Python : 从 CSV 构建分层 JSON
Python : build hierarchical JSON from CSV
我想从 CSV 构建一个 JSON 文件来表示我的数据的层次关系。关系是 parents 和 children ：一个 child 可以有一个或多个 parents 和一个 parent 可以有一个或多个 child仁。一个 child 也可以有 children 值，多个级别是可能的。我认为 D3 中的 dendrogram 可能是一个很好的可视化。
我的 CSV 源文件包含数千行，如下所示：
parent | children | date
---------------------------------------------
830010000C0419 | 830010000C1205 | 1993/09/15
830010000C0947 | 830010000C1205 | 1993/09/15
830010000C0948 | 830010000C1205 | 1993/09/15
830010000B0854 | 830010000B1196 | 1994/03/11
830010000B0854 | 830010000B1197 | 1994/03/11
830010000B0721 | 830010000B1343 | 1988/12/05
830010000B1343 | 830010000B1344 | 1988/12/05
830010000B0721 | 830010000B1345 | 1988/12/05
830010000B1345 | 830010000B1344 | 1986/12/05
...
我不想生成具有以下结构的 JSON 文件：
var treeData = [
{
"name": "Root",
"parent": "null",
"children": [
{
"name": "830010000B0854",
"parent": "Top Level",
"children": [
{
"name": "830010000B1196",
"parent": "830010000B0854"
},
{
"name": "830010000B1197",
"parent": "830010000B0854"
}
]
},
{
"name": "830010000B0721",
"parent": "Top Level",
"children": [
{
"name": "830010000B1343",
"parent": "830010000B0721",
"children": [
{
"name": "830010000B1344",
"parent": "830010000B1343"
}
]
}
]
},
{
"name": "830010000C0419",
"parent": "Top Level",
"children": [
{
"name": "830010000C1205",
"parent": "830010000C0419"
}
]
},
{
"name": "830010000C0947",
"parent": "Top Level",
"children": [
{
"name": "830010000C1205",
"parent": "830010000C0947"
}
]
},
{
"name": "830010000C0948",
"parent": "Top Level",
"children": [
{
"name": "830010000C1205",
"parent": "830010000C0948"
}
]
}
]
}
];
请注意，在这个例子中，我无法建立一个 child 有很多 parent 的关系，也许需要更复杂的树状图。
如何使用 Python 构建这种结构？
我首先想到的是下面这个。请注意，这还没有完成，您需要添加某种形式的 recursion/iteration 以深入到子节点，但我认为逻辑应该非常相似。
all_parents = df.parent
def get_children(parent_name):
children = [child for child in df[df.parent == parent_name].children]
return [{"name": name, "parent": parent_name} for name in children]
def get_node_representation(parent_name):
if parent_name in all_parents:
parent = "Top Level"
else:
# Your logic here
parent = "null"
return {"name": parent_name, "parent": parent, "children": get_children(parent_name)}
# this assumes all children are also parent which is not necessarily true of course, so you want to create some kind of recursion/iteration on calling node_representation on the children nodes
all_nodes = [get_node_representation(node) for node in df.parent]
我发现这个 method 允许父子之间的多重关系。
这是我的数据演示：
var width = 800,
height = 800,
boxWidth = 150,
boxHeight = 20,
gap = {
width: 150,
height: 12
},
margin = {
top: 16,
right: 16,
bottom: 16,
left: 16
},
svg;
var data = {
"Nodes": [
// Level 0
{
"lvl": 0,
"name": "830010000C0419"
},
{
"lvl": 0,
"name": "830010000C0947"
},
{
"lvl": 0,
"name": "830010000C0948"
},
{
"lvl": 0,
"name": "830010000B0854"
},
{
"lvl": 0,
"name": "830010000B0721"
},
// Level 1
{
"lvl": 1,
"name": "830010000C1205"
},
{
"lvl": 1,
"name": "830010000B1196"
},
{
"lvl": 1,
"name": "830010000B1197"
},
{
"lvl": 1,
"name": "830010000B1343"
},
{
"lvl": 1,
"name": "830010000B1345"
},
// Level 2
{
"lvl": 2,
"name": "830010000B1344"
}
],
"links": [
{
"source": "830010000C0419",
"target": "830010000C1205"
},
{
"source": "830010000C0947",
"target": "830010000C1205"
},
{
"source": "830010000C0948",
"target": "830010000C1205"
},
{
"source": "830010000B0854",
"target": "830010000B1196"
},
{
"source": "830010000B0854",
"target": "830010000B1197"
},
{
"source": "830010000B0721",
"target": "830010000B1343"
},
{
"source": "830010000B1343",
"target": "830010000B1344"
},
{
"source": "830010000B0721",
"target": "830010000B1345"
},
{
"source": "830010000B1345",
"target": "830010000B1344"
