#35 non obvious possibilities of python syntax 鲜为人知的python语法 (#56)

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+# Non-obvious possibilities of python syntax
+# 鲜为人知的 python 语法
+
+鲜为人知的python语法 [阅读原文 »](https://medium.com/@chipiga86/non-obvious-possibilities-of-python-syntax-a95a7210aaaf)
+
+
+Hi! All of us (ok not all :)) know [python](https://www.python.org/) as wide-used, simple-to-read, easy-to-start programming language.   
+
+所有人（好吧，不是所有人）都知道 [python](https://www.python.org/) 是一门用途广泛、易读、而且容易入门的编程语言。  
+
+But at the same time python syntax allows to make really strange things.    
+但同时 python 语法也允许我们做一些很奇怪的事情。    
+
+## Rewrite multiline function with lambda  
+## 使用 lambda 表达式重写多行函数  
+
+As it’s known python lambda doesn’t support multiline code. But it can be simulated.  
+众所周知 python 的 lambda 表达式不支持多行代码。但是可以模拟出多行代码的效果。  
+
+```python
+def f():
+    x = 'string'
+    if x.endswith('g'):
+        x = x[:-1]
+    r = ''
+    for i in xrange(len(x)):
+        if x[i] != 'i':
+            r += x[i]
+    return r
+f()
+-> 'strn'
+```
+
+Sounds strange but above function can be replaced with **lambda**-function:   
+虽然看起来很奇怪，但是上面的函数可以使用下面的 **lambda** 表达式函数代替：   
+
+```python
+(lambda: ([x for x in ['string']], x.endswith('g') and [x for x in [x[:-1]]], [r for r in ['']], [x[i] != 'i' and [r for r in [r+x[i]]] for i in xrange(len(x))], r)[-1])()
+-> 'strn'
+```  
+
+Never do that in production code :)    
+永远不要在生产环境写这样的代码 :)  
+
+## Ternary operator  
+## 三元运算符 
+
+Modern python gives you simple intuitive syntax:   
+现代的 python 提供了更简便的语法:   
+
+```python
+b if a else c   
+```   
+
+
+But it also can be rewritten via:   
+也可以通过下面的方式重写：  
+
+```python
+(a and [b] or [c])[0]
+
+(b, c)[not a]
+```  
+
+Btw, next variant is incorrect:    
+顺便说一下，下面的变体是错误的：   
+
+```python
+a and b or c
+True and [] or [1] -> [1], but: [] if True else [1] -> []
+``` 
+
+## Remove repeated symbols via list comprehension  
+## 通过列表推导式移除重复的元素 
+
+Let’s transform string `x = 'tteesstt'` to `'test'`.   
+让我们来把字符串  `x = 'tteesstt'` 转换成  `'test'` 吧。
+
+1. compare a symbol with previous in original string:   
+2. 在原字符串中和上一个字符比较：
+
+```python
+''.join(['' if i and j == x[i-1] else j for i,j in enumerate(x)])
+```   
+
+1.  save previous symbol in temporary variable: 
+2.  把前一个字符保存到临时变量中：
+
+```python
+''.join([('' if i == a else i, [a for a in [i]])[0] for a in [''] for i in x])
+''.join([('' if i == a.pop() else i, a.append(i))[0] for a in [['']] for i in x])
+```
+
+3. compare a symbol with previous in new string:    
+3. 在新字符串中和上一个字符比较：
+
+```python
+[(not r.endswith(i) and [r for r in [r+i]], r)[-1] for r in [''] for i in x][-1]
+```  
+
+4.  via reduce & lambda:     
+4. 通过 reduce 函数和 lambda 表达式：
+
+```python
+reduce(lambda a, b: a if a.endswith(b) else a + b, x)
+```
+
+## Fibonacci via list comprehension  
+## 通过列表推导式获得斐波拉契数列  
+
+1. save temporary values in list:    
+1. 把中间值保存在列表中 
+
+```python
+[(lambda: (l[-1], l.append(l[-1] + l[-2]))[0])() for l in [[1, 1]] for x in xrange(19)]
+[(l[-1], l.append(l[-1] + l[-2]))[0] for l in [[1, 1]] for x in xrange(19)]
+```   
+
+2. save temporary values in dict:   
+2. 把中间值保存到字典中: 
+
+```python
