在python中将嵌套列表转换为字典
它不是一个严格的嵌套列表,它是一个树结构,看起来像:在python中将嵌套列表转换为字典,python,list,parsing,dictionary,tree,Python,List,Parsing,Dictionary,Tree,它不是一个严格的嵌套列表,它是一个树结构,看起来像: A = [a, [b, c,[d,e]]] 对应的树是: a / \ b c / \ d e 每当一个元素后面有一个子列表时,该子列表对应于该元素的子节点。否则,元素位于同一层。我想生成一个字典,每个节点分别作为一个键,如: child[
A = [a, [b, c,[d,e]]]
对应的树是:
a
/ \
b c
/ \
d e
每当一个元素后面有一个子列表时,该子列表对应于该元素的子节点。否则,元素位于同一层。我想生成一个字典,每个节点分别作为一个键,如:
child[a] = [b,c,d,e]
child[c] = [d,e]
在python中如何实现这一点?或者对于树结构转换还有其他更好的建议吗?我仍然认为您应该使用现有的实现/从中获得灵感,但如果您需要这样做,这可能就是您想要的:
#!/usr/bin/env python
# added a test case
B = ['a', ['b', 'c',['d','e']], 'f', ['g', 'h']]
A = ['a', ['b', 'c',['d','e']]]
# found on stack overflow - flatten list of kids for parent
def flatten(iterable):
"""Recursively iterate lists and tuples.
"""
for elm in iterable:
if isinstance(elm, (list, tuple)):
for relm in flatten(elm):
yield relm
else:
yield elm
# add data to an existing tree (recursive)
def treeify(tree, l):
if isinstance(l, list):
# avoid looking back
l.reverse()
for index in range(len(l)):
if isinstance(l[index], list):
parent_name = l[index+1]
# flatten kids to a list
tree[parent_name] = list(flatten(l[index]))
# continue in deeper lists
treeify(tree, l[index])
tree = {}
treeify(tree, A)
print tree
tree = {}
treeify(tree, B)
print tree
这将反转
列表
,以避免在遍历它时向后看。如果当前成员是列表,Tt将名称设置为下一个成员,并立即(递归地)遍历子元素。我仍然认为您应该使用现有的实现/从现有的实现中获得灵感,但如果您需要此实现,这可能就是您想要的:
#!/usr/bin/env python
# added a test case
B = ['a', ['b', 'c',['d','e']], 'f', ['g', 'h']]
A = ['a', ['b', 'c',['d','e']]]
# found on stack overflow - flatten list of kids for parent
def flatten(iterable):
"""Recursively iterate lists and tuples.
"""
for elm in iterable:
if isinstance(elm, (list, tuple)):
for relm in flatten(elm):
yield relm
else:
yield elm
# add data to an existing tree (recursive)
def treeify(tree, l):
if isinstance(l, list):
# avoid looking back
l.reverse()
for index in range(len(l)):
if isinstance(l[index], list):
parent_name = l[index+1]
# flatten kids to a list
tree[parent_name] = list(flatten(l[index]))
# continue in deeper lists
treeify(tree, l[index])
tree = {}
treeify(tree, A)
print tree
tree = {}
treeify(tree, B)
print tree
这将反转列表
,以避免在遍历它时向后看。如果当前成员是列表,Tt将名称设置为下一个成员,并立即(递归地)遍历子元素。我仍然认为您应该使用现有的实现/从现有的实现中获得灵感,但如果您需要此实现,这可能就是您想要的:
#!/usr/bin/env python
# added a test case
B = ['a', ['b', 'c',['d','e']], 'f', ['g', 'h']]
A = ['a', ['b', 'c',['d','e']]]
# found on stack overflow - flatten list of kids for parent
def flatten(iterable):
"""Recursively iterate lists and tuples.
"""
for elm in iterable:
if isinstance(elm, (list, tuple)):
for relm in flatten(elm):
yield relm
else:
yield elm
# add data to an existing tree (recursive)
def treeify(tree, l):
if isinstance(l, list):
# avoid looking back
l.reverse()
for index in range(len(l)):
if isinstance(l[index], list):
parent_name = l[index+1]
# flatten kids to a list
tree[parent_name] = list(flatten(l[index]))
# continue in deeper lists
treeify(tree, l[index])
tree = {}
treeify(tree, A)
print tree
tree = {}
treeify(tree, B)
print tree
这将反转列表
,以避免在遍历它时向后看。如果当前成员是列表,Tt将名称设置为下一个成员,并立即(递归地)遍历子元素。我仍然认为您应该使用现有的实现/从现有的实现中获得灵感,但如果您需要此实现,这可能就是您想要的:
#!/usr/bin/env python
# added a test case
B = ['a', ['b', 'c',['d','e']], 'f', ['g', 'h']]
A = ['a', ['b', 'c',['d','e']]]
# found on stack overflow - flatten list of kids for parent
def flatten(iterable):
"""Recursively iterate lists and tuples.
