Support for log-transforms (and other arithmetic) in formula?

[martiningram]

Hi all,

I'm a big fan of bambi, thanks for all your great work!

I have a question about more complicated formulas. I'm currently trying to fit models taken from a textbook. One of the formula strings used there is:

log(weight) ~ log(I(diam1 * diam2 * canopy_height)) + log(I(diam1 * diam2)) + log(I(diam1 / diam2)) + log(total_height) + group

In other words, it includes log transformations, and also some other arithmetic operations. What I'm curious about is whether bambi supports this in some way, or whether all such transformations should be done beforehand, e.g. by creating a variable log_total_height = np.log(total_height)?

Thanks for your help!

[tomicapretto]

Hi @martiningram

Bambi has support for those.

• You can pass Python functions as your model terms.
• You can wrap mathematical operations between terms within I() or using the shorthand {}.

For example, the formula you're sharing would be

import numpy as np
formula = "np.log(weight) ~ np.log(diam1 * diam2 * canopy_height) + np.log(diam1 * diam2) + log(diam1 / diam2) + np.log(total_height) + group"

Note that I'm not using I() because the operations are carried within the arguments of a function, so they are interpreted as regular Python code, and not as model formula code.

If you want a term that is the result of doing some math operation between variables, you do need I() or `{}.

For example

formula = "y ~ I(x * z)"
formula = "y ~ I(x / z)" # 1
formula = "y ~ {x / z}"  # 2
formula = "y ~ {10 * x}"

where 1 and 2 are equivalent representations for the ratio between x and z

Update

You can also use custom functions (not only functions imported from NumPy or any other module)

def f(x):
    return (x - np.median(x)) / 10
formula = "y ~ f(x)"

is also valid

[martiningram]

Wow, this is working a treat, thanks Tomas! I was even able to use the formula as-is by defining

def log(x):
    return np.log(x)

as you mentioned in the Update.

Am I right in assuming that keeping the I is OK, even though it's actually not required as you say?

[tomicapretto]

@martiningram that's right. I() is literally an identity function. It returns whatever you pass in.

[martiningram]

Terrific, thanks Tomas! This is working very well, so I am closing this issue!

[aflaxman]

You can also perhaps save the annoyance of defining your own log by importing it explicitly instead with
from numpy import log --- very cool stuff!