并不完全漂亮，但是您可能会使用
def xpath_get(mydict, path):
elem = mydict
try:
for x in path.strip("/").split("/"):
elem = elem.get(x)
except:
pass
return elem
当然，这不支持xpath之类的东西，例如索引……更不用说/指出的关键陷阱unutbu了。
像Xpath一样查询嵌套的python字典
像Xpath一样查询嵌套的python字典
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