Python初学者的17个技巧
程序员文章站
2022-05-23 14:26:53
...
交换变量
x = 6 y = 5 x, y = y, x print x >>> 5 print y >>> 6
if 语句在行内
print "Hello" if True else "World" >>> Hello
连接
下面的最后一种方式在绑定两个不同类型的对象时显得很酷。
nfc = ["Packers", "49ers"] afc = ["Ravens", "Patriots"] print nfc + afc >>> ['Packers', '49ers', 'Ravens', 'Patriots'] print str(1) + " world" >>> 1 world print `1` + " world" >>> 1 world print 1, "world" >>> 1 world print nfc, 1 >>> ['Packers', '49ers'] 1
计算技巧
#向下取整 print 5.0//2 >>> 2 # 2的5次方 print 2**5 >> 32
注意浮点数的除法
print .3/.1 >>> 2.9999999999999996 print .3//.1 >>> 2.0
数值比较
x = 2 if 3 > x > 1: print x >>> 2 if 1 0: print x >>> 2
两个列表同时迭代
nfc = ["Packers", "49ers"] afc = ["Ravens", "Patriots"] for teama, teamb in zip(nfc, afc): print teama + " vs. " + teamb >>> Packers vs. Ravens >>> 49ers vs. Patriots
带索引的列表迭代
teams = ["Packers", "49ers", "Ravens", "Patriots"] for index, team in enumerate(teams): print index, team >>> 0 Packers >>> 1 49ers >>> 2 Ravens >>> 3 Patriots
列表推导
已知一个列表,刷选出偶数列表方法:
numbers = [1,2,3,4,5,6] even = [] for number in numbers: if number%2 == 0: even.append(number)
用下面的代替
numbers = [1,2,3,4,5,6] even = [number for number in numbers if number%2 == 0]
字典推导
teams = ["Packers", "49ers", "Ravens", "Patriots"] print {key: value for value, key in enumerate(teams)} >>> {'49ers': 1, 'Ravens': 2, 'Patriots': 3, 'Packers': 0}
初始化列表的值
items = [0]*3 print items >>> [0,0,0]
将列表转换成字符串
teams = ["Packers", "49ers", "Ravens", "Patriots"] print ", ".join(teams) >>> 'Packers, 49ers, Ravens, Patriots'
从字典中获取元素
不要用下列的方式
data = {'user': 1, 'name': 'Max', 'three': 4} try: is_admin = data['admin'] except KeyError: is_admin = False
替换为
data = {'user': 1, 'name': 'Max', 'three': 4} is_admin = data.get('admin', False)
获取子列表
x = [1,2,3,4,5,6] #前3个 print x[:3] >>> [1,2,3] #中间4个 print x[1:5] >>> [2,3,4,5] #最后3个 print x[-3:] >>> [4,5,6] #奇数项 print x[::2] >>> [1,3,5] #偶数项 print x[1::2] >>> [2,4,6]
60个字符解决FizzBuzz
前段时间Jeff Atwood 推广了一个简单的编程练习叫FizzBuzz,问题引用如下:
写一个程序,打印数字1到100,3的倍数打印“Fizz”来替换这个数,5的倍数打印“Buzz”,对于既是3的倍数又是5的倍数的数字打印“FizzBuzz”。
这里有一个简短的方法解决这个问题:
for x in range(101):print"fizz"[x%3*4::]+"buzz"[x%5*4::]or x
集合
用到Counter库
from collections import Counter print Counter("hello") >>> Counter({'l': 2, 'h': 1, 'e': 1, 'o': 1})
迭代工具
和collections库一样,还有一个库叫itertools
from itertools import combinations teams = ["Packers", "49ers", "Ravens", "Patriots"] for game in combinations(teams, 2): print game >>> ('Packers', '49ers') >>> ('Packers', 'Ravens') >>> ('Packers', 'Patriots') >>> ('49ers', 'Ravens') >>> ('49ers', 'Patriots') >>> ('Ravens', 'Patriots') False == True
在python中,True和False是全局变量,因此:
False = True if False: print "Hello" else: print "World" >>> Hello
上一篇: php中会话机制相关问题