hacktricks/generic-methodologies-and-resources/python/basic-python.md

基础 Python

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Python 基础

有用信息

list(xrange()) == range() --> 在 Python3 中，range 是 Python2 的 xrange (它不是一个列表而是一个生成器)
元组和列表的区别在于元组中值的位置赋予其含义，而列表只是有序值。元组有结构，而列表有顺序。

主要操作

要求一个数的幂，使用：3**2 (不是 3^2)
如果执行 2/3，会返回 1，因为你在除两个整数 (integers)。如果想要小数，应该除以浮点数 (2.0/3.0).
i >= j
i <= j
i == j
i != j
a and b
a or b
not a
float(a)
int(a)
str(d)
ord("A") = 65
chr(65) = 'A'
hex(100) = '0x64'
hex(100)[2:] = '64'
isinstance(1, int) = True
"a b".split(" ") = ['a', 'b']
" ".join(['a', 'b']) = "a b"
"abcdef".startswith("ab") = True
"abcdef".contains("abc") = True
"abc\n".strip() = "abc"
"apbc".replace("p","") = "abc"
dir(str) = 所有可用方法的列表
help(str) = 类 str 的定义
"a".upper() = "A"
"A".lower() = "a"
"abc".capitalize() = "Abc"
sum([1,2,3]) = 6
sorted([1,43,5,3,21,4])

连接字符
3 * ’a’ = ‘aaa’
‘a’ + ‘b’ = ‘ab’
‘a’ + str(3) = ‘a3’
[1,2,3]+[4,5]=[1,2,3,4,5]

列表的部分
‘abc’[0] = ‘a’
'abc’[-1] = ‘c’
'abc’[1:3] = ‘bc’ 从 [1] 到 [2]
"qwertyuiop"[:-1] = 'qwertyuio'

注释
# 单行注释
"""
多行注释
另一个
"""

循环

if a:
#somethig
elif b:
#something
else:
#something

while(a):
#comething

for i in range(0,100):
#something from 0 to 99

for letter in "hola":
#something with a letter in "hola"

元组

t1 = (1, '2', 'three')
t2 = (5, 6)
t3 = t1 + t2 = (1, '2', 'three', 5, 6)
(4,) = 单元素元组
d = () 空元组
d += (4,) --> 添加到元组
无法！ --> t1[1] == 'New value'
list(t2) = [5, 6] --> 从元组转换为列表

列表（数组）

d = [] 空
a = [1, 2, 3]
b = [4, 5]
a + b = [1, 2, 3, 4, 5]
b.append(6) = [4, 5, 6]
tuple(a) = (1, 2, 3) --> 从列表转换为元组

字典

d = {} 空
monthNumbers={1:'Jan', 2: 'feb','feb':2}—> monthNumbers ->{1:'Jan', 2: 'feb','feb':2}
monthNumbers[1] = 'Jan'
monthNumbers['feb'] = 2
list(monthNumbers) = [1, 2, 'feb']
monthNumbers.values() = ['Jan', 'feb', 2]
keys = [k for k in monthNumbers]
a={'9':9}
monthNumbers.update(a) = {'9':9, 1:'Jan', 2: 'feb','feb':2}
mN = monthNumbers.copy() #独立复制
monthNumbers.get('key',0) #检查键是否存在，如果存在则返回 monthNumbers["key"] 的值，否则返回 0

集合

集合中没有重复元素
myset = set(['a', 'b']) = {'a', 'b'}
myset.add('c') = {'a', 'b', 'c'}
myset.add('a') = {'a', 'b', 'c'} #没有重复
myset.update([1, 2, 3]) = set(['a', 1, 2, 'b', 'c', 3])
myset.discard(10) #如果存在则移除，否则不做任何操作
myset.remove(10) #如果存在则移除，否则引发异常
myset2 = set([1, 2, 3, 4])
myset.union(myset2) #myset 或 myset2 中的值
myset.intersection(myset2) #myset 和 myset2 中的值
myset.difference(myset2) #myset 中但不在 myset2 中的值
myset.symmetric_difference(myset2) #不在 myset 和 myset2 中的值（两者都不在）
myset.pop() #获取集合的第一个元素并将其移除
myset.intersection_update(myset2) #myset = myset 和 myset2 中的元素
myset.difference_update(myset2) #myset = myset 中但不在 myset2 中的元素
myset.symmetric_difference_update(myset2) #myset = 两者都不在的元素

类

__It__ 中的方法将用于排序，以比较此类对象是否大于其他对象

class Person(name):
def __init__(self,name):
self.name= name
self.lastName = name.split(‘ ‘)[-1]
self.birthday = None
def __It__(self, other):
if self.lastName == other.lastName:
return self.name < other.name
return self.lastName < other.lastName #Return True if the lastname is smaller

def setBirthday(self, month, day. year):
self.birthday = date tame.date(year,month,day)
def getAge(self):
return (date time.date.today() - self.birthday).days


class MITPerson(Person):
nextIdNum = 0	# Attribute of the Class
def __init__(self, name):
Person.__init__(self,name)
self.idNum = MITPerson.nextIdNum  —> Accedemos al atributo de la clase
MITPerson.nextIdNum += 1 #Attribute of the class +1

def __it__(self, other):
return self.idNum < other.idNum

map, zip, filter, lambda, sorted and one-liners

Map 的用法类似于：[f(x) for x in iterable] --> map(tutple,[a,b]) = [(1,2,3),(4,5)]
m = map(lambda x: x % 3 == 0, [1, 2, 3, 4, 5, 6, 7, 8, 9]) --> [False, False, True, False, False, True, False, False, True]

zip 在 foo 或 bar 较短时停止：

for f, b in zip(foo, bar):
print(f, b)

