x = 6
y = 5
 
x, y = y, x
 
print x
>>> 5
print y
>>> 6
 

if 語句在行內

print "Hello" if True else "World"
>>> Hello
 

連接

下面的最后一種方式在綁定兩個不同類型的對象時顯得很酷。

nfc = ["Packers", "49ers"]
afc = ["Ravens", "Patriots"]
printnfc + afc
>>> ['Packers', '49ers', 'Ravens', 'Patriots']
 
printstr(1) + " world"
>>> 1 world
 
print `1` + " world"
>>> 1 world
 
print 1, "world"
>>> 1 world
printnfc, 1
>>> ['Packers', '49ers'] 1
 

計算技巧

#向下取整
print 5.0//2
>>> 2

2的5次方
print 2*5
>> 32
 
</code></pre> 
  
注意浮點數的除法
 
  
print .3/.1
>>> 2.9999999999999996
print .3//.1
>>> 2.0
 
 
  
數值比較
 
  
x = 2
if 3 > x > 1:
  print x
>>> 2
if 1 < x > 0:
  print x
>>> 2
 
 
  
兩個列表同時迭代
 
  
nfc = ["Packers", "49ers"]
afc = ["Ravens", "Patriots"]
for teama, teambin zip(nfc, afc):
    printteama + " vs. " + teamb
>>> Packersvs. Ravens
>>> 49ers vs. Patriots
 
 
  
帶索引的列表迭代
 
  
teams = ["Packers", "49ers", "Ravens", "Patriots"]
for index, teamin enumerate(teams):
    printindex, team
>>> 0 Packers
>>> 1 49ers
>>> 2 Ravens
>>> 3 Patriots
 
 
  
列表推導
 
  
已知一個列表，刷選出偶數列表方法：
 
  
numbers = [1,2,3,4,5,6]
even = []
for numberin numbers:
    if number%2 == 0:
        even.append(number)
 
 
  
用下面的代替
 
  
numbers = [1,2,3,4,5,6]
even = [numberfor numberin numbersif number%2 == 0]
 
 
  
字典推導
 
  
teams = ["Packers", "49ers", "Ravens", "Patriots"]
print {key: valuefor value, keyin enumerate(teams)}
>>> {'49ers': 1, 'Ravens': 2, 'Patriots': 3, 'Packers': 0}
 
 
  
初始化列表的值
 
  
items = [0]3
printitems
>>> [0,0,0]
 
</code></pre> 
  
將列表轉換成字符串
 
  
teams = ["Packers", "49ers", "Ravens", "Patriots"]
print ", ".join(teams)
>>> 'Packers, 49ers, Ravens, Patriots'
 
 
  
從字典中獲取元素
 
  
不要用下列的方式
 
  
data = {'user': 1, 'name': 'Max', 'three': 4}
try:
  is_admin = data['admin']
exceptKeyError:
  is_admin = False
 
 
  
替換為
 
  
data = {'user': 1, 'name': 'Max', 'three': 4}
is_admin = data.get('admin', False)
 
 
  
獲取子列表
 
  
x = [1,2,3,4,5,6]

前3個
print x[:3]
>>> [1,2,3]
中間4個
print x[1:5]
>>> [2,3,4,5]
最后3個
print x[-3:]
>>> [4,5,6]
奇數項
print x[::2]
>>> [1,3,5]
偶數項
print x[1::2]
>>> [2,4,6]
 
</code></pre> 
  
60個字符解決FizzBuzz
 
  
前段時間Jeff Atwood 推廣了一個簡單的編程練習叫FizzBuzz，問題引用如下：
 
  
寫一個程序，打印數字1到100，3的倍數打印“Fizz”來替換這個數，5的倍數打印“Buzz”，對于既是3的倍數又是5的倍數的數字打印“FizzBuzz”。
 
 
  
這里有一個簡短的方法解決這個問題：
 
  
for x in range(101):print"fizz"[x%3*4::]+"buzz"[x%5*4::]or x
 
 
  
集合
 
  
用到Counter庫
 
  
fromcollectionsimportCounter
printCounter("hello")
>>> Counter({'l': 2, 'h': 1, 'e': 1, 'o': 1})
 
 
  
迭代工具
 
  
和collections庫一樣，還有一個庫叫itertools
 
  
fromitertoolsimportcombinations
teams = ["Packers", "49ers", "Ravens", "Patriots"]
for gamein combinations(teams, 2):
    printgame
>>> ('Packers', '49ers')
>>> ('Packers', 'Ravens')
>>> ('Packers', 'Patriots')
>>> ('49ers', 'Ravens')
>>> ('49ers', 'Patriots')
>>> ('Ravens', 'Patriots')
 
 
  
False == True
 
  
在python中，True和False是全局變量，因此：
 
  
False = True
if False:
  print "Hello"
else:
  print "World"
>>> Hello
 
 
  
 
 
  
來自：http://python.jobbole.com/85770/