标签：tool res lenovo 列表 lam 字符串 cti count 过滤
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)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)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)())#114.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]标签：tool res lenovo 列表 lam 字符串 cti count 过滤
原文地址：https://www.cnblogs.com/ciquankun/p/11227969.html
