Python面试题面经_python面经

def merge_dict(x,y):
    for k,v in x.items():
                if k in y.keys():
                    y[k] += v
                else:
                    y[k] = v

a,b=0,1
while i<100:
    print(b)
    a,b=b,a+b

def test(number):
    def test_in(number_in):
        print('number_in%d'%number_in)
        return number + number_in
    return test_in
ret = test(20)  #给test函数赋值，这个20就是给参数number
print(ret(200))  #这里的200给number_in   
print(ret(100))
#返回
#number_in200
#220
#number_in100
#120
 
def line_cof(a,b):
    def line(x):
        return a*x+b
    return line
line1 = line_cof(1,1)    #x+1
line2 = line_cof(4,5)    #4x+5
print(line1(5))
print(line2(5))
#返回值
#6
#25

def makebold(fn):
    def warp():
        return "<a>"+fn()+"<a>"
    return warp
def makeitalic(fn):
    def warp():
        return "<b>"+fn()+"<b>"
    return warp
@makebold         #这句话相当于makebold(test1),也就是把当前函数传过去
def test1():
    return "test1"
@makeitalic
def test2():
    return "test2"
    
@makebold
@makeitalic
def test3():     #函数和装饰器是倒着执行的，从下往上，相当于makeold(makeitalic(test3))
    return "test3"
    
print(test1())
print(test2())
print(test3())

DIGITS={'0':0,'1':1,'2':2,'3':3,'4':4,'5':5,'6':6,'7':7,'8':8,'9':9}
def str2float(s):
    s=s.split('.')
    if s[0]==0:
        return 0+reduce(lambda x,y:x/10+y , map(lambda x:DIGITS[x],s[1][::-1]))/10
    else:
        return reduce(lambda x,y:x*10+y,map(lambda x:DIGITS[x],s[0]))+reduce(lambda x,y:x/10+y , map(lambda x:DIGITS[x],s[1][::-1]))/10  
print(str2float('123.0456'))
print(str2float('0.0456'))
#注意上面小数点后面是逆置后元素然后进行x/10+y。最后返回各位是有值得，所以进行/10操作

1 li=[1,2,3,4,5,1,2,3]
2 new_li=[]
3 for i in li:
4     if i not in new_li:
5         new_li.append(i)
6 print(new_li)

1 li=[1,2,3,4,5,1,2,3]
2 new_li=list(set(li))
3 new_li.sort(key=li.index)
4 print(new_li)

#1.13通过公共键对字典列表排序,根据一个或多个字典中的值来对列表排序
#利用operator中的itemgetter函数对这类结构排序很简单
rows = [
{'fname': 'Brian', 'lname': 'Jones', 'uid': 1003},
{'fname': 'David', 'lname': 'Beazley', 'uid': 1002},
{'fname': 'John', 'lname': 'Cleese', 'uid': 1001},
{'fname': 'Big', 'lname': 'Jones', 'uid': 1004}
]
from operator import itemgetter
rows_by_name = sorted(rows,key=itemgetter('fname'))   #按照fname的大小对字典进行排序
rows_by_uid = sorted(rows,key=itemgetter('uid'))    #按照uid 对字典进行排序
print(rows_by_name)
print(rows_by_uid)
#itemgetter()函数可以接受多个键
rows_by_lfname = sorted(rows,key=itemgetter('lname','fname'))   #按照lname,fname同时排序
print(rows_by_lfname)

#也可以自定义，用lambda函数取代itemgetter()函数的功能
rows_by_name = sorted(rows,key=lambda x:x['fname'])   #按照fname的值排序
print(rows_by_name)
rows_by_lfname = sorted(rows,key=lambda x:(x['lname'],x['fname']))
print(rows_by_lfname)