假设您有一个数组def sortAnimal(llist):
while True:
print("\nChoose which parameter to sort: \n")
print("1. Name\n2. Age\n3. Species\n")
choice = int(input("Choice:"))
print("\n")
choices = {1: lambda animal: animal.name, 2: lambda animal: animal.age, 3: lambda animal: animal.species}
print("{:7s} {:7s} {:7s} {:7s}".format("Name:", "Age:", "Species:", "Gender:"))
sortedList = sorted(llist, key=choices[choice])
for obj in sortedList:
print(obj)
if input("\To reverse the list press enter. Otherwise press m + enter") == "":
print("\n")
print("{:7s} {:7s} {:7s} {:7s}".format("Name:", "Age:", "Species:", "Gender:"))
sortedList = sorted(llist, key=choices[choice], reverse=True)
for obj in sortedList:
print(obj)
else:
pass
if input("\To sort again press enter, back to menu press m + enter") == "":
continue
else:
break
，并且必须输出一个包含键-值对（其中键代表元素，值代表频率）的对象的排序数组，例如...
x = [1, 1, 2, 2, 2, 3]...什么是最优雅的方式？
答案 0 :(得分：1)
您可以创建一个临时对象并执行简单的@Bean
EmbeddedServletContainerCustomizer containerCustomizer() throws Exception {
return (ConfigurableEmbeddedServletContainer container) -> {
if (container instanceof TomcatEmbeddedServletContainerFactory) {
TomcatEmbeddedServletContainerFactory tomcat = (TomcatEmbeddedServletContainerFactory) container;
tomcat.addConnectorCustomizers(
(connector) -> {
connector.setMaxPostSize(10000000); // 10 MB
}
);
}
};
}
并检查对象中是否存在当前数字作为键，如果为true则将值加1，否则创建该键，然后使用简单的.forEach添加新数组中单独对象中的所有键值对
.map