Python初学者的17个技巧
zPatrick
8年前
<pre> <code class="language-python">x = 6 y = 5 x, y = y, x print x >>> 5 print y >>> 6 </code></pre> <h3>if 语句在行内</h3> <pre> <code class="language-python">print "Hello" if True else "World" >>> Hello </code></pre> <h3>连接</h3> <p>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。</p> <pre> <code class="language-python">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 </code></pre> <h3>计算技巧</h3> <pre> <code class="language-python">#向下取整 print 5.0//2 >>> 2 # 2的5次方 print 2**5 >> 32 </code></pre> <h3>注意浮点数的除法</h3> <pre> <code class="language-python">print .3/.1 >>> 2.9999999999999996 print .3//.1 >>> 2.0 </code></pre> <h3>数值比较</h3> <pre> <code class="language-python">x = 2 if 3 > x > 1: print x >>> 2 if 1 < x > 0: print x >>> 2 </code></pre> <h3>两个列表同时迭代</h3> <pre> <code class="language-python">nfc = ["Packers", "49ers"] afc = ["Ravens", "Patriots"] for teama, teambin zip(nfc, afc): printteama + " vs. " + teamb >>> Packersvs. Ravens >>> 49ers vs. Patriots </code></pre> <h3>带索引的列表迭代</h3> <pre> <code class="language-python">teams = ["Packers", "49ers", "Ravens", "Patriots"] for index, teamin enumerate(teams): printindex, team >>> 0 Packers >>> 1 49ers >>> 2 Ravens >>> 3 Patriots </code></pre> <h3>列表推导</h3> <p>已知一个列表，刷选出偶数列表方法：</p> <pre> <code class="language-python">numbers = [1,2,3,4,5,6] even = [] for numberin numbers: if number%2 == 0: even.append(number) </code></pre> <p>用下面的代替</p> <pre> <code class="language-python">numbers = [1,2,3,4,5,6] even = [numberfor numberin numbersif number%2 == 0] </code></pre> <h3>字典推导</h3> <pre> <code class="language-python">teams = ["Packers", "49ers", "Ravens", "Patriots"] print {key: valuefor value, keyin enumerate(teams)} >>> {'49ers': 1, 'Ravens': 2, 'Patriots': 3, 'Packers': 0} </code></pre> <h3>初始化列表的值</h3> <pre> <code class="language-python">items = [0]*3 printitems >>> [0,0,0] </code></pre> <h3>将列表转换成字符串</h3> <pre> <code class="language-python">teams = ["Packers", "49ers", "Ravens", "Patriots"] print ", ".join(teams) >>> 'Packers, 49ers, Ravens, Patriots' </code></pre> <h3>从字典中获取元素</h3> <p>不要用下列的方式</p> <pre> <code class="language-python">data = {'user': 1, 'name': 'Max', 'three': 4} try: is_admin = data['admin'] exceptKeyError: is_admin = False </code></pre> <p>替换为</p> <pre> <code class="language-python">data = {'user': 1, 'name': 'Max', 'three': 4} is_admin = data.get('admin', False) </code></pre> <h3>获取子列表</h3> <pre> <code class="language-python">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] </code></pre> <h3>60个字符解决FizzBuzz</h3> <p>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：</p> <pre> <code class="language-python">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”。 </code></pre> <p>这里有一个简短的方法解决这个问题：</p> <pre> <code class="language-python">for x in range(101):print"fizz"[x%3*4::]+"buzz"[x%5*4::]or x </code></pre> <h3>集合</h3> <p>用到Counter库</p> <pre> <code class="language-python">fromcollectionsimportCounter printCounter("hello") >>> Counter({'l': 2, 'h': 1, 'e': 1, 'o': 1}) </code></pre> <h3>迭代工具</h3> <p>和collections库一样，还有一个库叫itertools</p> <pre> <code class="language-python">fromitertoolsimportcombinations teams = ["Packers", "49ers", "Ravens", "Patriots"] for gamein combinations(teams, 2): printgame >>> ('Packers', '49ers') >>> ('Packers', 'Ravens') >>> ('Packers', 'Patriots') >>> ('49ers', 'Ravens') >>> ('49ers', 'Patriots') >>> ('Ravens', 'Patriots') </code></pre> <h3>False == True</h3> <p>在python中，True和False是全局变量，因此：</p> <pre> <code class="language-python">False = True if False: print "Hello" else: print "World" >>> Hello </code></pre> <p> </p> <p>来自：http://python.jobbole.com/85770/</p> <p> </p>