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<title>Beginner_Of_Python - Mike Tech</title>
<meta name="author" content="Mike Gao">
<meta name="description" content="变量交换 x = 6
y = 5 x, y = y, x
print x >>> 5
print y
>>> 6 if 语句再行内 print "Hello" if True else "World"
>>> Hello 连接 …">
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<article class="hentry" role="article">
<header>
<h1 class="entry-title">Beginner_Of_Python</h1>
<p class="meta">
<time datetime="2013-01-29T10:42:00+08:00" pubdate data-updated="true">Jan 29<span>th</span>, 2013</time>
</p>
</header>
<div class="entry-content"><h4>变量交换</h4>
<pre><code>x = 6
y = 5
x, y = y, x
print x
>>> 5
print y
>>> 6
</code></pre>
<h4>if 语句再行内</h4>
<pre><code>print "Hello" if True else "World"
>>> Hello
</code></pre>
<h4>连接</h4>
<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cool</p>
<pre><code>nfc = ["Packers", "49ers"]
afc = ["Ravens", "Patriots"]
print nfc + afc
>>> ['Packers', '49ers', 'Ravens', 'Patriots']
print str(1) + " world"
>>> 1 world
pring 1, "world"
>>> 1 world
print nfc, 1
>>> ['Packers', '49ers'] 1
</code></pre>
<h4>数字技巧</h4>
<pre><code># 除后向下取整
print 5.0//2
>>> 2
# 2的5次方
print 2**5
>>> 32
</code></pre>
<h4>注意浮点数的除法</h4>
<pre><code>print .3/.1
>>> 2.9999999999999996
print .3//.1
>>> 2.0
</code></pre>
<h4>数值比较</h4>
<pre><code>x = 2
if 3 > x > 1:
print x
>>> 2
if 1 < x > 0:
print x
>>> 2
</code></pre>
<h4>同时迭代两个列表</h4>
<pre><code>nfc = ['Packers', '49ers']
afc = ['Ravens', 'Patriots']
for teama, teamb in zip(nfc, afc)
print teama + " vs. " + teamb
>>> Packers vs. Ravens
>>> 49ers vs. Patriots
</code></pre>
<h4>带索引的列表迭代</h4>
<pre><code>teams = ["Packers", "49ers", "Ravens", "Patriots"]
for index, team in enumerate(teams):
print index, team
>>> 0 Packers
>>> 1 49ers
>>> 2 Ravens
>>> 3 Patriots
</code></pre>
<h4>列表推导式</h4>
<p> 已知一个列表，我们可以刷选出偶数列表方法</p>
<pre><code>numbers = [1,2,3,4,5]
even = [number for number in numbers if number%2 == 0]
</code></pre>
<h4>字典推导</h4>
<p> 和列表推导类似</p>
<pre><code>teams = ['Packers', '49ers', 'Ravens', 'Patriots']
print {key: value for value, key in enumerate(teams)}
>>> {'49ers': 1, 'Ravens': 2, 'Patriots': 3, 'Packers': 0}
</code></pre>
<h4>初始化列表的值</h4>
<pre><code>items = [0]*3
print items
>>> [0, 0, 0]
</code></pre>
<h4>列表转换为字符串</h4>
<pre><code>teams = ['Packers', '49ers', 'Ravens', 'Patriots']
print ", ".join(teams)
>>> 'Packers, 49ers, Ravens, Patriots'
</code></pre>
<h4>从字典中获取元素</h4>
<pre><code>data = {'user': 1, 'name': 'Max', 'three': 4}
try:
is_admin = data['admin']
except KeyError:
is_admin = False
</code></pre>
<p> 替换成这样：</p>
<pre><code>data = {'user': 1, 'name': 'Max', 'three': 4}
is_admin = data.get('admin', False)
</code></pre>
<h4>获取列表的子集</h4>
<pre><code>x= [1,2,3,4,5]
# 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>
<h4>60个字符解决FizzBuzz</h4>
<p> 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”</p>
<pre><code>for x in range(101):print'fizz'[x%3*4::]+'buzz'[x%5*4::]or x
</code></pre>
<h4>集合</h4>
<pre><code>from collections import Counter
print Counter("hello")
>>> Counter({'1': 2, 'h': 1, 'e': 1, 'o': 1})
</code></pre>
<h4>迭代工具</h4>
<p> 和collections库一样，还有一个库叫itertools，对某些问题真能高效地解决。其中一个用例是查找所有组合，他能告诉你在一个组中元素的所有不能的组合方式</p>
<pre><code>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')
</code></pre>
<h4>False == True</h4>
<pre><code>False = True
if False:
print "Hello"
else:
print "World"
>>> Hello
</code></pre>
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<span class="byline author vcard">Posted by <span class="fn">Mike Gao</span></span>
<time datetime="2013-01-29T10:42:00+08:00" pubdate data-updated="true">Jan 29<span>th</span>, 2013</time>
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