1、如何在列表，字典，集合种根据条件筛选数据？2、如何为元组中的每个元素命名，提高程序的可读性3、如何统计出序列中元素出现的频度4、如何根据字典中value的大小，对字典的key进行排序

 

一、数据筛选：

 处理方式：

 

1、filter函数在py3，返回的是个生成式。

from random import randint

data = [randint(-100,100) for i in range(10)]
data2 = [34, -59, -13, 96, -78, 38, 89, -96, -79, 98]

info = filter(lambda x:x>0,data2)
for i in info:
    print(i)

 

2、列表解析

from random import randint

data = [randint(-100,100) for i in range(10)]
data2 = [34, -59, -13, 96, -78, 38, 89, -96, -79, 98]

info = [i for i in data2 if i >0]
print(info)

filter与列表解析的比较：

在py2使用列表生成式效率高，在py3使用filter过滤器会效率高

from random import randint
import timeit
import pprint
data = [randint(-100,100) for i in range(10)]
data2 = [34, -59, -13, 96, -78, 38, 89, -96, -79, 98]

t1 = timeit.Timer('[i for i in [34, -59, -13, 96, -78, 38, 89, -96, -79, 98] if i >0]')
t2 = timeit.Timer('filter(lambda x:x>0,[34, -59, -13, 96, -78, 38, 89, -96, -79, 98])')
# t2 =
print(t1.timeit())
print(t2.timeit())


结果：
1.9095332647026366
0.6967581773661176

 

3、字典解析：

使用字典生成式来筛选数据

from random import randint
info = {x:randint(10,100) for x in range(1,11)}
print(info)
info2 = {key:value for key,value in info.items() if value >60}

 

 4、集合数据筛选：

结构看起来和字典生成式差不多

data = [34, -59, -13, 96, -78, 38, 89, -96, -79, 98]
data2 = set(data)
print(data2)
info = {x for x in data2 if x>0}
print(info)

 

 

 二、如何为元组中的每个元素命名，提高程序的可读性

 1、给index指定数值常量，类似C里的枚举

name,work,play,address = range(4)

people = ("Tom",35,"Teacher","swimming","shenzhen") print(people[name],people[work],people[address])

2、使用标准库中collections.namedtuple替代内置tuple，自定义一个tuple子类，这种方式开销仅仅比普通元组高一些。

from collections import namedtuple
people2 = namedtuple('people2',(['name','age','work','play','address']))
info = people2("Tom",35,"Teacher","swimming","shenzhen")
print(info)
print(info.name,info.age,info.work,info.play,info.address)

结果：
people2(name='Tom', age=35, work='Teacher', play='swimming', address='shenzhen')
Tom 35 Teacher swimming shenzhen

 

3、如何统计出序列中元素出现的频度

1、使用fromkey方法初始化一个dict，然后通过for循环迭代统计次数。

# from random import randint
#
# data = [randint(0,10) for x in range(30)]
# print(data)
data2 = [3, 0, 9, 1, 4, 1, 5, 7, 4, 7, 7, 3, 10, 4, 0, 6, 9, 2, 2, 4, 1, 1, 7, 8, 2, 7, 3, 1, 4, 9]
dict1 = dict.fromkeys(data2,0)
print(dict1)
for x in data2:
    dict1[x] += 1

print(dict1)

结果：

{0: 0, 1: 0, 2: 0, 3: 0, 4: 0, 5: 0, 6: 0, 7: 0, 8: 0, 9: 0, 10: 0}
{0: 2, 1: 5, 2: 3, 3: 3, 4: 5, 5: 1, 6: 1, 7: 5, 8: 1, 9: 3, 10: 1}

2、使用collections.Counter对象

# from random import randint
#
# data = [randint(0,10) for x in range(30)]
# print(data)
data2 = [3, 0, 9, 1, 4, 1, 5, 7, 4, 7, 7, 3, 10, 4, 0, 6, 9, 2, 2, 4, 1, 1, 7, 8, 2, 7, 3, 1, 4, 9]
from collections import Counter
dict1 = Counter(data2)
print(dict1)
print(dict1[1],dict1[4],dict1[7])
print(dict1.most_common(3))

结果：

Counter({1: 5, 4: 5, 7: 5, 2: 3, 3: 3, 9: 3, 0: 2, 5: 1, 6: 1, 8: 1, 10: 1})
5 5 5
[(1, 5), (4, 5), (7, 5)]

4、如何根据字典中value的大小，对字典的key进行排序

 

对于这种排序，一般选择用内置函数sorted，因为这些内置函数一般使用C编写，运算速度会快些

1、使用zip函数

from random import randint
dict1 = {x:randint(50,100) for x in 'abcdefg'}
print(dict1)
ret = sorted(zip(dict1.values(),dict1.keys()))
print(ret)


结果：
{'e': 77, 'd': 100, 'b': 51, 'c': 78, 'g': 55, 'f': 80, 'a': 87}
[(51, 'b'), (55, 'g'), (77, 'e'), (78, 'c'), (80, 'f'), (87, 'a'), (100, 'd')]

 2、sorted函数默认对每个迭代对象的第一个元素进行排序，可以通过指定key参数（传入一个函数，sorted每次迭代时会把选择的元素传入key中，然后让我们决定使用哪个元素作为排序对象）来排序

from random import randint
dict1 = {x:randint(50,100) for x in 'abcdefg'}
print(dict1)
print(dict1.items())
ret = sorted(dict1.items(),key=lambda x:x[1])
print(ret)

结果：
{'a': 64, 'f': 51, 'd': 67, 'e': 73, 'c': 57, 'g': 100, 'b': 71}
dict_items([('a', 64), ('f', 51), ('d', 67), ('e', 73), ('c', 57), ('g', 100), ('b', 71)])
[('f', 51), ('c', 57), ('a', 64), ('d', 67), ('b', 71), ('e', 73), ('g', 100)]