lambda map() filter() zip()练习

练习:

用map来处理字符串列表,把列表中所有人都变成sb,比方alex_sb
```
l=[{'name':'alex'},{'name':'y'}]
```
```
l=[{'name':'alex'},{'name':'y'}]
l = list(map(lambda x:{"name":x["name"]+"_sb"},l))
print(l)
```
用map来处理下述l，然后用list得到一个新的列表，列表中每个人的名字都是sb结尾

l=[{'name':'alex'},{'name':'y'}]

l=[{'name':'alex'},{'name':'y'}]
l1 = list(map(lambda x:{"name":x["name"]+"sb"},l))
print(l1)

用filter来处理,得到股票价格大于20的股票名字

shares={
 'IBM':36.6,
 'Lenovo':23.2,
 'oldboy':21.2,
 'ocean':10.2,
         }

shares={
 'IBM':36.6,
 'Lenovo':23.2,
 'oldboy':21.2,
 'ocean':10.2,
         }
lst = list(filter(lambda x:shares[x]>20,shares))
print(lst)

有下面字典，得到购买每只股票的总价格，并放在一个迭代器中。
结果：list一下[9110.0, 27161.0,......]

portfolio = [
  {'name': 'IBM', 'shares': 100, 'price': 91.1},
    {'name': 'AAPL', 'shares': 50, 'price': 543.22},
    {'name': 'FB', 'shares': 200, 'price': 21.09},
    {'name': 'HPQ', 'shares': 35, 'price': 31.75},
    {'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}]

portfolio = [
  {'name': 'IBM', 'shares': 100, 'price': 91.1},
    {'name': 'AAPL', 'shares': 50, 'price': 543.22},
    {'name': 'FB', 'shares': 200, 'price': 21.09},
    {'name': 'HPQ', 'shares': 35, 'price': 31.75},
    {'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}]
lst = map(lambda x: x["shares"]*x["price"],portfolio)
print(list(lst))

还是上面的字典，用filter过滤出单价大于100的股票。

portfolio = [
  {'name': 'IBM', 'shares': 100, 'price': 91.1},
    {'name': 'AAPL', 'shares': 50, 'price': 543.22},
    {'name': 'FB', 'shares': 200, 'price': 21.09},
    {'name': 'HPQ', 'shares': 35, 'price': 31.75},
    {'name': 'YHOO', 'shares': 45, 'price': 16.35},
{'name': 'ACME', 'shares': 75, 'price': 115.65}]
lst = filter(lambda x:x["shares"]>100,portfolio)
print(list(lst))

有下列三种数据类型，

 l1 = [1,2,3,4,5,6]
 l2 = ['oldboy','alex','wusir','太白','日天']
 tu = ('**','***','****','*******')

写代码，最终得到的是（每个元祖第一个元素>2,第三个*至少是4个。）

 [(3, 'wusir', '****'), (4, '太白', '*******')]

这样的数据。

l1 = [1, 2, 3, 4, 5, 6]
l2 = ['oldboy', 'alex', 'wusir', '太白', '日天']
tu = ('**', '***', '****', '*******')
lst = zip(l1,l2,tu)
print(list(filter(lambda x:x[0]>2 and len(x[2])>3,list(lst))))

)有如下数据类型(实战题)：

 l1 = [ {'sales_volumn': 0},
       {'sales_volumn': 108},
       {'sales_volumn': 337},
       {'sales_volumn': 475},
       {'sales_volumn': 396},
       {'sales_volumn': 172},
       {'sales_volumn': 9},
       {'sales_volumn': 58}, 
       {'sales_volumn': 272}, 
       {'sales_volumn': 456}, 
       {'sales_volumn': 440},
       {'sales_volumn': 239}]