+[i for x in [(lambda: (l['a'], l.update({'a': l['a'] + l['b']}), l['b'], l.update({'b': l['a'] + l['b']}))[::2])() for l in [{'a': 1, 'b': 1}] for x in xrange(10)] for i in x]
+[i for x in [(l['a'], l.update({'a': l['a'] + l['b']}), l['b'], l.update({'b': l['a'] + l['b']}))[::2] for l in [{'a': 1, 'b': 1}] for x in xrange(10)] for i in x]
+```  
+
+3. via reduce & lambda:  
+3. 通过 reduce 函数和 lambda 表达式：
+
+```python
+reduce(lambda a, b: a + [a[-1] + a[-2]], xrange(10), [1, 1])
+reduce(lambda a, b: a.append(a[-1] + a[-2]) or a, xrange(10), [1, 1])
+``` 
+
+4. the quickest variant:   
+4. 速度最快的变体：  
+
+```python
+[l.append(l[-1] + l[-2]) or l for l in [[1, 1]] for x in xrange(10)][0]
+```
+
+## Eternal cycle with list comprehension  
+## 使用列表推导式产生死循环
+`[a.append(b) for a in [[None]] for b in a]`  
+
+## List slice tricks   
+## 列表切片技巧 
+
+1. copy list:   
+1. 复制列表： 
+
+```python
+l = [1, 2, 3]
+m = l[:]
+m
+-> [1, 2, 3]
+```
+
+2. remove/replace any number of elements:  
+2. 移除/替换 列表中的任意元素：
+
+```python
+l = [1, 2, 3]
+l[1:-1] = [4, 5, 6, 7]
+l
+-> [1, 4, 5, 6, 7, 3]
+```
+
+3. add elements to begin of list:   
+3. 在列表的开头添加元素：
+
+```python
+l = [1, 2, 3]
+l[:0] = [4, 5, 6]
+l
+-> [4, 5, 6, 1, 2, 3]
+```
+
+4. add elements to end of list:
+4. 在列表的尾部添加元素： 
+
+```python
+l = [1, 2, 3]
+l[-1:] = [l[-1], 4, 5, 6]
+l
+-> [1, 2, 3, 4, 5, 6]
+```
+
+5. reverse list:  
+5. 反转列表：
+
+```python
+l = [1, 2, 3]
+l[:] = l[::-1]
+``` 
+
+## Replace method byte code  
+## 替换方法字节码
+
+Python prohibits to replace instance method, making it as read-only property:   
+Python 阻止替换类实例中的方法，因为 python 给类实例中的方法赋予了只读属性：   
+
+```python
+class A(object):
+    def x(self):
+        print "hello"
+a = A()
+def y(self):
+    print "world"
+a.x.im_func = y
+-> TypeError: readonly attribute
+``` 
+
+But it can be replaced on byte-code level:  
+但是可以在字节码的层面上进行替换：   
+
+```python
+a.x.im_func.func_code = y.func_code
+a.x()
+-> 'world'
+```   
+
+**Note!** It has influence not only to current instance but to class (if to be precise to function which is bound with class) and all other instances too:   
+**注意！** 这不仅对当前的实例有影响，而且对整个类都有影响（准确的说是与这个类绑定的函数）（译者注:此处应该是笔误，推测作者原意是:准确的说是与这个函数绑定的所有类），并且所有其他的实例也会受到影响：   
+
+```python
+new_a = A()
+new_a.x()
+-> 'world'
+```
+
+## Mutable object as default function argument  
+## 让可变元素作为函数参数默认值   
+
+To assign mutable object as default value to function argument is very dangerous and there are a lot of tricky questions on interviews about that. But it can be helpful as cache mechanism.   
+把可变对象作为函数参数的默认值是非常危险的一件事，并且在面试中有大量关于这方面棘手的面试问题。但这一点对于缓存机制非常有帮助。
+
+1. Factorial: 
+2. 阶乘函数：
+
+```python
+def f(n, c={}):
+    if n in c:
+        return c[n]
+    if (n < 2):
+        r = 1
+    else:
+        r = n * f(n - 1)
+    c[n] = r
+    return r
+f(10)
+-> 3628800
+f.func_defaults
+({1: 1,
+  2: 2,
+  3: 6,
+  4: 24,
+  5: 120,
+  6: 720,
+  7: 5040,
+  8: 40320,
+  9: 362880,
+  10: 3628800},)
+```
+
+
+2. Fibonacci: 
+2. 斐波拉契数列：   
+
+```python
+def fib(n, c={}):
+    if n in c:
+        return c[n]
+    if (n < 2):
+        r = 1
+    else:
+        r = fib(n - 2) + fib(n - 1)
+    c[n] = r
+    return r
+fib(10)
+-> 89
+fib.func_defaults[0].values()
+-> [1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89]
+```