我是python的新手,正在尝试读取文件并从中创建字典。
格式如下:
.1.3.6.1.4.1.14823.1.1.27 {
TYPE = Switch
VENDOR = Aruba
MODEL = ArubaS3500-48T
CERTIFICATION = CERTIFIED
CONT = Aruba-Switch
HEALTH = ARUBA-Controller
VLAN = Dot1q INSTRUMENTATION:
Card-Fault = ArubaController:DeviceID
CPU/Memory = ArubaController:DeviceID
Environment = ArubaSysExt:DeviceID
Interface-Fault = MIB2
Interface-Performance = MIB2
Port-Fault = MIB2
Port-Performance = MIB2
}
第一行OID(.1.3.6.1.4.1.14823.1.1.27 {)我希望这是键,其余行是直到}的值
我尝试了几种组合,但无法获得正确的正则表达式来匹配这些组合
有什么帮助吗?
我已经尝试过类似的东西
lines = cache.readlines()
for line in lines:
searchObj = re.search(r'(^.\d.*{)(.*)$', line)
if searchObj:
(oid, cert ) = searchObj.groups()
results[searchObj(oid)] = ", ".join(line[1:])
print("searchObj.group() : ", searchObj.group(1))
print("searchObj.group(1) : ", searchObj.group(2))
最佳答案
您可以尝试以下方法:
import re
data = open('filename.txt').read()
the_key = re.findall("^\n*[\.\d]+", data)
values = [re.split("\s+\=\s+", i) for i in re.findall("[a-zA-Z0-9]+\s*\=\s*[a-zA-Z0-9]+", data)]
final_data = {the_key[0]:dict(values)}
输出:
{'\n.1.3.6.1.4.1.14823.1.1.27': {'VENDOR': 'Aruba', 'CERTIFICATION': 'CERTIFIED', 'Fault': 'MIB2', 'VLAN': 'Dot1q', 'Environment': 'ArubaSysExt', 'HEALTH': 'ARUBA', 'Memory': 'ArubaController', 'Performance': 'MIB2', 'CONT': 'Aruba', 'MODEL': 'ArubaS3500', 'TYPE': 'Switch'}}
关于python - 从文件python创建字典,我们在Stack Overflow上找到一个类似的问题:https://stackoverflow.com/questions/46798445/