The default input is decimal
Binary 0b, Octal 0o, Hexadecimal 0x
16 == 0b10000 == 0o20 == 0x10
Conversion between decimal and other bases
True
a = bin(16) # 转二进制 b = oct(16) # 转八进制 c = hex(16) # 转十六进制 print(a, b, c)
Note: The result after the above conversion is of string type, so if an equality comparison is performed, the output is False result
0b10000 0o20 0x10
a == b == c
False
type(a)
Convert other bases to decimal
str
d = int(a, 2) # 二进制转十进制 e = int(b, 8) # 八进制转十进制 f = int(c, 16) # 十六进制转十进制 print(d, e, f)
Uncertain decimal problem
16 16 16
(0.1+0.2) == 0.3
False
0.1+0.2
Computers use binary decimals to represent the decimal part of floating point numbers
Reason: Some decimals cannot be completely represented by binary decimals
0.30000000000000004
0.00011001100110011001 0.09999942779541016 0.0011001100110011 0.1999969482421875 0.01001100110011001 0.29999542236328125 0.01100110011001101 0.40000152587890625 0.1 === === 0.5
Normally it will not affect the calculation accuracy
二进制 十进制
0.1 + 0.7
Round to get the accuracy Solution
can be solved by rounding: round(parameter, retain the number of decimal places)
0.7999999999999999
a = 3*0.1 print(a)
0.30000000000000004
b = round(a, 1) print(b) b == 0.3
0.3 True
3+4j 2+5J
(2+5j)
2+1j
(1+3-4*2)/5
-0.8
x = 1 -x
-1
2**3
8
13//5 # 整数商 x/y 向下取整数
2
A few notes
13 % 5 # 模运算 余数 13=2*5+3
1+1.5
2.5
2/5
0.4
8/4
2.0
abs(-5)
5
abs(3+4j) # 对复数a+bj 执行的是求模运算(a^2+b^2)^0.5
5.0
pow(2, 5) # pow(x,n) x的n次方 等价于x**n
32
pow(2, 5, 3) # 2^5 % 3 更快速
2
a = 1.618 print(round(a)) # 默认四舍五入为整数
2
print(round(a, 2)) # 参数2表示四舍五入后保留2位小数
1.62
print(round(a, 5)) # 位数不足,无需补齐
1.618
divmod(13, 5) # 较(x//y,x % y)更快,只执行了一次x/y
(2, 3)
max(3, 2, 3, 6, 9, 4, 5)
9
a = [3, 2, 3, 6, 9, 4, 5] print("max:", max(a)) print("min:", min(a))
max: 9 min: 2
sum((1, 2, 3, 4, 5))
15
import math # 导入库 print(math.exp(1)) # 指数运算 e^x print(math.log2(2)) # 对数运算 print(math.sqrt(4)) # 开平方运算 等价于4^0.5
2.718281828459045 1.0 2.0
import numpy as np a = [1, 2, 3, 4, 5] print(np.mean(a)) # 求均值 print(np.median(a)) # 求中位数 print(np.std(a)) # 求标准差
3.0 3.0 1.4142135623730951
print("Python") print('Python')
There is an odd in the double
Python Python
print("I'm 18 years old")
There is a double in the odd
print('"Python" is good')
"Python" is good
双中有双,单中有单——转义符 \
# print(""Python" is good") print("\"Python\" is good") # \ 字符
"Python" is good
转义符可以用来换行继续输入
s = "py\ thon" print(s)
python
s = "My name is Peppa Pig"
变量名[位置编号]
正向索引——从零开始递增
位置编号不能超过字符串的长度
空格也是一个位置
print(s[0]) print(s[2]) print(s[5])
M m
s = "My name is Peppa Pig"
反向索引——从-1开始递减
print(s[-1]) print(s[-3]) print(s[-5])
g P a
索引只能获得一个字符,如何获得多个字符?
