Python 内置函数

print(10 * 10)
print(pow(10, 3))
print(divmod(5, 2))

x = 3.1415926
print(round(x, 3))

print(chr(65))
print(chr(97))
print(ord("A"))
print(ord("a"))

# s = {2, 4, 5, 6, 78, 2, 5}
# print(type(s))
# print(s)

l = [1, 2, 3, 4, 3, 2]
print(list(set(l)))

print(all([2 > 1, 2 == 3, 4 < 7]))
print(any([2 > 1, 2 == 3, 4 < 7]))

names = ["yuan", "rain", "alvin"]
ages = [22, 23, 32]
# [("yuan",22),("rain",23),(“alvin”,32)]
# print([(name, age) for name in names for age in ages])
print(list(zip(names, ages)))  # [('yuan', 22), ('rain', 23), ('alvin', 32)]

l = [2, 4, 1, 7, 5, 10]
l.sort()
print(l)

l2 = [('yuan', 22, 180), ('rain', 13, 175), ('alvin', 32, 198)]

l3 = ({"name": "yuan", "age": 23}, {"name": "rain", "age": 33}, {"name": "alvin", "age": 25})


# l2.sort()
# def foo(item):
#     return item[2]
#
# l3 = sorted(l2, key=foo, reverse=True)
# print(l3)  # [('rain', 13), ('yuan', 22), ('alvin', 32)]


# def bar(item):
#     return item["age"]
# print(sorted(l3, key=bar))  # [{'name': 'yuan', 'age': 23}, {'name': 'alvin', 'age': 25}, {'name': 'rain', 'age': 33}]

# print(sorted(l3, key=lambda item: item["age"]))