Python如何减少元组列表?

I am able to use map and sum to achieve this functionality, but how to use reduce?

我可以使用map和sum来实现这个功能，但是如何使用reduce呢?

There are 2 lists: a, b, they have same number of values. I want to calculate

有两个列表:a, b，它们有相同数量的值。我要计算

a[0]*b[0]+a[1]*b[1]+...+a[n]*b[n]

The working version I wrote using map is

我使用map编写的工作版本是。

value =  sum(map(lambda (x,y): x*y, zip(a, b)))

How to use reduce then? I wrote:

那么如何减少使用呢?我写:

value =  reduce(lambda (x,y): x[0]*y[0] + x[1]*y[1], zip(a, b)))

I got the error "TypeError: 'float' object is unsubscriptable".

我得到了错误的“类型错误:‘float’对象是不可订阅的”。

Can anyone shed some light on this?

有人能解释一下吗?

5 个解决方案

#1

The first argument of the lambda function is the sum so far and the second argument is the next pair of elements:

lambda函数的第一个参数是迄今为止的和，第二个参数是下一对元素:

value = reduce(lambda sum, (x, y): sum + x*y, zip(a, b), 0)

#2

I would do it this way (I don't think you need lambda)...

我这样做(我认为你不需要)……

sum(x*y for x, y in zip(a, b))

This also seems slightly more explicit. Zip AB, multiply them, and sum up the terms.

这似乎也更加明确。压缩AB，乘以它们，然后求和。

#3

A solution using reduce and map,

使用reduce和map的解决方案，

from operator import add,mul

a = [1,2,3]
b = [4,5,6]

print reduce(add,map(mul,a,b))

#4

Difficulties with reduce happen when you have incorrect map.

当你有不正确的地图时，会有减少的困难。

Let's take expression: value = sum(map(lambda (x,y): x*y, zip(a, b)))

我们取表达式:value = sum(映射(x,y): x*y, zip(a, b))

Map is transformation. We need it to convert tuples into simple flat values. In your case it will look like:

映射转换。我们需要它将元组转换成简单的平值。在你的情况下，它会是:

map(lambda x: x[0]*x[1], zip(a,b))

And then, if you want to express sum via reduce - it will look like:

然后，如果你想通过reduce来表示求和，它会是:

reduce(lambda x,y: x + y, map)

So, here is example:

这是例子:

a = [1,2,3]
b = [4,5,6] 
l = zip(a,b)
m = map(lambda x: x[0]*x[1], l)
r = reduce(lambda x,y: x + y, m)

#5

it looks like you want an inner product. use an inner product. https://docs.scipy.org/doc/numpy/reference/generated/numpy.inner.html

看起来你想要一个内积。使用一个内积。https://docs.scipy.org/doc/numpy/reference/generated/numpy.inner.html

np.inner(a, b) = sum(a[:]*b[:])

Ordinary inner product for vectors:

向量的普通内积:

a = np.array([1,2,3])
b = np.array([0,1,0])
np.inner(a, b)

output: 2

输出:2

A multidimensional example:

一个多维的例子:

a = np.arange(24).reshape((2,3,4))
b = np.arange(4)
np.inner(a, b)

output: array([[ 14, 38, 62],[ 86, 110, 134]])

输出:数组([[14,38,62]，[86,110,134])

#1