麻痹的倒数 [英] Reciprocals in patsy

问题描述

Patsy的幂不允许负整数，因此，如果我们有一些序列数据X，

Patsy's power doesn't allow for negative integers, so, if we have some series data X,

patsy.dmatrices('X + X**(-1)', X)

返回错误.我该如何将X的倒数添加到这样的小数公式中?

returns an error. How would I add the reciprocal of X to such a patsy formula?

推荐答案

在嵌入式函数调用中关闭了运算符的特殊含义；因此，如果您编写X + 1 / x，则patsy会将其解释为特殊的patsy +和/运算符，但是如果您编写类似X + sin(1 / X)的内容，则patsy会继续将+解释为特殊的patsy运算符，但是整个sin(1 / X)表达式将传递给Python进行评估，Python会将/评估为常规除法.

The special patsy meaning of operators gets switched off inside embedded function calls; so if you write X + 1 / x then patsy interprets that as the special patsy + and / operators, but if you write something like X + sin(1 / X), then patsy continues to interpret the + as a special patsy operator, but the whole sin(1 / X) expression gets passed to Python to evaluate, and Python will evaluate the / as regular division.

因此，如果我们要计算sin(1 / X)，那就很好.但是我们不(为什么?).我们只想要普通的1 / X.那我们该怎么办呢?

So that's fine if we wanted to compute sin(1 / X). But we don't (why would we?). We just want plain 1 / X. So how can we do that?

好吧，我们可能会很棘手:我们需要一个函数调用来欺骗patsy的解析器以忽略/并将其提供给Python-但没有任何东西表明该函数必须要做 .我们可以定义一个识别函数:

Well, we can be tricky: we need a function call to trick patsy's parser into ignoring the / and giving it to Python -- but there's nothing that says that function has to do anything. We could just define an identify function:

def identity(value):
    return value

，然后在类似X + identity(1 / X)的公式中使用它.

and then use that in a formula like X + identity(1 / X).

实际上，这个技巧非常方便，patsy已经为您预定义了一个功能，并将其作为

And in fact, this trick is so handy that patsy has already predefined an function for you, and provides it as a built-in called I(...). Generally, you can think of I(...) as a kind of quoting operator -- it's a way to say "hey patsy, please do not try to interpret anything in this region, just pass it through to Python kthx".

因此，请回答您的原始问题:尝试编写dmatrix("X + I(1 / X)", data)

So to answer your original question: try writing dmatrix("X + I(1 / X)", data)

(下一个问题:为什么使用功能I和所有这些奇怪的技巧?答案是，这就是30年前R的工作方式，我想不到有什么更好的东西值得值得打破了兼容性.)

(Next question: why this weird hack with the function I and everything? The answer to that is that this is how R did it 30 years ago, and I couldn't think of anything sufficiently better to be worth breaking compatibility.)

这篇关于麻痹的倒数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！