Python高效编程

2018-01-20 本篇共537字 | 阅读需要3分钟

本篇记录汇总实际编程和面试中都可能会遇到的典型问题。

在列表,字典,集合中根据条件筛选数据

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from random import randint
from timeit import timeit
​
data = [randint(-10,10) for _ in range(10)]
filter(lambda x:x >= 0,data)
#首选列表解析 时间比filter少一半左右
[x for x in data if x >= 0]
​
data = {x:randint(60,100) for x in range(1,21)}
{k:v for k,v in data.items() if v > 90}
​
data = set(data)
{x for x in data if x%3==0}

为元组/列表中每个元素命名,提高程序的可读性

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#第一种
NAME,AGE,SEX,EMAIL = range(4)
student = ('Jan',14,'male','jan@jan.net')
print(student[NAME])
#第二种
from collections import namedtuple
Student = namedtuple('Student',['name','age','sex','email'])
student= Student('Jan',14,'male','jan@jan.net')
print(student[NAME])
student.age
isinstance(student,tuple)

统计序列中元素的出现频度

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from random import randint
data = [randint(1,10) for x in range(20)]
c = dict.fromkeys(data,0)
#第一种
for x in data:
    c[x]+=1
#第二种
from collections import Counter
c2 = Counter(data)
#出现频度最高的三个
c2.most_common(3)
#统计一篇文章中单词出现次数
import re
#导入Python之禅
import this
c3 = Counter(re.split('\W+',this.s))

根据字典中值的大小,对字典中的项排序

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from random import randint
data = {x:randint(60,100) for x in 'abcdefg'}
#sorted(data.values())
#第一种
t = zip(data.values(),data.keys())
sorted(t)
#第二种
sorted(data.items(),key=lambda x: x[1])

快速找到多个字典中的公共键(key)

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from random import randint,sample
s1 = {x: randint(1,4) for x in sample('abcdefgh',randint(3,6))}
s2 = {x: randint(1,4) for x in sample('abcdefgh',randint(3,6))}
s3 = {x: randint(1,4) for x in sample('abcdefgh',randint(3,6))}
#第一种
res = []
for k in s1:
    if k in s2 and k in s3:
        res.append(k)
#第二种
from functools import reduce
s1.keys()&s2.keys()&s3.keys()
map(dict.keys,[s1,s2,s3])
reduce(lambda a, b:a & b,map(dict.keys,[s1,s2,s3]))

实现用户的历史记录功能(最多n条)

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from collections import deque
from random import randint
N = randint(0,100)
## 一个容纳5个值的队列
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:
    input_number = input('please input a number: ')
    if input_number.isdigit():
        k = int(input_number)
        history.append(k)
        if guess(k):
            break
    elif input_number == 'history':
        print(list(history))

未完待续…

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