否则，是否有人深入了解如何使用示例字符串中的其余帐户进行此操作？

Answer 1

我花了很多时间在脑海中梳理。字典很简单，一个键总是有一个列表作为值 - 它们习惯于有一个命名列表。

列表内部将是一个字符串或另一个字典（带有列表的键）。

这意味着我们可以拆分'资产：银行：汽车'并在匹配{"Assets":[<whatever>]}的根列表中查找字典或添加一个字典 - 然后再向两个级别跳转到[<whatever>]列表。接下来循环，查找匹配{"Bank":[<whatever>]}的字典，或添加一个字典，跳转到更深层次的[<whatever>]列表。继续这样做，直到我们到达最后一个节点Car。我们必须在 a 列表上，因为我们总是跳转到现有列表或创建新列表，因此将Car放在当前列表中。

NB。如果你有

，这种方法会破裂

Assets:Reserved
Assets:Reserved:Painting

但这将是一个无意义的冲突输入，要求“保留”既是叶节点又是容器，在那种情况下你只会：

Assets:Reserved:Painting

正确？

data = """
Assets:Bank:Car
Assets:Bank:House
Assets:Savings:Emergency
Assets:Savings:Goals:Roof
Assets:Reserved
"""
J = []

for line in data.split('\n'):
    if not line: continue

    # split the line into parts, start at the root list
    # is there a dict here for this part?
    #   yes? cool, dive into it for the next loop iteration
    #   no? add one, with a list, ready for the next loop iteration
    #    (unless we're at the final part, then stick it in the list 
    #     we made/found in the previous loop iteration)

    parts = line.split(':')
    parent_list, current_list = J, J

    for index, part in enumerate(parts):
        for item in current_list:
            if part in item:
                parent_list, current_list = current_list, item[part]
                break
        else:
            if index == len(parts) - 1:
                # leaf node, add part as string
                current_list.append(part)
            else:
                new_list = []
                current_list.append({part:new_list})
                parent_list, current_list = current_list, new_list      

print J

- ＆GT;

[{'Assets': [{'Bank': ['Car', 'House']}, {'Savings': ['Emergency', {'Goals': ['Roof']}]}, 'Reserved']}]

在线试用：https://repl.it/Ci5L

从Python

1 个答案: