Python multi-level nested dictionary
Change a list to a python3 multi-level nested dictionary with a new dictionary in the value of each layer.
I have data now：
list = [
('First', 'Second', 'Third'),
('First_2', 'Second_2', 'Third_3','Fourth_4')
]The result I want：
[
{"name": "First", "children": [{"name": "Second", "children": [{"name": "Third", "children": [{"name": "Fourth"}]}]}]},
{"name": "First", "children": [{"name": "Second_1", "children": [{"name": "Second_2", "children": [{"name": "Third_3", "children": [{"name": "Fourth_4"}]}]}]}]}
]
As you can see, a big dictionary.
The key value of the first dictionary is 'name', and the value is the subscript of the list.
In the second dictionary, the key value is 'children', the value is a list, and there is a dictionary in the list. The dictionary has the same structure as the first level dictionary. The dictionary key value of children is +1 for the list. By analogy, all the elements in the list are looped.
Thank you for your answer or suggestion
python python-3.x dictionary
edited Nov 16 '18 at 2:45
ecg8
8761515
asked Nov 15 '18 at 9:08
张嘉斌张嘉斌
1
add a comment |
Change a list to a python3 multi-level nested dictionary with a new dictionary in the value of each layer.
I have data now：
list = [
('First', 'Second', 'Third'),
('First_2', 'Second_2', 'Third_3','Fourth_4')
]The result I want：
[
{"name": "First", "children": [{"name": "Second", "children": [{"name": "Third", "children": [{"name": "Fourth"}]}]}]},
{"name": "First", "children": [{"name": "Second_1", "children": [{"name": "Second_2", "children": [{"name": "Third_3", "children": [{"name": "Fourth_4"}]}]}]}]}
]
As you can see, a big dictionary.
The key value of the first dictionary is 'name', and the value is the subscript of the list.
In the second dictionary, the key value is 'children', the value is a list, and there is a dictionary in the list. The dictionary has the same structure as the first level dictionary. The dictionary key value of children is +1 for the list. By analogy, all the elements in the list are looped.
Thank you for your answer or suggestion
python python-3.x dictionary
edited Nov 16 '18 at 2:45
ecg8
8761515
asked Nov 15 '18 at 9:08
张嘉斌张嘉斌
1
add a comment |
Change a list to a python3 multi-level nested dictionary with a new dictionary in the value of each layer.
I have data now：
list = [
('First', 'Second', 'Third'),
('First_2', 'Second_2', 'Third_3','Fourth_4')
]The result I want：
[
{"name": "First", "children": [{"name": "Second", "children": [{"name": "Third", "children": [{"name": "Fourth"}]}]}]},
{"name": "First", "children": [{"name": "Second_1", "children": [{"name": "Second_2", "children": [{"name": "Third_3", "children": [{"name": "Fourth_4"}]}]}]}]}
]
As you can see, a big dictionary.
The key value of the first dictionary is 'name', and the value is the subscript of the list.
In the second dictionary, the key value is 'children', the value is a list, and there is a dictionary in the list. The dictionary has the same structure as the first level dictionary. The dictionary key value of children is +1 for the list. By analogy, all the elements in the list are looped.
Thank you for your answer or suggestion
python python-3.x dictionary
edited Nov 16 '18 at 2:45
ecg8
8761515
asked Nov 15 '18 at 9:08
张嘉斌张嘉斌
1
Change a list to a python3 multi-level nested dictionary with a new dictionary in the value of each layer.
I have data now：
list = [
('First', 'Second', 'Third'),
('First_2', 'Second_2', 'Third_3','Fourth_4')
]The result I want：
[
{"name": "First", "children": [{"name": "Second", "children": [{"name": "Third", "children": [{"name": "Fourth"}]}]}]},
{"name": "First", "children": [{"name": "Second_1", "children": [{"name": "Second_2", "children": [{"name": "Third_3", "children": [{"name": "Fourth_4"}]}]}]}]}
]
As you can see, a big dictionary.
The key value of the first dictionary is 'name', and the value is the subscript of the list.
In the second dictionary, the key value is 'children', the value is a list, and there is a dictionary in the list. The dictionary has the same structure as the first level dictionary. The dictionary key value of children is +1 for the list. By analogy, all the elements in the list are looped.