}
]
};
// test layout
var Nodes = [];
var links = [];
var lvlCount = 0;
var diagonal = d3.svg.diagonal()
.projection(function (d) {
"use strict";
return [d.y, d.x];
});
function find(text) {
"use strict";
var i;
for (i = 0; i < Nodes.length; i += 1) {
if (Nodes[i].name === text) {
return Nodes[i];
}
}
return null;
}
function mouse_action(val, stat, direction) {
"use strict";
d3.select("#" + val.id).classed("active", stat);
links.forEach(function (d) {
if (direction == "root") {
if (d.source.id === val.id) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
if (d.target.lvl < val.lvl)
mouse_action(d.target, stat, "left");
else if (d.target.lvl > val.lvl)
mouse_action(d.target, stat, "right");
}
if (d.target.id === val.id) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
if (direction == "root") {
if(d.source.lvl < val.lvl)
mouse_action(d.source, stat, "left");
else if (d.source.lvl > val.lvl)
mouse_action(d.source, stat, "right");
}
}
}else if (direction == "left") {
if (d.source.id === val.id && d.target.lvl < val.lvl) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
mouse_action(d.target, stat, direction);
}
if (d.target.id === val.id && d.source.lvl < val.lvl) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
mouse_action(d.source, stat, direction);
}
}else if (direction == "right") {
if (d.source.id === val.id && d.target.lvl > val.lvl) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
mouse_action(d.target, stat, direction);
}
if (d.target.id === val.id && d.source.lvl > val.lvl) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
mouse_action(d.source, stat, direction);
}
}
});
}
function unvisite_links() {
"use strict";
links.forEach(function (d) {
d.visited = false;
});
}
function renderRelationshipGraph(data) {
"use strict";
var count = [];
data.Nodes.forEach(function (d) {
count[d.lvl] = 0;
});
lvlCount = count.length;
data.Nodes.forEach(function (d, i) {
d.x = margin.left + d.lvl * (boxWidth + gap.width);
d.y = margin.top + (boxHeight + gap.height) * count[d.lvl];
d.id = "n" + i;
count[d.lvl] += 1;
Nodes.push(d);
});
data.links.forEach(function (d) {
links.push({
source: find(d.source),
target: find(d.target),
id: "l" + find(d.source).id + find(d.target).id
});
});
unvisite_links();
svg.append("g")
.attr("class", "nodes");
var node = svg.select(".nodes")
.selectAll("g")
.data(Nodes)
.enter()
.append("g")
.attr("class", "unit");
node.append("rect")
.attr("x", function (d) { return d.x; })
.attr("y", function (d) { return d.y; })
.attr("id", function (d) { return d.id; })
.attr("width", boxWidth)
.attr("height", boxHeight)
.attr("class", "node")
.attr("rx", 6)
.attr("ry", 6)
.on("mouseover", function () {
mouse_action(d3.select(this).datum(), true, "root");
unvisite_links();
})
.on("mouseout", function () {
mouse_action(d3.select(this).datum(), false, "root");
unvisite_links();
});
node.append("text")
.attr("class", "label")
.attr("x", function (d) { return d.x + 14; })
.attr("y", function (d) { return d.y + 15; })
.text(function (d) { return d.name; });
links.forEach(function (li) {
svg.append("path", "g")
.attr("class", "link")
.attr("id", li.id)
.attr("d", function () {
var oTarget = {
x: li.target.y + 0.5 * boxHeight,
y: li.target.x
};
var oSource = {
x: li.source.y + 0.5 * boxHeight,
y: li.source.x
};
if (oSource.y < oTarget.y) {
oSource.y += boxWidth;
} else {
oTarget.y += boxWidth;
}
return diagonal({
source: oSource,
target: oTarget
});
});
});
}
svg = d3.select("#tree").append("svg")
.attr("width", width)
.attr("height", height)
.append("g");
renderRelationshipGraph(data);
rect {
fill: #CCC;
cursor: pointer;
}
.active {
fill: orange;
stroke: orange;
}
.activelink {
