- 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])