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python - 将文件解析为 JSON 文件的父/子格式

转载 作者:行者123 更新时间:2023-12-01 04:26:50 27 4
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我想要一些关于如何解析此文件 Gene ontology (.obo) 的帮助/建议

我正在 D3 中创建可视化,需要创建一个 JSON 格式的“树”文件 -

{
"name": "flare",
"description": "flare",
"children": [
{
"name": "analytic",
"description": "analytics",
"children": [
{
"name": "cluster",
"description": "cluster",
"children": [
{"name": "Agglomer", "description": "AgglomerativeCluster", "size": 3938},
{"name": "Communit", "description": "CommunityStructure", "size": 3812},
{"name": "Hierarch", "description": "HierarchicalCluster", "size": 6714},
{"name": "MergeEdg", "description": "MergeEdge", "size": 743}
]
}, etc..

这种格式似乎很容易在 python 字典中复制,每个条目有 3 个字段:名称、描述和子项[]。

我的问题实际上是如何提取数据。上面链接的文件的“对象”结构如下:

[Term]
id: GO:0000001
name: mitochondrion inheritance
namespace: biological_process
def: "The distribution of mitochondria, including the mitochondrial genome, into daughter cells after mitosis or meiosis, mediated by interactions between mitochondria and the cytoskeleton." [GOC:mcc, PMID:10873824, PMID:11389764]
synonym: "mitochondrial inheritance" EXACT []
is_a: GO:0048308 ! organelle inheritance
is_a: GO:0048311 ! mitochondrion distribution

我需要 id、is_a 和 name 字段。我尝试使用 python 来解析它,但我似乎找不到找到每个对象的方法。

有什么想法吗?

最佳答案

这是解析“.obo”文件中对象的相当简单的方法。它将对象数据保存到一个 dict 中,其中 id 作为键,nameis_a 数据保存在一个列表。然后它使用标准 json 模块的 .dumps 函数来漂亮地打印它。

出于测试目的,我在链接中使用了文件的截断版本,最多仅包含 id: GO:0000006

此代码会忽略包含 is_obsolete 字段的任何对象。它还从 is_a 字段中删除描述信息;我想您可能想要这个,但禁用该功能很容易。

#!/usr/bin/env python

''' Parse object data from a .obo file

From http://stackoverflow.com/q/32989776/4014959

Written by PM 2Ring 2015.10.07
'''

from __future__ import print_function, division

import json
from collections import defaultdict

fname = "go-basic.obo"
term_head = "[Term]"

#Keep the desired object data here
all_objects = {}

def add_object(d):
#print(json.dumps(d, indent = 4) + '\n')
#Ignore obsolete objects
if "is_obsolete" in d:
return

#Gather desired data into a single list,
# and store it in the main all_objects dict
key = d["id"][0]
is_a = d["is_a"]
#Remove the next line if you want to keep the is_a description info
is_a = [s.partition(' ! ')[0] for s in is_a]
all_objects[key] = d["name"] + is_a


#A temporary dict to hold object data
current = defaultdict(list)

with open(fname) as f:
#Skip header data
for line in f:
if line.rstrip() == term_head:
break

for line in f:
line = line.rstrip()
if not line:
#ignore blank lines
continue
if line == term_head:
#end of term
add_object(current)
current = defaultdict(list)
else:
#accumulate object data into current
key, _, val = line.partition(": ")
current[key].append(val)

if current:
add_object(current)

print("\nall_objects =")
print(json.dumps(all_objects, indent = 4, sort_keys=True))

输出

all_objects =
{
"GO:0000001": [
"mitochondrion inheritance",
"GO:0048308",
"GO:0048311"
],
"GO:0000002": [
"mitochondrial genome maintenance",
"GO:0007005"
],
"GO:0000003": [
"reproduction",
"GO:0008150"
],
"GO:0000006": [
"high-affinity zinc uptake transmembrane transporter activity",
"GO:0005385"
]
}

关于python - 将文件解析为 JSON 文件的父/子格式,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/32989776/

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