Python 如何改进此find（）函数（或类似函数）

如果在[table_t1.keys]中找到四个停止字中的一个，我想询问如何改进此工作代码以跳过[I]的执行

for i in table_t1.keys():
    if i.find("data") == -1:
        if i.find("split") == -1:
            if i.find("loss") == -1:
                if i.find("prob") == -1:
                    #do something

---

不需要使用find函数。

定义单词，如下所示

words = ("data", "split", "loss", "prob")

if all(word not in i for word in words):
    ...

if not any(word in i for word in words):
    ...

for i in table_t1:

现在，您可以使用或函数，如下所示

words = ("data", "split", "loss", "prob")

if all(word not in i for word in words):
    ...

if not any(word in i for word in words):
    ...

for i in table_t1:

它们基本上都检查单词元组中的单词是否不存在于i中

注意：如果table_t1实际上是一个字典，那么您不必调用.keys，它将创建一个键列表。您可以像这样简单地迭代它

words = ("data", "split", "loss", "prob")

if all(word not in i for word in words):
    ...

if not any(word in i for word in words):
    ...

for i in table_t1: