我需要使用這個字串串列來創建具有某些值的嵌套字典 ['C/A', 'C/B/A', 'C/B/B']
輸出將采用以下格式 {'C': {'A': [1, 2, 3], 'B': {'A': [1, 2, 3], 'B': [1, 2, 3]}}}
我嘗試使用下面的代碼來創建嵌套字典并更新值，但我得到{'C': {'A': [1, 2, 3], 'C': {'B': {'A': [1, 2, 3], 'C': {'B': {'B': [1, 2, 3]}}}}}}的輸出格式不正確。我還在想辦法。有任何想法嗎？
s = ['C/A', 'C/B/A', 'C/B/B']
new = current = dict()
for each in s:
lst = each.split('/')
for i in range(len(lst)):
current[lst[i]] = dict()
if i != len(lst)-1:
current = current[lst[i]]
else:
current[lst[i]] = [1,2,3]
print(new)
uj5u.com熱心網友回復：
您可以創建自定義Tree類：
class Tree(dict):
'''
Create arbitrarily nested dicts.
>>> t = Tree()
>>> t[1][2][3] = 4
>>> t
{1: {2: {3: 4}}}
>>> t.set_nested_item('a', 'b', 'c', value=5)
>>> t
{1: {2: {3: 4}}, 'a': {'b': {'c': 5}}}
'''
def __missing__(self, key):
self[key] = type(self)()
return self[key]
def set_nested_item(self, *keys, value):
head, *rest = keys
if not rest:
self[head] = value
else:
self[head].set_nested_item(*rest, value=value)
>>> s = ['C/A', 'C/B/A', 'C/B/B']
>>> output = Tree()
>>> default = [1, 2, 3]
>>> for item in s:
... output.set_nested_item(*item.split('/'), value=list(default))
>>> output
{'C': {'A': [1, 2, 3], 'B': {'A': [1, 2, 3], 'B': [1, 2, 3]}}}
uj5u.com熱心網友回復：
您不需要 numpy 來解決這個問題，但您可能想使用遞回。這是一個遞回函式add，它將字串鍵lst串列和數字串列添加到字典中d：
def add(lst, d):
key = lst[0]
if len(lst) == 1: # if the list has only 1 element
d[key] = [1, 2, 3] # That element is the last key
return
if key not in d: # Haven't seen that key before
d[key] = dict()
add(lst[1:], d[key]) # The recursive part
要使用該函式，請創建一個新字典并將該函式應用于每個拆分器字串：
d = dict()
for each in s:
add(each.split("/"), d)
# d
# {'C': {'A': [1, 2, 3], 'B': {'A': [1, 2, 3], 'B': [1, 2, 3]}}}
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