fill: none;
stroke: orange;
stroke-width: 2.5px;
}
.label {
fill: white;
font-family: sans-serif;
pointer-events: none;
}
.link {
fill: none;
stroke: #ccc;
stroke-width: 2.5px;
}
<script src="https://d3js.org/d3.v3.min.js"></script>
<div id="tree"></div>
我需要知道一个脚本来生成节点和链接结构
我会首先构建一个节点字典，其中键是节点名称，值是一个包含父列表和子列表的元组。为了有一个更简单的方法来构建树，我还会保留所有顶级节点的集合（没有父节点）。
从那个字典，然后可以递归地构建一个 json 类数据，可以用来构建一个真正的 json 字符串。
但是你显示的是 不是 的 csv 格式，我使用 re.split 来解析输入：
import re
# First the data
t = '''parent | children | date
---------------------------------------------
830010000C0419 | 830010000C1205 | 1993/09/15
830010000C0947 | 830010000C1205 | 1993/09/15
830010000C0948 | 830010000C1205 | 1993/09/15
830010000B0854 | 830010000B1196 | 1994/03/11
830010000B0854 | 830010000B1197 | 1994/03/11
830010000B0721 | 830010000B1343 | 1988/12/05
830010000B1343 | 830010000B1344 | 1988/12/05
'''
rx = re.compile(r'\s*\|\s*')
# nodes is a dictionary of nodes, nodes[None] is the set of top-level names
nodes = {None: set()}
with io.StringIO(t) as fd:
_ = next(fd) # skip initial lines
_ = next(fd)
for linenum, line in enumerate(fd, 1):
p, c = rx.split(line.strip())[:2] # parse a line
if p == c: # a node cannot be its parent
raise ValueError(f'Same node as parent and child {p} at line {linenum}')
# process the nodes
if c not in nodes:
nodes[c] = ([], [])
elif c in nodes[None]:
nodes[None].remove(c)
if p not in nodes:
nodes[p] = ([], [c])
nodes[None].add(p)
else:
nodes[p][1].append(c)
nodes[c][0].append(p)
def subtree(node, nodes, parent=None, seen = None):
"""Builds a dict with the subtree of a node.
node is a node name, nodes the dict, parent is the parent name,
seen is a list of all previously seen node to prevent cycles
"""
if seen is None:
seen = [node]
elif node in seen: # special processing to break possible cycles
return {'name': node, 'parent': parent, 'children': '...'}
else:
seen.append(node)
return {'name': node, 'parent': parent, 'children':
[subtree(c, nodes, node, seen) for c in nodes[node][1]]}
# We can now build the json data
js = {node: subtree(node, nodes) for node in nodes[None]}
pprint.pprint(js)
它给出：
{'830010000B0721': {'children': [{'children': [{'children': [],
'name': '830010000B1344',
'parent': '830010000B1343'}],
'name': '830010000B1343',
'parent': '830010000B0721'}],
'name': '830010000B0721',
'parent': None},
'830010000B0854': {'children': [{'children': [],
'name': '830010000B1196',
'parent': '830010000B0854'},
{'children': [],
'name': '830010000B1197',
'parent': '830010000B0854'}],
'name': '830010000B0854',
'parent': None},
'830010000C0419': {'children': [{'children': [],
'name': '830010000C1205',
'parent': '830010000C0419'}],
'name': '830010000C0419',
'parent': None},
'830010000C0947': {'children': [{'children': [],
'name': '830010000C1205',
'parent': '830010000C0947'}],
'name': '830010000C0947',
'parent': None},
'830010000C0948': {'children': [{'children': [],
'name': '830010000C1205',
'parent': '830010000C0948'}],
'name': '830010000C0948',
'parent': None}}
我想从 CSV 构建一个 JSON 文件来表示我的数据的层次关系。关系是 parents 和 children ：一个 child 可以有一个或多个 parents 和一个 parent 可以有一个或多个 child仁。一个 child 也可以有 children 值，多个级别是可能的。我认为 D3 中的 dendrogram 可能是一个很好的可视化。
我的 CSV 源文件包含数千行，如下所示：
parent | children | date
---------------------------------------------
830010000C0419 | 830010000C1205 | 1993/09/15
830010000C0947 | 830010000C1205 | 1993/09/15
830010000C0948 | 830010000C1205 | 1993/09/15
830010000B0854 | 830010000B1196 | 1994/03/11
830010000B0854 | 830010000B1197 | 1994/03/11
830010000B0721 | 830010000B1343 | 1988/12/05
830010000B1343 | 830010000B1344 | 1988/12/05
830010000B0721 | 830010000B1345 | 1988/12/05
830010000B1345 | 830010000B1344 | 1986/12/05
...
