交換變數
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x = 6 y = 5 x, y = y, x print x >>> 5 print y >>> 6 |
if 語句在行內
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print "Hello" if True else "World" >>> Hello |
連線
下面的最後一種方式在繫結兩個不同型別的物件時顯得很酷。
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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
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#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
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print .3/.1 >>> 2.9999999999999996 print .3//.1 >>> 2.0 |
數值比較
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x = 2 if 3 > x > 1: print x >>> 2 if 1 < x > 0: print x >>> 2 |
兩個列表同時迭代
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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
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teams = ["Packers", "49ers", "Ravens", "Patriots"] for index, team in enumerate(teams): print index, team >>> 0 Packers >>> 1 49ers >>> 2 Ravens >>> 3 Patriots |
列表推導
已知一個列表，刷選出偶數列表方法：
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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
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numbers = [1,2,3,4,5,6] even = [number for number in numbers if number%2 == 0] |
字典推導
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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
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items = [0]*3 print items >>> [0,0,0] |
將列表轉換成字串
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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
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data = {'user': 1, 'name': 'Max', 'three': 4} try: is_admin = data['admin'] except KeyError: is_admin = False |
替換為
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data = {'user': 1, 'name': 'Max', 'three': 4} is_admin = data.get('admin', False) |
獲取子列表
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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] #最後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] |
60個字元解決FizzBuzz
前段時間Jeff Atwood 推廣了一個簡單的程式設計練習叫FizzBuzz，問題引用如下：
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寫一個程式，列印數字1到100，3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解決這個問題：
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for x in range(101):print"fizz"[x%3*4::]+"buzz"[x%5*4::]or x |
集合
用到Counter庫
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from collections import Counter print Counter("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
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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是全域性變數，因此：
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False = True if False: print "Hello" else: print "World" >>> Hello |