Python将等长的元组相乘 [英] Python Multiply tuples of equal length

问题描述

我希望能找到一种优雅或有效的方法来乘以整数序列(或浮点数).

I was hoping for an elegant or effective way to multiply sequences of integers (or floats).

我的第一个想法是尝试(1、2、3)*(1、2、2)会产生(1、4、6)，即产品各个乘法的数量.

My first thought was to try (1, 2, 3) * (1, 2, 2) would result (1, 4, 6), the products of the individual multiplications.

尽管python尚未预先设置为执行序列操作.很好，我真的不希望如此.那么，将两个系列中的每个项目与各自的索引相乘并相乘的Python方法是什么?

Though python isn't preset to do that for sequences. Which is fine, I wouldn't really expect it to. So what's the pythonic way to multiply (or possibly other arithmetic operations as well) each item in two series with and to their respective indices?

第二个示例(0.6，3.5)*(4，4) = (2.4，14)

推荐答案

最简单的方法是使用 zip 函数，带有生成器表达式，像这样

The simplest way is to use zip function, with a generator expression, like this

tuple(l * r for l, r in zip(left, right))

例如，

>>> tuple(l * r for l, r in zip((1, 2, 3), (1, 2, 3)))
(1, 4, 9)
>>> tuple(l * r for l, r in zip((0.6, 3.5), (4, 4)))
(2.4, 14.0)

在Python 2.x中， zip 返回一个元组列表.如果要避免创建临时列表，可以使用 itertools.izip ，就像这样

In Python 2.x, zip returns a list of tuples. If you want to avoid creating the temporary list, you can use itertools.izip, like this

>>> from itertools import izip
>>> tuple(l * r for l, r in izip((1, 2, 3), (1, 2, 3)))
(1, 4, 9)
>>> tuple(l * r for l, r in izip((0.6, 3.5), (4, 4)))
(2.4, 14.0)

您可以在zip 和 itertools.izip 之间的区别.>这个问题.

You can read more about the differences between zip and itertools.izip in this question.

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