内置函数

- 第1组(5个)

- abs,绝对值

v = abs(-10)

- pow,指数

v1 = pow(2,5) # 2的5次方 2**5
print(v1)

- sum,求和

v1 = sum([-11, 22, 33, 44, 55]) # 可以被迭代-for循环
print(v1)

- divmod,求商和余数

v1, v2 = divmod(9, 2)
print(v1, v2)

- round,小数点后n位(四舍五入)

v1 = round(4.11786, 2)
print(v1) # 4.12

- 第2组:(4个)

- min,最小值

v1 = min(11, 2, 3, 4, 5, 56)
print(v1) # 2

v2 = min([11, 22, 33, 44, 55]) # 迭代的类型(for循环)
print(v2)

v3 = min([-11, 2, 33, 44, 55], key=lambda x: abs(x))
print(v3) # 2

- max,最大值

v1 = max(11, 2, 3, 4, 5, 56)
print(v1)

v2 = max([11, 22, 33, 44, 55])
print(v2)

v3 = max([-11, 22, 33, 44, 55], key=lambda x: x * 10)
print(v3) # 55

- all,是否全部为True

v1 = all( [11,22,44,""] ) # False

- any,是否存在True

v2 = any([11,22,44,""]) # True



- 第3组(3个)

- bin,十进制转二进制
- oct,十进制转八进制
- hex,十进制转十六进制

- 第4组(2个)

- ord,获取字符对应的unicode码点(十进制)

v1 = ord("武")
print(v1, hex(v1))

- chr,根据码点(十进制)获取对应字符

v1 = chr(27494)
print(v1)

- 第5组(9个)

- int

- foat

- str,unicode编码

- bytes,utf-8、gbk编码

v1 = "武沛齐" # str类型

v2 = v1.encode('utf-8') # bytes类型

v3 = bytes(v1,encoding="utf-8") # bytes类型

- bool

- list

- dict

- tuple

- set

- 第6组(13个)

- len

- print

- input

- open

- type,获取数据类型

v1 = "123"

if type(v1) == str:
pass
else:
pass

- range

range(10)

- enumerate

v1 = ["武沛齐", "alex", 'root']

for num, value in enumerate(v1, 1):
print(num, value)

- id

- hash

v1 = hash("武沛齐")

- help,帮助信息

- pycharm,不用
- 终端,使用

- zip

v1 = [11, 22, 33, 44, 55, 66]
v2 = [55, 66, 77, 88]
v3 = [10, 20, 30, 40, 50]

result = zip(v1, v2, v3)
for item in result:
print(item)

- callable,是否可执行,后面是否可以加括号。

v1 = "武沛齐"
v2 = lambda x: x
def v3():
pass


print( callable(v1) ) # False
print(callable(v2))
print(callable(v3))

- sorted,排序

v1 = sorted([11,22,33,44,55])
posted @ 2021-03-15 16:50  A熙  阅读(51)  评论(0)    收藏  举报