![](/img/trans.png)
[英]Python mypy: float and int are incompatible types with numbers.Real
[英]Python mypy incompatible types
所以,我有一些代码看起来像这样:
def get_tree(args):
sensor_ids = argToList(args.get("sensor_ids"))
process_ids = argToList(args.get("process_ids"))
all_results = {}
for sensor_id in sensor_ids:
sensor_id_str = str(sensor_id)
for process_id in process_ids:
process_id_str = str(process_id)
main_process = query_process(sensor_id_str, process_id_str)
results = {}
json_response = main_process.json()
for vertex in json_response["resources"]:
if vertex["vertex_type"] == "process":
results["main_process"] = {"sensor_id": sensor_id_str,
"process_id": process_id_str,
"properties": vertex["properties"]}
for edge in vertex["edges"]:
if edge == "child_process":
results["child_processes"] = []
MyPy 正在创建此错误,我不确定如何修复,flake 非常高兴:
赋值中的不兼容类型(表达式的类型为“List[]”,目标的类型为“Dict[str, Any]”)[assignment] results["child_processes"] = [] ^
为了那原始代码mypy
(用 0.950 和 0.961 测试)应该告诉你
“结果”需要类型注释(提示:“
results: Dict[<type>, <type>] =...
”)
所以它可以判断你是否在做正确的事。
使用增强代码,如果您调用reveal_type(results)
,您会看到
a.py:27: note: Revealed type is "builtins.dict[builtins.str, builtins.dict[builtins.str, Any]]"
即根据你对results
做的第一个任务,mypy 推断你有一个 dict of dicts,因此,在那里放一个列表是行不通的。
因此,如果您想将results
注释为带有字符串键和任意值的基本“包”,
results: Dict[str, Any] = {}
或者如果你想要更严格的输入,请使用TypedDict
:
class Results(TypedDict):
child_processes: Optional[List[Any]] # or be more specific...
results: Results = {}
声明:本站的技术帖子网页,遵循CC BY-SA 4.0协议,如果您需要转载,请注明本站网址或者原文地址。任何问题请咨询:yoyou2525@163.com.