并行算术？ [英] Parallel arithmetic?
问题描述
给定a和b，两个相等长度的整数列表，我希望c为
[a1-b1，a2-b2，...，an-bn]。我可以这样做：
c = [0] * len（a）
为ndx，项目为枚举（a）：
c [ndx] = item - b [ndx]
但我想知道是否有更好的方法，或许可以避免循环？
尼克。
（我似乎记得从我遥远的过去，这种事情已经死了
轻松使用APL .. .c = ab，或多或少。）
N
" Terrance N. Phillip" ; <我************* @ hotmail.com>写道：给定a和b，两个相等长度的整数列表，我希望c为
[a1-b1，a2-b2，...，an-bn]。
c = [a [i] - b [i] for x in xrange（len（a））]
< blockquote> Terrance N. Phillip写道：
给定a和b，两个相等长度的整数列表，我希望c为
[a1-b1，a2-b2，...，一个-BN。我可以这样做：
c = [0] * len（a）
对于ndx，枚举项目（a）：
c [ndx] = item - b [ndx]
但我想知道是否有更好的方法，或许可以避免循环？
这里''一种方式：
c = [a_item - b_item为a_item，b_item为zip（a，b）]
和另一个：
导入运算符
c = map（operator.sub，a，b）
-
迈克尔霍夫曼
有很多方法可以做到这一点。它们都没有避免循环，技术上是
，尽管你可以很容易地避免使用for。语法。
- 简单但浪费一些记忆
c = [ij for i，j in zip（a，b）]
- 使用itertools.izip（python 2.3）
c = [ij for i，j in itertools.izip（a，b）]
- 生成器表达式（python 2.4）
c =（ij for i，j in itertools.izip（a，b））
Given a and b, two equal length lists of integers, I want c to be
[a1-b1, a2-b2, ... , an-bn]. I can do something like:
c = [0] * len(a)
for ndx, item in enumerate(a):
c[ndx] = item - b[ndx]
But I''m wondering if there''s a better way, perhaps that avoids a loop?
Nick.
(I seem to recall from my distant past that this sort of thing was dead
easy with APL... c = a-b, more or less.)
N
"Terrance N. Phillip" <me*************@hotmail.com> writes:Given a and b, two equal length lists of integers, I want c to be
[a1-b1, a2-b2, ... , an-bn].
c = [a[i] - b[i] for i in xrange(len(a))]
Terrance N. Phillip wrote:Given a and b, two equal length lists of integers, I want c to be
[a1-b1, a2-b2, ... , an-bn]. I can do something like:
c = [0] * len(a)
for ndx, item in enumerate(a):
c[ndx] = item - b[ndx]
But I''m wondering if there''s a better way, perhaps that avoids a loop?
Here''s one way:
c = [a_item - b_item for a_item, b_item in zip(a, b)]
And another:
import operator
c = map(operator.sub, a, b)
--
Michael Hoffman
There are many ways to do this. None of them avoids looping,
technically, although you can easily avoid the "for" syntax.
-- Simple but wastes some memory
c = [i-j for i,j in zip(a,b)]
-- Using itertools.izip (python 2.3)
c = [i-j for i,j in itertools.izip(a,b) ]
-- Generator expression (python 2.4)
c = ( i-j for i,j in itertools.izip(a,b) )
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