计算加权和 [英] computing a weighted sum

问题描述

假设我有一个n浮点数列表x和一个n浮点数列表w我想要

来计算x [0] * w [0] + .. + x [n-1 ] * w [n-1]。


是否有一些优雅的表达式（可能使用lambda）在一个语句中完成

？如：

y = lambda x，w：...


我问，因为我现在正在这样做：

y = 0

for i in range（0，n）：y + = x [i] * w [i]


似乎没有非常pythonic :)


谢谢，

Andrei

解决方案

< blockquote> an*****@mail.dntis.ro 写道：

假设我有一个n浮点数列表x和一个n浮点数列表w我想要
来计算x [0] * w [0] + .. + x [n-1] * w [n-1] 。

是否有一些优雅的表达（也许使用lambda）在一个声明中完成它？如下：
y = lambda x，w：...

我问，因为我现在这样做的方式：
y = 0
我在范围（0，n）：y + = x [i] * w [i]

看起来不是非常pythonic：）

我'我会抓住它！


在Python 2.3中

sum（[_x * _w表示_x，_w in zip（x， w）]）


或2.4

总和（_x * _w代表_x，_w代表zip（x，w））


如果列表很大，您可能需要使用itertools.izip代替zip。


将McGugan

写道：

假设我有一个n浮点数列表x和一个n浮点数列表w我想要
来计算x [0] * w [0] + .. + x [n-1] * w [n-1]。

是否有一些优雅的表达（也许使用lambda）来完成它
一个声明？如：
y = lambda x，w：...

x = [1,2,3]
w = [4,5,6]
sum（a * b为（a，b）in zip（x，w））

32

谢谢Will，2.4表达式看起来非常好。

Suppose I have a list of n floats x and a list of n floats w and I want
to compute x[0]*w[0] + .. + x[n-1]*w[n-1].

Is there some elegant expression (perhaps using lambda) to have it done
in one statement ? As in :
y = lambda x,w : ...

I ask because the way I am doing it now :
y = 0
for i in range(0,n): y += x[i]*w[i]

doesn''t seem very pythonic :)

Thanks,
Andrei

解决方案

an*****@mail.dntis.ro wrote:

Suppose I have a list of n floats x and a list of n floats w and I want
to compute x[0]*w[0] + .. + x[n-1]*w[n-1].

Is there some elegant expression (perhaps using lambda) to have it done
in one statement ? As in :
y = lambda x,w : ...

I ask because the way I am doing it now :
y = 0
for i in range(0,n): y += x[i]*w[i]

doesn''t seem very pythonic :)

I''ll take a stab at that!

In Python 2.3

sum( [ _x * _w for _x, _w in zip( x, w ) ] )

or in 2.4

sum( _x * _w for _x, _w in zip( x, w ) )

You may want to use itertools.izip in place of zip if the lists are large.

Will McGugan

wrote:

Suppose I have a list of n floats x and a list of n floats w and I want
to compute x[0]*w[0] + .. + x[n-1]*w[n-1].

Is there some elegant expression (perhaps using lambda) to have it done
in one statement ? As in :
y = lambda x,w : ...

x = [1, 2, 3]
w = [4, 5, 6]
sum(a*b for (a,b) in zip(x, w))

32

Thanks Will, the 2.4 expression looks really nice.

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