1. 问:嘿, 你可以在0.007kb以内的代码里转置一个矩阵吗？

Q: Hey, can you write code to transpose a matrix in 0.007KB or less?

我以为你永远也不会问这种问题。如果你把矩阵当做序列的序列的话,只要用zip就可以搞定:

I thought you'd never ask. If you represent a matrix as a sequence of sequences, then zip can do the job:

   1 >>> m = [(1,2,3), (4,5,6)]
   2 >>> zip(*m)
   3 [(1, 4), (2, 5), (3, 6)]

如果你知道 f(*m) 和 apply(f, m) 是等价的,你就可以理解上面的代码。这个提问源于一个古老的 Lisp 的问题, 答案等价于 Python 的 map(None,*m) 函数, 不过 Chih-Chung 建议的 zip 版本甚至可以做到更短。你可能会想这只不过是一个无聊的最短程序把戏. 可是有一天我遇到了类似的问题: 提供一个数据库记录列表, 每一条记录都是一个有序的值列表, 要你找出每一列出现的唯一值(无重复值)的列表, 于是我写下这行代码:

   1 possible_values = map(unique, zip(*db))

To understand this, you need to know that f(*m) is like apply(f, m). This is based on an old Lisp question, the answer to which is Python's equivalent of map(None,*m), but the zip version, suggested by Chih-Chung Chang, is even shorter. You might think this is only useful for an appearance on Letterman's Stupid Programmer's Tricks, but just the other day I was faced with this problem: given a list of database rows, where each row is a list of ordered values, find the list of unique values that appear in each column. So I wrote:

   1 possible_values = map(unique, zip(*db))

-  ⇤ ← Revision 1 as of 2006-04-22 16:08:15 → 
  Size: 2114
  Editor: HuangYi
  Comment: 有几句实在翻不出来了, 哪位兄弟有空来看下
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  Editor: localhost
  Comment: converted to 1.6 markup
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= 问:Hey, can you write code to transpose a matrix in 0.007KB or less?？ =
''Q: Hey, can you write code to transpose a matrix in 0.007KB or less??''
+::-- ZoomQuiet [<<DateTime(2005-09-06T04:10:30Z)>>] 

::-- WeiZhong [<<DateTime(2006-04-24T16:45:30Z)>>] 

= 问:嘿, 你可以在0.007kb以内的代码里转置一个矩阵吗？ =
''Q: Hey, can you write code to transpose a matrix in 0.007KB or less? ''

我以为你永远也不会问这种问题。如果你把矩阵当做序列的序列的话,只要用zip就可以搞定:
-Line 16:
+Line 19:
-{{{#!python
+{{{
#!python
-Line 22:
+Line 26:
-To understand this, you need to know that f(*m) is like apply(f, m). This is based on an old Lisp question, the answer to which is Python's equivalent of map(None,*m), but the zip version, suggested by Chih-Chung Chang, is even shorter. You might think this is only useful for an appearance on Letterman's Stupid Programmer's Tricks, but just the other day I was faced with this problem: given a list of database rows, where each row is a list of ordered values, find the list of unique values that appear in each column. So I wrote:

{{{#!python
+如果你知道 f(*m) 和 apply(f, m) 是等价的,你就可以理解上面的代码。这个提问源于一个古老的 Lisp 的问题, 答案等价于 Python 的 map(None,*m) 函数, 不过 Chih-Chung 建议的 zip 版本甚至可以做到更短。 你可能会想这只不过是一个无聊的最短程序把戏. 可是有一天我遇到了类似的问题: 提供一个数据库记录列表, 每一条记录都是一个有序的值列表, 要你找出每一列出现的唯一值(无重复值)的列表, 于是我写下这行代码:
{{{
#!python
-Line 28:
+Line 32:
-我以为你永远也不会问这种问题。如果你把矩阵当做序列的序列的话,那么zip就可以完成这个任务:
+To understand this, you need to know that f(*m) is like apply(f, m). This is based on an old Lisp question, the answer to which is Python's equivalent of map(None,*m), but the zip version, suggested by Chih-Chung Chang, is even shorter. You might think this is only useful for an appearance on Letterman's Stupid Programmer's Tricks, but just the other day I was faced with this problem: given a list of database rows, where each row is a list of ordered values, find the list of unique values that appear in each column. So I wrote:
-Line 30:
+Line 34:
-{{{#!python
>>> m = [(1,2,3), (4,5,6)]
>>> zip(*m)
[(1, 4), (2, 5), (3, 6)]
}}}

要理解这个, 你要先搞清楚 f(*m) 和 apply(f, m) 是一样的。这是基于一个古老的 Lisp 的问题, 答案就是这个和 map(None,*m) 等价的东西, 不过用 zip 的这段程序甚至更短一点。 你可能会想这个只不过对 Letterman's Stupid Programmer's Tricks 的出现有点用而已, 但是有一天, 我遇到了这个问题: 提供一个数据库行的列表, 每一行都是一个排好序了的值的列表, 要你找出这么一个列表, 他里面的值在每 (翻不出来了啊, 这一句, 哪位兄弟来帮一把啊), 于是我这么写下:

{{{#!python
+{{{
#!python

Diff for "PyIAQ/Q11"

1. 问:嘿, 你可以在0.007kb以内的代码里转置一个矩阵吗？