Thank you for your answer or suggestion
python python-3.x dictionary
python python-3.x dictionary
edited Nov 16 '18 at 2:45
ecg8
8761515
asked Nov 15 '18 at 9:08
张嘉斌张嘉斌
1
edited Nov 16 '18 at 2:45
ecg8
8761515
asked Nov 15 '18 at 9:08
张嘉斌张嘉斌
1
edited Nov 16 '18 at 2:45
ecg8
8761515
edited Nov 16 '18 at 2:45
ecg8
8761515
edited Nov 16 '18 at 2:45
ecg8
8761515
8761515
asked Nov 15 '18 at 9:08
张嘉斌张嘉斌
1
asked Nov 15 '18 at 9:08
张嘉斌张嘉斌
1
asked Nov 15 '18 at 9:08
张嘉斌张嘉斌
1
1
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
I have found the answer, recursive method
list_1 = [('First', 'Second', 'Third'), ('First_1', 'Second_2', 'Third_3','Fourth_4')]
def all_road(children, count, item):
if count < len(item):
load_dict = {
'name': item[count],
'children': ,
}
if count == (len(item)-1):
load_dict = {
'name': item[count]
}
children.append(load_dict)
return children
children.append(load_dict)
count += 1
all_road(load_dict['children'], count,item)
return children
if __name__ == "__main__":
result =
for item in list_1:
children =
result.append(all_road(children, 0, item=item)[0])
print(result)answered Nov 16 '18 at 1:06
张嘉斌张嘉斌
1
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
I have found the answer, recursive method
list_1 = [('First', 'Second', 'Third'), ('First_1', 'Second_2', 'Third_3','Fourth_4')]
def all_road(children, count, item):
if count < len(item):
load_dict = {
'name': item[count],
'children': ,
}
if count == (len(item)-1):
load_dict = {
'name': item[count]
}
children.append(load_dict)
return children
children.append(load_dict)
count += 1
all_road(load_dict['children'], count,item)
return children
if __name__ == "__main__":
result =
for item in list_1:
children =
result.append(all_road(children, 0, item=item)[0])
print(result)answered Nov 16 '18 at 1:06
张嘉斌张嘉斌
1
add a comment |
I have found the answer, recursive method
list_1 = [('First', 'Second', 'Third'), ('First_1', 'Second_2', 'Third_3','Fourth_4')]
def all_road(children, count, item):
if count < len(item):
load_dict = {
'name': item[count],
'children': ,
}
if count == (len(item)-1):
load_dict = {
'name': item[count]
}
children.append(load_dict)
return children
children.append(load_dict)
count += 1
all_road(load_dict['children'], count,item)
return children
if __name__ == "__main__":
result =
for item in list_1:
children =
result.append(all_road(children, 0, item=item)[0])
print(result)answered Nov 16 '18 at 1:06
张嘉斌张嘉斌
1
add a comment |
I have found the answer, recursive method
list_1 = [('First', 'Second', 'Third'), ('First_1', 'Second_2', 'Third_3','Fourth_4')]
def all_road(children, count, item):
if count < len(item):
load_dict = {
'name': item[count],
'children': ,
}
if count == (len(item)-1):
load_dict = {
'name': item[count]
}
children.append(load_dict)
return children
children.append(load_dict)
count += 1
all_road(load_dict['children'], count,item)
return children
if __name__ == "__main__":
result =
for item in list_1:
children =
result.append(all_road(children, 0, item=item)[0])
print(result)answered Nov 16 '18 at 1:06
张嘉斌张嘉斌
1
I have found the answer, recursive method
list_1 = [('First', 'Second', 'Third'), ('First_1', 'Second_2', 'Third_3','Fourth_4')]
def all_road(children, count, item):
if count < len(item):
load_dict = {
'name': item[count],
'children': ,
}
if count == (len(item)-1):
load_dict = {
'name': item[count]
}
children.append(load_dict)
return children
children.append(load_dict)
count += 1
all_road(load_dict['children'], count,item)
return children
if __name__ == "__main__":
result =
for item in list_1:
children =
result.append(all_road(children, 0, item=item)[0])
print(result)list_1 = [('First', 'Second', 'Third'), ('First_1', 'Second_2', 'Third_3','Fourth_4')]
def all_road(children, count, item):
if count < len(item):
load_dict = {
'name': item[count],
'children': ,
}
if count == (len(item)-1):
load_dict = {
'name': item[count]
}
children.append(load_dict)
return children
children.append(load_dict)
count += 1
all_road(load_dict['children'], count,item)
return children
if __name__ == "__main__":
result =
for item in list_1:
children =
result.append(all_road(children, 0, item=item)[0])
print(result)list_1 = [('First', 'Second', 'Third'), ('First_1', 'Second_2', 'Third_3','Fourth_4')]
def all_road(children, count, item):
if count < len(item):
load_dict = {
'name': item[count],
'children': ,
}
if count == (len(item)-1):
load_dict = {
'name': item[count]
}
children.append(load_dict)
return children
children.append(load_dict)
count += 1
all_road(load_dict['children'], count,item)
return children
if __name__ == "__main__":
result =
for item in list_1:
children =
result.append(all_road(children, 0, item=item)[0])
print(result)answered Nov 16 '18 at 1:06
张嘉斌张嘉斌
1
answered Nov 16 '18 at 1:06
张嘉斌张嘉斌
1
answered Nov 16 '18 at 1:06
张嘉斌张嘉斌
1
answered Nov 16 '18 at 1:06
张嘉斌张嘉斌
1
1
add a comment |
add a comment |
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