Lambda 用于定义一个函数
(lambda x,y: x+y)(5,3) = 8 --> 使用 lambda 作为简单的 函数
sorted(range(-5,6), key=lambda x: x** 2) = [0, -1, 1, -2, 2, -3, 3, -4, 4, -5, 5] --> 使用 lambda 对列表进行排序
m = filter(lambda x: x % 3 == 0, [1, 2, 3, 4, 5, 6, 7, 8, 9]) = [3, 6, 9] --> 使用 lambda 进行过滤
reduce (lambda x,y: x*y, [1,2,3,4]) = 24

def make_adder(n):
return lambda x: x+n
plus3 = make_adder(3)
plus3(4) = 7 # 3 + 4 = 7

class Car:
crash = lambda self: print('Boom!')
my_car = Car(); my_car.crash() = 'Boom!'

<h2>异常</h2>

mult1 = \[x for x in \[1, 2, 3, 4, 5, 6, 7, 8, 9] if x%3 == 0 ]

def divide(x,y):
try:
result = x/y
except ZeroDivisionError, e:
print “division by zero!” + str(e)
except TypeError:
divide(int(x),int(y))
else:
print “result i”, result
finally
print “executing finally clause in any case”

Assert()

如果条件为假，则字符串将被打印在屏幕上

def avg(grades, weights):
assert not len(grades) == 0, 'no grades data'
assert len(grades) == 'wrong number grades'

生成器，yield

一个生成器，不是返回某个东西，而是"产出"某个东西。当你访问它时，它会"返回"第一个生成的值，然后，你可以再次访问它，它将返回下一个生成的值。因此，所有的值不是同时生成的，使用这种方法而不是包含所有值的列表可以节省大量内存。

def myGen(n):
yield n
yield n + 1

g = myGen(6) --> 6\
next(g) --> 7\
next(g) --> Error

### 正则表达式

import re\
re.search("\w","hola").group() = "h"\
re.findall("\w","hola") = \['h', 'o', 'l', 'a']\
re.findall("\w+(la)","hola caracola") = \['la', 'la']

**特殊含义:**\
. --> Everything\
\w --> \[a-zA-Z0-9\_]\
\d --> Number\
\s --> WhiteSpace char\[ \n\r\t\f]\
\S --> Non-whitespace char\
^ --> Starts with\
$ --> Ends with\
\+ --> One or more\
\* --> 0 or more\
? --> 0 or 1 occurrences

**选项:**\
re.search(pat,str,re.IGNORECASE)\
IGNORECASE\
DOTALL --> Allow dot to match newline\
MULTILINE --> Allow ^ and $ to match in different lines

re.findall("<.\*>", "\<b>foo\</b>and\<i>so on\</i>") = \['\<b>foo\</b>and\<i>so on\</i>']\
re.findall("<.\*?>", "\<b>foo\</b>and\<i>so on\</i>") = \['\<b>', '\</b>', '\<i>', '\</i>']

IterTools\
**product**\
from **itertools** import product --> Generates combinations between 1 or more lists, perhaps repeating values, cartesian product (distributive property)\
print list(**product**(\[1,2,3],\[3,4])) = \[(1, 3), (1, 4), (2, 3), (2, 4), (3, 3), (3, 4)]\
print list(**product**(\[1,2,3],repeat = 2)) = \[(1, 1), (1, 2), (1, 3), (2, 1), (2, 2), (2, 3), (3, 1), (3, 2), (3, 3)]

**permutations**\
from **itertools** import **permutations** --> Generates combinations of all characters in every position\
print list(permutations(\['1','2','3'])) = \[('1', '2', '3'), ('1', '3', '2'), ('2', '1', '3'),... Every posible combination\
print(list(permutations('123',2))) = \[('1', '2'), ('1', '3'), ('2', '1'), ('2', '3'), ('3', '1'), ('3', '2')] Every possible combination of length 2

**combinations**\
from itertools import **combinations** --> Generates all possible combinations without repeating characters (if "ab" existing, doesn't generate "ba")\
print(list(**combinations**('123',2))) --> \[('1', '2'), ('1', '3'), ('2', '3')]

**combinations\_with\_replacement**\
from itertools import **combinations\_with\_replacement** --> Generates all possible combinations from the char onwards(for example, the 3rd is mixed from the 3rd onwards but not with the 2nd o first)\
print(list(**combinations\_with\_replacement**('1133',2))) = \[('1', '1'), ('1', '1'), ('1', '3'), ('1', '3'), ('1', '1'), ('1', '3'), ('1', '3'), ('3', '3'), ('3', '3'), ('3', '3')]

### 装饰器

Decorator that size the time that a function needs to be executed (from [here](https://towardsdatascience.com/decorating-functions-in-python-619cbbe82c74)):

from functools import wraps
import time
def timeme(func):
@wraps(func)
def wrapper(*args, **kwargs):
print("Let's call our decorated function")
start = time.time()
result = func(*args, **kwargs)
print('Execution time: {} seconds'.format(time.time() - start))
return result
return wrapper

@timeme
def decorated_func():
print("Decorated func!")

如果你运行它，你会看到类似以下内容：

Let's call our decorated function
Decorated func!
Execution time: 4.792213439941406e-05 seconds

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支持 HackTricks

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• 加入 💬 Discord 群组 或 电报群组 或 关注我们的 Twitter 🐦 @hacktricks_live.
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