day16作业

声明式编程：

1、将names = ['egon','alex_sb','wupeiqi','yuanhu']中的名字全部大写

names = ['egon','alex_sb','wupeiqi','yuanhu']
names = map(lambda x:x.upper(),names)
print(list(names))

#['EGON', 'ALEX_SB', 'WUPEIQI', 'YUANHU']

2、将names = ['egon','alex_sb','wupeiqi','yuanhu']中的名字全部大写

names = ['egon','alex_sb','wupeiqi','yuanhu']
res = filter(lambda name:not name.endswith('sb'),names)
print(list(res))

#['egon', 'wupeiqi', 'yuanhu']

3、求文件a.txt中最长的行的长度，（长度以字符个数计算，需要使用max函数）

f = open(r'a.txt','rt',encoding='utf-8')
res = len(max(f,key=lambda x:len(x)))
print(res)
f.close()
#30

4、求a.txt文件中字符的个数，思考在第一次后的n次求和中sum求和的结果都为0，（需要使用sum函数）

f = open(r'a.txt','rt',encoding='utf-8')
res = sum(map(lambda x:len(x),f))
print(res)
#90

5、思考题

with open(r'a.txt','r',encoding='utf-8') as f:
    g = (len(line) for line in f)
print(sum(g))
#为何出错

求sum(g)的时候，文件已经关闭，没有办法是文件迭代生成想要的内容，可以改为：

f= open(r'a.txt','r',encoding='utf-8')
g = (len(line) for line in f)
print(sum(g))
f.close()

6、文件shopping.txt内容如下：

mac,20000,3
lenovo,3000,10
tesla,1000000,10
chicken,200,1

求总价：

f = open(r'shopping.txt','r',encoding='utf-8')
res = map(lambda x:x.strip('
').split(','),f)
total = sum(map(lambda x:float(x[1])*float(x[2]),res))
print(total)
f.close()
#10090200.0

打印商品信息 格式为：
 [{'name':'xxx','price':'xxxx','count':'xx'}]

f = open(r'shopping.txt','r',encoding='utf-8')
# [{'name':'xxx','price':'xxxx','count':'xx'}]
res = map(lambda x:x.strip('
').split(','),f)
msg = map(lambda x:{'name':x[0],'price':x[1],'count':x[2]},res)
msg_list=[i for i in msg]
print(msg_list)
#[{'name': 'mac', 'price': '20000', 'count': '3'}, {'name': 'lenovo', 'price': '3000', 'count': '10'}, {'name': 'tesla', 'price': '1000000', 'count': '10'}, {'name': 'chicken', 'price': '200', 'count': '1'}]

求单价大于10000的商品格式同上

f = open(r'shopping.txt','r',encoding='utf-8')
res = map(lambda x:x.strip('
').split(','),f)
res10000 = filter(lambda x:float(x[1])>10000,res)
msg = map(lambda x:{'name':x[0],'price':x[1],'count':x[2]},res10000)
msg_list=[i for i in msg]
print(msg_list)
f.close()
#[{'name': 'mac', 'price': '20000', 'count': '3'}, {'name': 'tesla', 'price': '1000000', 'count': '10'}]

函数的递归：

二分法：

想从一个从小到大排列的数字列表中找到指定的数字，遍历的效率太低，用二分法（算法的一种）

实现in的效果

def in2(num,li):
    if len(li) ==0:
        print('no exit')
        return False
    mid_ind = len(li)//2
    if num>li[mid_ind]:
        li = li[mid_ind+1:]
        return in2(num,li)
    elif num < li[mid_ind]:
        li = li[:mid_ind]
        return in2(num,li)
    else:
        return True

print(in2(33,l))
#True

实现l.index(30)效果

l=[1,2,3,4,5,6,7,8,11,33,44,55,66,77,88,99]
def index2(num,li,start =0,stop=len(l)-1):
    if start<stop:
        mid_index = start+(stop-start)//2
        if num > li[mid_index]:
            start = mid_index+1
            return index2(num,li,start,stop)
        elif num < li[mid_index]:
            stop=mid_index-1
            return index2(num,li,start,stop)
        else:
            return mid_index
    else:
        print('no exist')
        return None

res =index2(33,l)
print(res)
#9