交換變數
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
下面的最後一種方式在綁定兩個不同類型的物件時顯得很酷。
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
計算技巧
#向下取整 print 5.0//2 >>> 2 # 2的5次方 print 2**5 >> 32
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, teamb in zip(nfc, afc): print teama + " vs. " + teamb >>> Packers vs. Ravens >>> 49ers vs. Patriots
帶索引的列表迭代列表,c列表方法:
teams = ["Packers", "49ers", "Ravens", "Patriots"] for index, team in enumerate(teams): print index, team >>> 0 Packers >>> 1 49ers >>> 2 Ravens >>> 3 Patriots
用下面的代替
numbers = [1,2,3,4,5,6] even = [] for number in numbers: if number%2 == 0: even.append(number)
字典推導
numbers = [1,2,3,4,5,6] even = [number for number in numbers if number%2 == 0]
初始化列表的值
teams = ["Packers", "49ers", "Ravens", "Patriots"] print {key: value for value, key in enumerate(teams)} >>> {'49ers': 1, 'Ravens': 2, 'Patriots': 3, 'Packers': 0}
將列表轉換成字串
items = [0]*3 print items >>> [0,0,0]
teams = ["Packers", "49ers", "Ravens", "Patriots"] print ", ".join(teams) >>> 'Packers, 49ers, Ravens, Patriots'
data = {'user': 1, 'name': 'Max', 'three': 4} try: is_admin = data['admin'] except KeyError: is_admin = False
寫一個程序,打印數字3的倍數列印“Fizz”來替換這個數,5的倍數列印“Buzz”,對於既是3的倍數又是5的倍數的數字列印“FizzBuzz”。
這裡有一個簡短的方法來解決這個問題:
data = {'user': 1, 'name': 'Max', 'three': 4} is_admin = data.get('admin', False)
集合
用到Counter庫
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]
是全域變量,因此:
for x in range(101):print"fizz"[x%3*4::]+"buzz"[x%5*4::]or x