我不想生成具有以下结构的 JSON 文件：
var treeData = [
{
"name": "Root",
"parent": "null",
"children": [
{
"name": "830010000B0854",
"parent": "Top Level",
"children": [
{
"name": "830010000B1196",
"parent": "830010000B0854"
},
{
"name": "830010000B1197",
"parent": "830010000B0854"
}
]
},
{
"name": "830010000B0721",
"parent": "Top Level",
"children": [
{
"name": "830010000B1343",
"parent": "830010000B0721",
"children": [
{
"name": "830010000B1344",
"parent": "830010000B1343"
}
]
}
]
},
{
"name": "830010000C0419",
"parent": "Top Level",
"children": [
{
"name": "830010000C1205",
"parent": "830010000C0419"
}
]
},
{
"name": "830010000C0947",
"parent": "Top Level",
"children": [
{
"name": "830010000C1205",
"parent": "830010000C0947"
}
]
},
{
"name": "830010000C0948",
"parent": "Top Level",
"children": [
{
"name": "830010000C1205",
"parent": "830010000C0948"
}
]
}
]
}
];
请注意，在这个例子中，我无法建立一个 child 有很多 parent 的关系，也许需要更复杂的树状图。
如何使用 Python 构建这种结构？
我首先想到的是下面这个。请注意，这还没有完成，您需要添加某种形式的 recursion/iteration 以深入到子节点，但我认为逻辑应该非常相似。
all_parents = df.parent
def get_children(parent_name):
children = [child for child in df[df.parent == parent_name].children]
return [{"name": name, "parent": parent_name} for name in children]
def get_node_representation(parent_name):
if parent_name in all_parents:
parent = "Top Level"
else:
# Your logic here
parent = "null"
return {"name": parent_name, "parent": parent, "children": get_children(parent_name)}
# this assumes all children are also parent which is not necessarily true of course, so you want to create some kind of recursion/iteration on calling node_representation on the children nodes
all_nodes = [get_node_representation(node) for node in df.parent]
我发现这个 method 允许父子之间的多重关系。
这是我的数据演示：
var width = 800,
height = 800,
boxWidth = 150,
boxHeight = 20,
gap = {
width: 150,
height: 12
},
margin = {
top: 16,
right: 16,
bottom: 16,
left: 16
},
svg;
var data = {
"Nodes": [
// Level 0
{
"lvl": 0,
"name": "830010000C0419"
},
{
"lvl": 0,
"name": "830010000C0947"
},
{
"lvl": 0,
"name": "830010000C0948"
},
{
"lvl": 0,
"name": "830010000B0854"
},
{
"lvl": 0,
"name": "830010000B0721"
},
// Level 1
{
"lvl": 1,
"name": "830010000C1205"
},
{
"lvl": 1,
"name": "830010000B1196"
},
{
"lvl": 1,
"name": "830010000B1197"
},
{
"lvl": 1,
"name": "830010000B1343"
},
{
"lvl": 1,
"name": "830010000B1345"
},
// Level 2
{
"lvl": 2,
"name": "830010000B1344"
}
],
"links": [
{
"source": "830010000C0419",
"target": "830010000C1205"
},
{
"source": "830010000C0947",
"target": "830010000C1205"
},
{
"source": "830010000C0948",
"target": "830010000C1205"
},
{
"source": "830010000B0854",
"target": "830010000B1196"
},
{
"source": "830010000B0854",
"target": "830010000B1197"
},
{
"source": "830010000B0721",
"target": "830010000B1343"
},
{
"source": "830010000B1343",
"target": "830010000B1344"
},
{
"source": "830010000B0721",
"target": "830010000B1345"
},
{
"source": "830010000B1345",
"target": "830010000B1344"
}
]
};
// test layout
var Nodes = [];
var links = [];
var lvlCount = 0;
var diagonal = d3.svg.diagonal()
.projection(function (d) {
"use strict";
return [d.y, d.x];
});
function find(text) {
"use strict";
var i;
for (i = 0; i < Nodes.length; i += 1) {
if (Nodes[i].name === text) {
return Nodes[i];