"""
for elm in iterable:
if isinstance(elm, (list, tuple)):
for relm in flatten(elm):
yield relm
else:
yield elm
# add data to an existing tree (recursive)
def treeify(tree, l):
if isinstance(l, list):
# avoid looking back
l.reverse()
for index in range(len(l)):
if isinstance(l[index], list):
parent_name = l[index+1]
# flatten kids to a list
tree[parent_name] = list(flatten(l[index]))
# continue in deeper lists
treeify(tree, l[index])
tree = {}
treeify(tree, A)
print tree
tree = {}
treeify(tree, B)
print tree
这将反转列表
,以避免在遍历它时向后看。TT如果当前的代码是<代码>列表>代码,将其命名为下一个成员,并立即(递归地)遍历子元素。 如果要进行大量的图形操作,我会考虑导入<代码> NETWorkX,因为它会使事情变得更容易。要将嵌套列表解析为networkx
树,请执行以下操作:
import networkx as nx
def parse_tree(node_list):
"""Parses a nested list into a networkx tree."""
tree = nx.DiGraph()
root = node_list[0]
tree.add_node(root)
queue = [(root, node_list[1])]
while queue:
parent, nodes = queue.pop(0)
prev = None
for node in nodes:
if isinstance(node, list):
queue.append((prev, node))
else:
tree.add_node(node)
tree.add_edge(parent, node)
prev = node
return tree
使用此函数,很容易获得每个节点的后代的字典:
>>> l = ["a", ["b", "c",["d","e"]]]
>>> tree = parse_tree(l)
>>> {node: nx.descendants(tree, node) for node in tree}
{'a': {'b', 'c', 'd', 'e'},
'b': set(),
'c': {'d', 'e'},
'd': set(),
'e': set()}
如果你要做很多图表操作,我会考虑导入<代码> NETWorkX,因为它会使事情变得更容易。要将嵌套列表解析为
networkx
树,请执行以下操作:
import networkx as nx
def parse_tree(node_list):
"""Parses a nested list into a networkx tree."""
tree = nx.DiGraph()
root = node_list[0]
tree.add_node(root)
queue = [(root, node_list[1])]
while queue:
parent, nodes = queue.pop(0)
prev = None
for node in nodes:
if isinstance(node, list):
queue.append((prev, node))
else:
tree.add_node(node)
tree.add_edge(parent, node)
prev = node
return tree
使用此函数,很容易获得每个节点的后代的字典:
>>> l = ["a", ["b", "c",["d","e"]]]
>>> tree = parse_tree(l)
>>> {node: nx.descendants(tree, node) for node in tree}
{'a': {'b', 'c', 'd', 'e'},
'b': set(),
'c': {'d', 'e'},
'd': set(),
'e': set()}
如果你要做很多图表操作,我会考虑导入<代码> NETWorkX,因为它会使事情变得更容易。要将嵌套列表解析为
networkx
树,请执行以下操作:
import networkx as nx
def parse_tree(node_list):
"""Parses a nested list into a networkx tree."""
tree = nx.DiGraph()
root = node_list[0]
tree.add_node(root)
queue = [(root, node_list[1])]
while queue:
parent, nodes = queue.pop(0)
prev = None
for node in nodes:
if isinstance(node, list):
queue.append((prev, node))
else:
tree.add_node(node)
tree.add_edge(parent, node)
prev = node
return tree
使用此函数,很容易获得每个节点的后代的字典:
>>> l = ["a", ["b", "c",["d","e"]]]
>>> tree = parse_tree(l)
>>> {node: nx.descendants(tree, node) for node in tree}
{'a': {'b', 'c', 'd', 'e'},
'b': set(),
'c': {'d', 'e'},
'd': set(),
'e': set()}
如果你要做很多图表操作,我会考虑导入<代码> NETWorkX,因为它会使事情变得更容易。要将嵌套列表解析为
networkx
树,请执行以下操作:
import networkx as nx
def parse_tree(node_list):
"""Parses a nested list into a networkx tree."""
tree = nx.DiGraph()
root = node_list[0]
tree.add_node(root)
queue = [(root, node_list[1])]
while queue:
parent, nodes = queue.pop(0)
prev = None
for node in nodes:
if isinstance(node, list):
queue.append((prev, node))
else:
tree.add_node(node)
tree.add_edge(parent, node)
prev = node
return tree
使用此函数,很容易获得每个节点的后代的字典:
>>> l = ["a", ["b", "c",["d","e"]]]
>>> tree = parse_tree(l)
>>> {node: nx.descendants(tree, node) for node in tree}
{'a': {'b', 'c', 'd', 'e'},
'b': set(),
'c': {'d', 'e'},
'd': set(),
'e': set()}
重复的键会发生什么情况,比如[a[b,a[d,e]]]
?结果如何?尝试使用树的现有实现并使用它。@ReutSharabani感谢Reut。所有的元素都是不同的。在重复的键上会发生什么,比如[a[b,a[d,e]]]
?结果如何?尝试使用树的现有实现并使用它。@ReutSharabani感谢Reut。所有的元素都是不同的。在重复的键上会发生什么,比如[a[b,a[d,e]]]
?结果如何?尝试使用树的现有实现并使用它。@ReutSharabani感谢Reut。所有的元素都是不同的。在重复的键上会发生什么,比如[a[b,a[d,e]]]
?结果如何?尝试使用树的现有实现并使用它。@ReutSharabani感谢Reut。所有元素都是不同的。