列表，元组，字点，集合，队列

1：在字点、列表、集合中根据条件筛选数据

     过滤掉列表 data1 = [1,5,-3,-2,9,0,6] 中的负数，

     筛选出字典 data2 = {'LIlei':'79','Jim':'88','Lucy':'92'} 中值高于90的项，

     筛出集合 data3 = {77,89,32,20，153}中能背3整除的元素。

问题一的三种解法（列表解析式最快，filter其次，第一种最慢）

result = []
for i in data1:
    if i>=0:
        result.append(i)
print(result)

result = filter(lambda x:x>=0,data)#python2中返回列表，python3中返回可迭代对象
for x in result:
    print(x,end=',')

from random import randint
data =[randint(-10,10) for _ in range(10)]
a = [x for x in data if x>=0]
print(a)

问题二的解法

from random import randint
data ={x : randint(60,100) for x in range(1,11)}
print(data)

a = {k:v  for k,v in data.items() if v >= 90}
print(a)

问题三的解法

from random import randint
data =[randint(10,30) for _ in range(10)]
print(data)

a = set(data)
print(a)

b = {x for x in a if x % 3 == 0}
print(b)

2 ：如何为元组中的每个元素命名，提高程序可读性（枚举法定义变量作为索引坐标，）

student = ('jim',18,'male','[email protected]')

name,age,sex,email = range(4)
print(student[name])
print(student[age])
print(student[sex])
print(student[email])

from collections import namedtuple

student = namedtuple('student',['name','age','sex','email'])

s = student('jim',16,'male','[email protected]')

print(s,s.name ,s.age,s.sex,s.email)

3：如何统计序列中元素的出现频度

        随机序列[12，5,5,6,4,6,7,5...]中，找出出现次数最高的3个元素，它们出现次数是多少

        某英文文章的单词，进行词频统计，找出出现次数最高的10个单词，它们出现次数是多少

    问题一  

import random
data = [random.randint(0,10) for _ in range(50)]
print(data)

d = dict.fromkeys(data,0)
#print(d)
for x in data:
    d[x] += 1

print(d)

c = sorted(d.items(),key=lambda x:x[1])  #lambda 是匿名函数，key是sorted（min，max都可以）的参数，告诉它按找第几个元素来排序
print(c)

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

from collections import Counter

data = [9, 3, 8, 3, 10, 3, 1, 2, 4, 10, 2, 0, 4, 8, 1, 10, 8, 6, 7, 1, 2, 0, 6, 2, 9, 6, 2, 6, 4, 8, 2, 7, 4, 0, 3, 9, 0, 10, 4, 2, 0, 1, 9, 0, 2, 1, 9, 6, 4, 7]

c = Counter(data)
print(c)

#x.most_common(n) x是Counter()对象，此方法得到频度最高的n个元素
print(c.most_common(3))

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

[(2, 8), (4, 6), (0, 6)]

问题二

from collections import Counter
import re

txt = 'When people ask me what kind of job I want to work on in the future, my answer is definitely being a teacher. I can pass my knowledge to the students and make contribution to the society, what a great job. I also like to share new things with others, so I am so eager to search the world and collect the valuable information. In order to realize my dream, I must study hard. No matter what kind of difficulties I meet, I will never give up. The successful people’ experiences show me the importance of insistence, and I know I can make it some day.'

txt_list = re.split('\W+',txt)
print(txt_list)

d = Counter(txt_list)
print(d)

print(d.most_common(5))


