Reputation: 187
Essentially I'm trying to use the brentq method row-wise on a pandas dataframe to get the root of a function that takes as arguments columns and constants as well.
Something like the following
import pandas as pd
import numpy as np
from scipy.optimize import brentq
np.random.seed(0)
df = pd.DataFrame(np.random.rand(10, 3), columns=list('ABC'))
CONST = 0.5
def fcn(x, y, k):
return x**2 + y - k
def objective_function(x, y, k, delta):
return fcn(x, y, k) - delta
def get_root(x, k, delta, a=-10.0, b=10.0, xtol=1e-6):
# avoid mirroring outer scope
_x, _k, _delta = x, k, delta
# nested function that takes the target param as the input
def nfcn(y):
# get y that makes fcn and delta equal
return objective_function(_x, y, _k, _delta)
result = brentq(nfcn, a=a, b=b, xtol=xtol)
return result
I got this to work with apply and lambda function, on rows
df['D'] = df.apply(lambda x: get_root(x['A'], CONST, x['C'], a=-10., b=10.), axis=1)
but when the dataframe is really big is very slow, as expected.
Any ideas on how to vectorize this?
Many thanks
Upvotes: 0
Views: 448
Reputation: 109546
Perhaps you can use a partial. I get a speed-up of almost 4x for a dataframe of 100k rows.
from functools import partial
# Rearrange parameters (swap `delta` and constant `k`).
def get_root(x, delta, k, a=-10.0, b=10.0, xtol=1e-6):
...
p = partial(get_root, k=CONST, a=-10., b=10.)
df['D'] = [p(*vals) for vals in df[['A', 'C']].values]
Upvotes: 1