将l1按照列表中的每个字典的values大小进行排序，形成一个新的列表。

l1 = [{'sales_volumn': 0},
      {'sales_volumn': 108},
      {'sales_volumn': 337},
      {'sales_volumn': 475},
      {'sales_volumn': 396},
      {'sales_volumn': 172},
      {'sales_volumn': 9},
      {'sales_volumn': 58},
      {'sales_volumn': 272},
      {'sales_volumn': 456},
      {'sales_volumn': 440},
      {'sales_volumn': 239}]
lst = sorted(l1,key=lambda x:x["sales_volumn"])
print(lst)

有如下数据结构,通过过滤掉年龄大于16岁的字典

lst = [{'id':1,'name':'alex','age':18},
        {'id':1,'name':'wusir','age':17},
        {'id':1,'name':'taibai','age':16},]

lst = [{'id':1,'name':'alex','age':18},
        {'id':1,'name':'wusir','age':17},
        {'id':1,'name':'taibai','age':16},]
new_lst =filter(lambda x:x["age"]>16,lst)
print(list(new_lst))

9.有如下列表,按照元素的长度进行升序

lst = ['天龙八部','西游记','红楼梦','三国演义']

lst = ['天龙八部','西游记','红楼梦','三国演义']
new_lst = sorted(lst,key=lambda x:len(x))
print(new_lst)

10.有如下数据,按照元素的年龄进行升序

lst = [{'id':1,'name':'alex','age':18},
    {'id':2,'name':'wusir','age':17},
    {'id':3,'name':'taibai','age':16},]

lst = [{'id':1,'name':'alex','age':18},
    {'id':2,'name':'wusir','age':17},
    {'id':3,'name':'taibai','age':16},]
new_lst = sorted(lst,key=lambda x:x["age"])
print(new_lst)

11..看代码叙说,两种方式的区别

lst = [1,2,3,5,9,12,4]
lst.reverse()
print(lst)

print(list(reversed(lst)))

lst = [1,2,3,5,9,12,4]
lst.reverse()
print(lst)
#直接修改了原列表

print(list(reversed(lst)))
#原列表没有改变,只是新建了一个列表

求结果(面试题)

v = [lambda :x for x in range(10)]
print(v)
print(v[0])
print(v[0]())

v = [lambda :x for x in range(10)]
print(v)#10个lambda函数地址
print(v[0])#第一个lambda函数的地址<function <listcomp>.<lambda> at 0x000002846C5EDB70>
print(v[0]())#9

13.求结果(面试题)

v = (lambda :x for x in range(10))
print(v)
print(v[0])
print(v[0]())
print(next(v))
print(next(v)())

v = (lambda :x for x in range(10))
print(v)#生成器地址<generator object <genexpr> at 0x00000257F81E4C50>
print(v[0])#报错
print(v[0]())#报错
print(next(v))#生成器第一个函数的地址<function <genexpr>.<lambda> at 0x00000257F827DAE8>
print(next(v)())#1

14.map(str,[1,2,3,4,5,6,7,8,9])输出是什么? (面试题)

map(str,[1,2,3,4,5,6,7,8,9])
#什么也不输出
print(map(str,[1,2,3,4,5,6,7,8,9]))
#得到一个迭代器的地址
print(list(map(str,[1,2,3,4,5,6,7,8,9])))
#['1','2','3','4','5','6,'7','8','9']

15.有一个数组[34,1,2,5,6,6,5,4,3,3]请写一个函数，找出该数组中没有重复的数的总和（上面数据的么有重复的总和为1+2=3)(面试题)

#方法一

lst = [34,1,2,5,6,6,5,4,3,3]
l = 2*sum(set(lst)) - sum(lst)
print(l)
#方法二
lst = [34,1,2,5,6,6,5,4,3,3]
from functools import reduce
lst = sum(filter(lambda x: lst.count(x) < 2, lst))
print(lst)

16.求结果：（面试题，比较难，先做其他题）

def num():
    return [lambda x:i*x for i in range(4)]
print([m(2)for m in num()])

结果
#[6,6,6,6]

lambda map() filter() zip()练习

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