['When', 'people', 'ask', 'me', 'what', 'kind', 'of', 'job', 'I', 'want', 'to', 'work', 'on', 'in', 'the', 'future', 'my', 'answer', 'is', 'definitely', 'being', 'a', 'teacher', 'I', 'can', 'pass', 'my', 'knowledge', 'to', 'the', 'students', 'and', 'make', 'contribution', 'to', 'the', 'society', 'what', 'a', 'great', 'job', 'I', 'also', 'like', 'to', 'share', 'new', 'things', 'with', 'others', 'so', 'I', 'am', 'so', 'eager', 'to', 'search', 'the', 'world', 'and', 'collect', 'the', 'valuable', 'information', 'In', 'order', 'to', 'realize', 'my', 'dream', 'I', 'must', 'study', 'hard', 'No', 'matter', 'what', 'kind', 'of', 'difficulties', 'I', 'meet', 'I', 'will', 'never', 'give', 'up', 'The', 'successful', 'people', 'experiences', 'show', 'me', 'the', 'importance', 'of', 'insistence', 'and', 'I', 'know', 'I', 'can', 'make', 'it', 'some', 'day', '']
Counter({'I': 9, 'to': 6, 'the': 6, 'what': 3, 'of': 3, 'my': 3, 'and': 3, 'people': 2, 'me': 2, 'kind': 2, 'job': 2, 'a': 2, 'can': 2, 'make': 2, 'so': 2, 'When': 1, 'ask': 1, 'want': 1, 'work': 1, 'on': 1, 'in': 1, 'future': 1, 'answer': 1, 'is': 1, 'definitely': 1, 'being': 1, 'teacher': 1, 'pass': 1, 'knowledge': 1, 'students': 1, 'contribution': 1, 'society': 1, 'great': 1, 'also': 1, 'like': 1, 'share': 1, 'new': 1, 'things': 1, 'with': 1, 'others': 1, 'am': 1, 'eager': 1, 'search': 1, 'world': 1, 'collect': 1, 'valuable': 1, 'information': 1, 'In': 1, 'order': 1, 'realize': 1, 'dream': 1, 'must': 1, 'study': 1, 'hard': 1, 'No': 1, 'matter': 1, 'difficulties': 1, 'meet': 1, 'will': 1, 'never': 1, 'give': 1, 'up': 1, 'The': 1, 'successful': 1, 'experiences': 1, 'show': 1, 'importance': 1, 'insistence': 1, 'know': 1, 'it': 1, 'some': 1, 'day': 1, '': 1})
[('I', 9), ('to', 6), ('the', 6), ('what', 3), ('of', 3)]

4：如何根据字点中值的大小，对字典进行排序

from random import randint

data = {x:randint(60,100) for x in "abcdef"} # 构造随机成绩字典

# 方式一
print(sorted(zip(data.values(),data.keys()))) # 利用元组的比较


# 方式二
print(sorted(data.items(),key=lambda x:x[1])) # 利用key参数

5：如何找到多个字典中的公共键

方法一执行效率不算好，稍微啰嗦

import random

a = random.sample('abcdefg',random.randint(3,6)) #进球的球员，sapmle是取样，有两个参数
print(a)

#第s1,s2,s3轮：进球的人和进球数(随机1-4)
s1 = {x:random.randint(1,4) for x in random.sample('abcdefg',random.randint(3,6))}
s2 = {x:random.randint(1,4) for x in random.sample('abcdefg',random.randint(3,6))}
s3 = {x:random.randint(1,4) for x in random.sample('abcdefg',random.randint(3,6))}

print(s1,'\n',s2,'\n',s3)

for k in s1:
    if k  in s2 and k in s3:
        print(k,end=',')



['d', 'c', 'b', 'e']
{'g': 1, 'f': 4, 'd': 3, 'c': 3} 
 {'d': 2, 'a': 2, 'g': 3, 'f': 2, 'c': 4, 'b': 2} 
 {'e': 3, 'a': 3, 'f': 2, 'g': 4, 'd': 4}
g,f,d,

方法二

a = s1.keys()     #python3是dict.keys()方法
b = s2.keys()
c = s3.keys()
f = a & b & c    #集合取交集，这是前三轮的，如果是前n轮，则如下所示

print(f)

map(function, iterable, ...) 函数会根据提供的函数对指定序列做映射。

第一个参数 function 以参数序列中的每一个元素调用function 函数，返回包含每次 function 函数返回值的新列表。

# 计算平方数
def square(x) :            
   return x ** 2
map(square, [1,2,3,4,5])   

# 使用 lambda 匿名函数
map(lambda x: x ** 2, [1, 2, 3, 4, 5])  
# [1, 4, 9, 16, 25]

# 提供了两个列表，对相同位置的列表数据进行相加
map(lambda x, y: x + y, [1, 3, 5, 7, 9], [2, 4, 6, 8, 10])
# [3, 7, 11, 15, 19]

reduce(function, iterable) 函数会对参数序列中元素进行累积。
函数将一个数据集合（链表，元组等）中的所有数据进行下列操作：用传给 reduce中的函数 function（有两个参数）先对集合中的第 1、2 个元素进行操作，得到的结果再与第三个数据用 function 函数运算，最后得到一个结果。

