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#35 non obvious possibilities of python syntax 鲜为人知的python语法 #56
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translation/Non-obvious-possibilities-of-python-syntax-鲜为人知的python语法.md
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# Non-obvious possibilities of python syntax | ||
# 鲜为人知的 python 语法 | ||
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鲜为人知的python语法 [阅读原文 »](https://medium.com/@chipiga86/non-obvious-possibilities-of-python-syntax-a95a7210aaaf) | ||
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Hi! All of us (ok not all :)) know [python](https://www.python.org/) as wide-used, simple-to-read, easy-to-start programming language. | ||
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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。 | ||
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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。 | ||
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## Rewrite multiline function with lambda | ||
## 使用 lambda 表达式重写多行函数 | ||
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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。 | ||
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``` | ||
There was a problem hiding this comment. Choose a reason for hiding this commentThe reason will be displayed to describe this comment to others. Learn more. 代码块后面加上语言类型说明 |
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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' | ||
``` | ||
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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: | ||
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``` | ||
(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' | ||
``` | ||
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Never do that in production code :) | ||
永远不要在生产环境写这样的代码 :) | ||
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## Ternary operator | ||
## 三元运算符 | ||
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Modern python gives you simple intuitive syntax: | ||
现代的 python 提供了更简便的语法: | ||
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``` | ||
b if a else c | ||
``` | ||
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But it also can be rewritten via: | ||
也可以通过下面的方式重写: | ||
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``` | ||
(a and [b] or [c])[0] | ||
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(b, c)[not a] | ||
``` | ||
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Btw, next variant is incorrect: | ||
顺便说一下,下面的变体是错误的: | ||
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``` | ||
a and b or c | ||
True and [] or [1] -> [1], but: [] if True else [1] -> [] | ||
``` | ||
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## Remove repeated symbols via list comprehension | ||
## 通过列表推导式移除重复的元素 | ||
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Let’s transform string `x = 'tteesstt'` to `'test'`. | ||
让我们来把字符串 `x = 'tteesstt'` 转换成 `'test'` 吧。 | ||
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1. compare a symbol with previous in original string: | ||
1. 在原字符串中和上一个字符比较: | ||
`''.join(['' if i and j == x[i-1] else j for i,j in enumerate(x)])` | ||
2. save previous symbol in temporary variable: | ||
2. 把前一个字符保存到临时变量中: | ||
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``` | ||
''.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]) | ||
``` | ||
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3. compare a symbol with previous in new string: | ||
3. 在新字符串中和上一个字符比较: | ||
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``` | ||
[(not r.endswith(i) and [r for r in [r+i]], r)[-1] for r in [''] for i in x][-1] | ||
``` | ||
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4. via reduce & lambda: | ||
4. 通过 reduce 函数和 lambda 表达式: | ||
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``` | ||
reduce(lambda a, b: a if a.endswith(b) else a + b, x) | ||
``` | ||
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## Fibonacci via list comprehension | ||
## 通过列表推导式获得斐波拉契数列 | ||
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1. save temporary values in list: | ||
1. 把中间值保存在列表中 | ||
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``` | ||
[(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)] | ||
``` | ||
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2. save temporary values in dict: | ||
2. 把中间值保存到字典中: | ||
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``` | ||
[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] | ||
``` | ||
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3. via reduce & lambda: | ||
3. 通过 reduce 函数和 lambda 表达式: | ||
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``` | ||
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]) | ||
``` | ||
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4. the quickest variant: | ||
4. 速度最快的变体: | ||
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``` | ||
[l.append(l[-1] + l[-2]) or l for l in [[1, 1]] for x in xrange(10)][0] | ||
``` | ||
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## 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]` | ||
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## List slice tricks | ||
## 列表切片技巧 | ||
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1. copy list: | ||
1. 复制列表: | ||
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``` | ||
l = [1, 2, 3] | ||
m = l[:] | ||
m | ||
-> [1, 2, 3] | ||
``` | ||
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2. remove/replace any number of elements: | ||
2. 移除/替换 列表中的任意元素: | ||
``` | ||
l = [1, 2, 3] | ||
l[1:-1] = [4, 5, 6, 7] | ||
l | ||
-> [1, 4, 5, 6, 7, 3] | ||
``` | ||
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3. add elements to begin of list: | ||
3. 在列表的开头添加元素: | ||
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``` | ||
l = [1, 2, 3] | ||
l[:0] = [4, 5, 6] | ||
l | ||
-> [4, 5, 6, 1, 2, 3] | ||
``` | ||
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4. add elements to end of list: | ||
4. 在列表的尾部添加元素: | ||
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``` | ||
l = [1, 2, 3] | ||
l[-1:] = [l[-1], 4, 5, 6] | ||
l | ||
-> [1, 2, 3, 4, 5, 6] | ||
``` | ||
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5. reverse list: | ||
5. 反转列表: | ||
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``` | ||
l = [1, 2, 3] | ||
l[:] = l[::-1] | ||
``` | ||
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## Replace method byte code | ||
## 替换方法字节码 | ||
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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: | ||
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``` | ||
class A(object): | ||
def x(self): | ||
print "hello" | ||
a = A() | ||
def y(self): | ||
print "world" | ||
a.x.im_func = y | ||
-> TypeError: readonly attribute | ||
``` | ||
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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: | ||
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``` | ||
a.x.im_func.func_code = y.func_code | ||
a.x() | ||
-> 'world' | ||
``` | ||
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**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: | ||
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``` | ||
new_a = A() | ||
new_a.x() | ||
-> 'world' | ||
``` | ||
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## Mutable object as default function argument | ||
## 让可变元素作为函数参数默认值 | ||
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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。 | ||
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1. Factorial: | ||
2. 阶乘函数: | ||
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``` | ||
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},) | ||
``` | ||
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2. Fibonacci: | ||
2. 斐波拉契数列: | ||
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``` | ||
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] | ||
``` |
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顿号是不是好一些
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ok搞定