变量名[开始位置:结束位置:切片间隔]
切片间隔如不设置默认为1,可省略
切片范围不包含结束位置(前闭后开)
s = "Python" print(s[0:3:1])
Pyt
print(s[0:3])
Pyt
print(s[0:3:2])
Pt
起始位置是0 可以省略
结束位置省略,代表可以取到最后一个字符
可以使用反向索引
s = "Python" print(s[0:6])
Python
print(s[:6])
Python
print(s[:])
Python
print(s[-6:])
Python
反向切片
起始位置是-1也可以省略
结束位置省略,代表可以取到第一个字符
关键点在于-1,代表前一个位置比后一个位置大-1
s = "123456789" print(s[-1:-10:-1])
987654321
print(s[:-10:-1])
987654321
print(s[::-1])
987654321
字符串1+字符串2
a = "I love " b = "my wife " a+b
'I love my wife '
字符串 * n n * 字符串
c = a+b print(c*3) print(3*c)
I love my wife I love my wife I love my wife I love my wife I love my wife I love my wife
子集in全集 任何一个连续的切片都是原字符串的子集
folk_singers = "Peter, Paul and Mary" "Peter" in folk_singers
True
"PPM" in folk_singers
False
遍历字符串字符 for 字符 in 字符串
for s in "Python": print(s)
P y t h o n
所含字符的个数
s = "python" len(s)
6
将中文字库,英文字母、数字、特殊字符等转化成计算机可识别的二进制数
每个单一字符对应一个唯一的互不重复的二进制编码
Python 中使用的是Unicode编码
将字符转化为Unicode码——ord(字符)
print(ord("1")) print(ord("a")) print(ord("*")) print(ord("中")) print(ord("国"))
49 97 42 20013 22269
将Unicode码转化为字符——chr(Unicode码)
print(chr(1010)) print(chr(10000)) print(chr(12345)) print(chr(23456))
ϲ ✐ 〹 宠
返回一个列表
原字符串不变
上述特性适合以下所有字符串处理方法
languages = "Python C C++ Java PHP R" languages_list = languages.split(" ")#括号里的参数就是我们希望对目标字符串进行分割的标记 print(languages_list) print(languages)
['Python', 'C', 'C++', 'Java', 'PHP', 'R'] Python C C++ Java PHP R
可迭代类型 如:字符串、列表
s = "12345" s_join = ",".join(s) #把可迭代的对象每一个都取出来,相邻两个之间加上聚合字符 s_join
'1,2,3,4,5'
序列类型的元素必须是字符类型
# s = [1, 2, 3, 4, 5] 无法使用聚合 s = ["1", "2", "3", "4", "5"] "*".join(s)
'1*2*3*4*5'
strip从两侧开始搜索,遇到指定字符执行删除,遇到非指定字符,搜索停止
类似的还有左删除lstrip和右删除rstrip
s = " I have many blanks " print(s.strip(" ")) #从两端进行搜索,遇到指定字符后删除空格,然后停止 print(s.lstrip(" ")) print(s.rstrip(" ")) print(s)
I have many blanks I have many blanks I have many blanks I have many blanks
s = "Python is coming" s1 = s.replace("Python","Py") print(s1)
Py is coming
s = "Python is an excellent language" print("an:", s.count("an")) print("e:", s.count("e"))
an: 2 e: 4
字符串.upper() 字母全部大写
s = "Python" s.upper()
'PYTHON'
字符串.lower() 字母全部小写
print(s.lower()) print(s)
python Python
字符串.title()首字母大写
s.title()
'Python'
a = 10 print(a > 8) print(a == 12) print(a < 5)
True False False
any() 数据有一个是非零就为True
all() 数据有一个是零就为False (元素都是非零的)
print(any([False,1,0,None])) # 0 False None 都是无 print(all([False,1,0,None]))
True False
n = 2800 while True: m = eval("请输入一个正整数:")) if m == n: print("正确") break elif m > n: print("太大了") else: print("太小了")
请输入一个正整数:280 太小了 请输入一个正整数:2800 正确
import numpy as np x = np.array([[1, 3, 2, 5, 7]]) # 定义 numpy数组 print(x > 3) x[x > 3]
[[False False False True True]] array([5, 7])
type(变量)
age = 20 name = "Ada" print(type(age)) print(type(name))
<class 'int'> <class 'str'>
isinstance(变量,预判类型) 承认继承
变量类型是预判类型的子类型,则为真,否则为假
print(isinstance(age, int)) # 承认继承 这里的int就相当于是一个类
True
print(isinstance(age, object)) print(isinstance(name, object)) # object 是所有类的老祖宗
True True
字符串检查方法
字符串.isdigit()字符是否只有数字组成
age = "20" name = "Ada"
age.isdigit()
True
name.isdigit()
False
字符串.isalpha()字符是否只有字母组成
name.isalpha()
True
age.isalpha()
False
字符串.isalnum()字符是否只有数字和字母组成
"Ada20".isalnum() # 比如可用于判断用户名是否合法
True
数字类型转字符串 str(数字类型)
age = 20 print("My age is "+str(age))
My age is 20
仅有数字组成的字符串转数字 int() float() eval(232, 232, 232); background: rgb(249, 249, 249);">
s1 = "20" s2 = "10.1"
int(s1) # 仅整型 # int(s2) 会错误
20
float(s1)
20.0
float(s2)
10.1
eval(232, 232, 232); background: rgb(249, 249, 249);">20
eval(232, 232, 232); background: rgb(249, 249, 249);">10.1
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