# 求和
reduce(lambda x, y: x+y, [1,2,3,4,5]) 
# 15

from random import randint, sample
from functools import reduce
# 随机产生abcdefg球员中的3-6个人随机进1-4个球
s1 = {x: randint(1, 4) for x in sample("abcdefg", randint(3, 6))}
s2 = {x: randint(1, 4) for x in sample("abcdefg", randint(3, 6))}
s3 = {x: randint(1, 4) for x in sample("abcdefg", randint(3, 6))}

res = reduce(lambda a, b: a & b, map(dict.keys, [s1, s2, s3]))
print(res)

6：如何让字典保持有序

from collections import OrderedDict

d = OrderedDict()
d['Jim'] = (1, 35)
d['Leo'] = (2, 37)
d['Bob'] = (3, 40)
for k in d:
    print(k)

OrderedDict([('Jim', (1, 35)), ('Leo', (2, 37)), ('Bob', (3, 40))])
Jim
Leo
Bob

import time
from random import randint
from collections import OrderedDict

d = OrderedDict()

players = list('ABCDEFGH')
start = time.time()# 考试开始时间

for i in range(8):#在8个人中
    input()# 等待输入表示每输入一次就有一个人比赛做完题目
    person = players.pop(randint(0,7-i))#随机选择完成的人，从列表中删除
    end =time.time()#完成的人的时间

    print(i+1,person,end-start) #打印 名词，姓名，时间
    d[person] = (i+1,end-start) #按 姓名:(名次,时间)传入OrderDict()的字典对象中

for k in d:
    print(k,d[k])


1 B 1.9241104125976562

2 E 3.323190450668335

3 C 4.003229141235352

4 H 4.50725793838501

5 D 5.011286735534668

6 A 5.531316518783569

7 F 6.428367853164673

8 G 7.011401176452637
B (1, 1.9241104125976562)
E (2, 3.323190450668335)
C (3, 4.003229141235352)
H (4, 4.50725793838501)
D (5, 5.011286735534668)
A (6, 5.531316518783569)
F (7, 6.428367853164673)
G (8, 7.011401176452637)

7：如何实现用户的历史记录功能（队列，pickle.dump(obj,file)）

from random import randint

N = randint(0, 100)

def guess(k):
    if k == N:
        print 'right'
        return True
    if k < N:
        print '%s is less-than N' % k
    else:
        print '%s is greater-than N' % k
    return False

while True:
    line = raw_input("please input a number: ")
    # 判断用户输入的数据是否为数字
    if line.isdigit():
        k = int(line)
        if guess(k):
            break

用户猜数字，一遍猜中的概论是很低很低的，所以用户猜数字要猜很多很多遍才可能猜中。在猜数字的过程中，由于用户猜的数字次数很多，以至于用户会忘记猜过的数字。因此，我们是有必要添加历史记录功能的。这里为了简化操作，只显示用户最近输入的5个数字。

那如何实现这一功能呢？我们可以使用容量为n的队列存储历史记录：

• 使用标准库collections中的deque，它是一个双端循环队列
• 程序退出前，可以使用pickle将队列对象存入文件，再次运行程序时将其导入

import pickle

from random import randint
from collections import deque

N = randint(0, 100)

history = deque([], 5)

def guess(k):
    if k == N:
        print 'right'
        return True
    if k < N:
        print '%s is less-than N' % k
    else:
        print '%s is greater-than N' % k
    return False

while True:
    try:
        # 导入用户猜数字的历史记录
        history = pickle.load(open('history'))
        line = raw_input("please input a number: ")
    except:
        line = raw_input("please input a number: ")
    # 判断用户输入的数据是否为数字
    if line.isdigit():
        k = int(line)
        history.append(k)
        # 将用户历史记录保存至history文件中
        pickle.dump(history, open('history', 'w'))
        if guess(k):
            break
    # 用户查看历史记录
    elif line == 'history' or line == 'his?':
        print list(history)