1 import copy 2 #淺拷貝 3 n1={'k1':'wu','k2':123,'k3':['carl',852]} 4 n2=n1 5 n3=copy.copy(n1) 6 print(id(n1)) 7 print(id(n2)) 8 print(id(n3)) 9 print(id(n1['k3'])) 10 print(id(n3['k3'])) 11 #深拷貝 12 n4=copy.deepcopy(n1) 13 print(id(n4)) 14 print(id(n1['k3'])) 15 print(id(n4['k3']))
返回值:
10787656
10787656
11532848
20277688
20277688
11455064
20277688
20276328python
二.函數的基本定義數組
1 def mail(): 2 def func(name, age = 18): 3 print"%s:%s" %(name,age) 4 # 指定參數 5 func('wupeiqi', 19) 6 # 使用默認參數 7 func('alex')
1 def func(*args): 2 print args 3 # 執行方式一 4 func(11,33,4,4454,5) 5 # 執行方式二 6 li = [11,2,2,3,3,4,54] 7 func(*li
1 def func(**kwargs): 2 print args 3 # 執行方式一 4 func(name='wupeiqi',age=18) 5 # 執行方式二 6 li = {'name':'wupeiqi', age:18, 'gender':'male'} 7 func(**li)
def show(*arg,**kwargs): print(arg,type(arg)) print(kwargs,type(kwargs)) show(64,56,99,w=76,p=33)
1 s1 ="{0} is {1}" 2 l=['alex','sb'] 3 result=s1.format(*l) 4 print(result) 5 s1 = "{name} is {a}" 6 result=s1.format(name='helen',a=19) 7 print(result) 8 9 s1 = "{name} is {a}" 10 d={"name":"helen",'a':19} 11 #result=s1.format(name='helen',a=19) 12 result=s1.format(**d) 13 print(result)
1 def func(a): 2 b=a+1 3 return b 4 等於 5 func=lambda a:a+1 6 ret=func(5) 7 print(ret)
三.內置函數app
1 print(ascii(10), ascii(9000000), ascii('b\31'), ascii('0x\1000')) 2 返回結果: 3 10 9000000 'b\x19' '0x@0'
1 #!usr/bin/env python 2 #coding:utf-8 3 namespace = {'name':'wupeiqi','data':[18,73,84]} 4 code = '''def hellocute():return "name %s ,age %d" %(name,data[0],) ''' 5 func = compile(code, '<string>', "exec") 6 exec func in namespace 7 result = namespace['hellocute']() 8 print result
divmod(a,b)方法返回的是a//b(除法取整)以及a對b的餘數less
返回結果類型爲tuple函數
在python2.3版本以前不容許處理複數,這個你們要注意一下spa
Take two (non complex) numbers as arguments and return a pair of numbers consisting of their quotient and remainder when using long division. With mixed operand types, the rules for binary arithmetic operators apply. For plain and long integers, the result is the same as (a // b, a % b). For floating point numbers the result is (q, a % b), where q is usually math.floor(a / b) but may be 1 less than that. In any case q * b + a % b is very close to a, if a % b is non-zero it has the same sign as b, and 0 <= abs(a % b) < abs(b).3d
Changed in version 2.3: Using divmod() with complex numbers is deprecated.code
1 >>> divmod(9,2) 2 (4, 1) 3 >>> divmod(11,3) 4 (3, 2) 5 >>> divmod(1+2j,1+0.5j) 6 ((1+0j), 1.5j)