}
}
return null;
}
function mouse_action(val, stat, direction) {
"use strict";
d3.select("#" + val.id).classed("active", stat);
links.forEach(function (d) {
if (direction == "root") {
if (d.source.id === val.id) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
if (d.target.lvl < val.lvl)
mouse_action(d.target, stat, "left");
else if (d.target.lvl > val.lvl)
mouse_action(d.target, stat, "right");
}
if (d.target.id === val.id) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
if (direction == "root") {
if(d.source.lvl < val.lvl)
mouse_action(d.source, stat, "left");
else if (d.source.lvl > val.lvl)
mouse_action(d.source, stat, "right");
}
}
}else if (direction == "left") {
if (d.source.id === val.id && d.target.lvl < val.lvl) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
mouse_action(d.target, stat, direction);
}
if (d.target.id === val.id && d.source.lvl < val.lvl) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
mouse_action(d.source, stat, direction);
}
}else if (direction == "right") {
if (d.source.id === val.id && d.target.lvl > val.lvl) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
mouse_action(d.target, stat, direction);
}
if (d.target.id === val.id && d.source.lvl > val.lvl) {
d3.select("#" + d.id).classed("activelink", stat); // change link color
d3.select("#" + d.id).classed("link", !stat); // change link color
mouse_action(d.source, stat, direction);
}
}
});
}
function unvisite_links() {
"use strict";
links.forEach(function (d) {
d.visited = false;
});
}
function renderRelationshipGraph(data) {
"use strict";
var count = [];
data.Nodes.forEach(function (d) {
count[d.lvl] = 0;
});
lvlCount = count.length;
data.Nodes.forEach(function (d, i) {
d.x = margin.left + d.lvl * (boxWidth + gap.width);
d.y = margin.top + (boxHeight + gap.height) * count[d.lvl];
d.id = "n" + i;
count[d.lvl] += 1;
Nodes.push(d);
});
data.links.forEach(function (d) {
links.push({
source: find(d.source),
target: find(d.target),
id: "l" + find(d.source).id + find(d.target).id
});
});
unvisite_links();
svg.append("g")
.attr("class", "nodes");
var node = svg.select(".nodes")
.selectAll("g")
.data(Nodes)
.enter()
.append("g")
.attr("class", "unit");
node.append("rect")
.attr("x", function (d) { return d.x; })
.attr("y", function (d) { return d.y; })
.attr("id", function (d) { return d.id; })
.attr("width", boxWidth)
.attr("height", boxHeight)
.attr("class", "node")
.attr("rx", 6)
.attr("ry", 6)
.on("mouseover", function () {
mouse_action(d3.select(this).datum(), true, "root");
unvisite_links();
})
.on("mouseout", function () {
mouse_action(d3.select(this).datum(), false, "root");
unvisite_links();
});
node.append("text")
.attr("class", "label")
.attr("x", function (d) { return d.x + 14; })
.attr("y", function (d) { return d.y + 15; })
.text(function (d) { return d.name; });
links.forEach(function (li) {
svg.append("path", "g")
.attr("class", "link")
.attr("id", li.id)
.attr("d", function () {
var oTarget = {
x: li.target.y + 0.5 * boxHeight,
y: li.target.x
};
var oSource = {
x: li.source.y + 0.5 * boxHeight,
y: li.source.x
};
if (oSource.y < oTarget.y) {
oSource.y += boxWidth;
} else {
oTarget.y += boxWidth;
}
return diagonal({
source: oSource,
target: oTarget
});
});
