英文原文：http://www.maxburstein.com/blog/python-shortcuts-for-the-python-beginner/
转贴原文：http://segmentfault.com/a/1190000002506498#articleHeader16
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 < x > 0:
print x
>>> 2
两个列表同时迭代
zip函数所有条件都满足时才会进行循环，任意个列表遍历完都会跳出循环
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