});
}
svg = d3.select("#tree").append("svg")
.attr("width", width)
.attr("height", height)
.append("g");
renderRelationshipGraph(data);
rect {
fill: #CCC;
cursor: pointer;
}
.active {
fill: orange;
stroke: orange;
}
.activelink {
fill: none;
stroke: orange;
stroke-width: 2.5px;
}
.label {
fill: white;
font-family: sans-serif;
pointer-events: none;
}
.link {
fill: none;
stroke: #ccc;
stroke-width: 2.5px;
}
<script src="https://d3js.org/d3.v3.min.js"></script>
<div id="tree"></div>
我需要知道一个脚本来生成节点和链接结构
我会首先构建一个节点字典，其中键是节点名称，值是一个包含父列表和子列表的元组。为了有一个更简单的方法来构建树，我还会保留所有顶级节点的集合（没有父节点）。
从那个字典，然后可以递归地构建一个 json 类数据，可以用来构建一个真正的 json 字符串。
但是你显示的是 不是 的 csv 格式，我使用 re.split 来解析输入：
import re
# First the data
t = '''parent | children | date
---------------------------------------------
830010000C0419 | 830010000C1205 | 1993/09/15
830010000C0947 | 830010000C1205 | 1993/09/15
830010000C0948 | 830010000C1205 | 1993/09/15
830010000B0854 | 830010000B1196 | 1994/03/11
830010000B0854 | 830010000B1197 | 1994/03/11
830010000B0721 | 830010000B1343 | 1988/12/05
830010000B1343 | 830010000B1344 | 1988/12/05
'''
rx = re.compile(r'\s*\|\s*')
# nodes is a dictionary of nodes, nodes[None] is the set of top-level names
nodes = {None: set()}
with io.StringIO(t) as fd:
_ = next(fd) # skip initial lines
_ = next(fd)
for linenum, line in enumerate(fd, 1):
p, c = rx.split(line.strip())[:2] # parse a line
if p == c: # a node cannot be its parent
raise ValueError(f'Same node as parent and child {p} at line {linenum}')
# process the nodes
if c not in nodes:
nodes[c] = ([], [])
elif c in nodes[None]:
nodes[None].remove(c)
if p not in nodes:
nodes[p] = ([], [c])
nodes[None].add(p)
else:
nodes[p][1].append(c)
nodes[c][0].append(p)
def subtree(node, nodes, parent=None, seen = None):
"""Builds a dict with the subtree of a node.
node is a node name, nodes the dict, parent is the parent name,
seen is a list of all previously seen node to prevent cycles
"""
if seen is None:
seen = [node]
elif node in seen: # special processing to break possible cycles
return {'name': node, 'parent': parent, 'children': '...'}
else:
seen.append(node)
return {'name': node, 'parent': parent, 'children':
[subtree(c, nodes, node, seen) for c in nodes[node][1]]}
# We can now build the json data
js = {node: subtree(node, nodes) for node in nodes[None]}
pprint.pprint(js)
它给出：
{'830010000B0721': {'children': [{'children': [{'children': [],
'name': '830010000B1344',
'parent': '830010000B1343'}],
'name': '830010000B1343',
'parent': '830010000B0721'}],
'name': '830010000B0721',
'parent': None},
'830010000B0854': {'children': [{'children': [],
'name': '830010000B1196',
'parent': '830010000B0854'},
{'children': [],
'name': '830010000B1197',
'parent': '830010000B0854'}],
'name': '830010000B0854',
'parent': None},
'830010000C0419': {'children': [{'children': [],
'name': '830010000C1205',
'parent': '830010000C0419'}],
'name': '830010000C0419',
'parent': None},
'830010000C0947': {'children': [{'children': [],
'name': '830010000C1205',
'parent': '830010000C0947'}],
'name': '830010000C0947',
'parent': None},
'830010000C0948': {'children': [{'children': [],
'name': '830010000C1205',
'parent': '830010000C0948'}],
'name': '830010000C0948',
'parent': None}}