I have a simple question, I run a linear model using below
model1 <- lm(output ~. , data=df)
I get some output, I want to extract linear equation directly instead of seeing output in raw form
I know broom package what I do best is
library(broom)
tidy(mod1)
I want output in form
output=1.1*var1 + 1.2*var2+....
You should have a look at equatiomatic package (https://github.com/datalorax/equatiomatic)
Below code should solve your problem, I am using iris as an example
remotes::install_github("datalorax/equatiomatic")
library(equatiomatic)
mod1 <- lm(Petal.Width~.-Species, data= iris)
extract_eq(mod1, use_coefs = TRUE, wrap=TRUE,fix_signs = FALSE,ital_vars = TRUE,terms_per_line = 4)
Related
I am estimating a model using the pyMC3 library in python. In my "real" model, there are four parameter arrays, two of which have over 170,000 parameters in them. Summarising this array of parameters is too computationally intensive on my computer. I have been trying to figure out if the summary function in arviz will allow me to only summarise one (or a small number) of parameters in the array. Below is a reprex where the same problem is present, though the model is a lot simpler. In the linear regression model below, the parameter array b has three parameters in it b[0], b[1], b[2]. I would like to know how to get the summary for just b[0] and b[1] or alternatively for just a single parameter, e.g., b[0].
import pandas as pd
import pymc3 as pm
import arviz as az
d = pd.read_csv("https://quantoid.net/files/mtcars.csv")
mpg = d['mpg'].values
hp = d['hp'].values
weight = d['wt'].values
with pm.Model() as model:
b = pm.Normal("b", mu=0, sigma=10, shape=3)
sig = pm.HalfCauchy("sig", beta=2)
mu = pm.Deterministic('mu', b[0] + b[1]*hp + b[2]*weight)
like = pm.Normal('like', mu=mu, sigma=sig, observed=mpg)
fit = pm.fit(10000, method='advi')
samp = fit.sample(1500)
with model:
smry = az.summary(samp, var_names = ["b"])
It looked like the coords argument to the summary() function would do it, but after googling around and finding a few examples, like the one here with plot_posterior() instead of summary(), I was unable to get something to work. In particular, I tried the following in the hopes that it would return the summary for b[0] and b[1].
with model:
smry = az.summary(samp, var_names = ["b"], coords={"b_dim_0": range(1)})
or this to return the summary of b[0]:
with model:
smry = az.summary(samp, var_names = ["b"], coords={"b_dim_0": [0]})
I suspect I am missing something simple (I'm an R user who dabbles occasionally with Python). Any help is greatly appreciated.
(BTW, I am using Python 3.8.0, pyMC3 3.9.3, arviz 0.10.0)
To use coords for this, you need to update to the development (which will still show 0.11.2 but has the code from github or any >0.11.2 release) version of ArviZ. Until 0.11.2, the coords argument in summary was not used to subset the data (like it did in all plotting functions) but instead it was only taken into account if the input was not already InferenceData in which case it was passed to the converter.
With older versions, you need to use xarray to subset the data before passing it to summary. Therefore you need to explicitly convert the trace to inferencedata beforehand. In the example above it would look like:
with model:
...
samp = fit.sample(1500)
idata = az.from_pymc3(samp)
az.summary(idata.posterior[["b"]].sel({"b_dim_0": [0]}))
Moreover, you may also want to indicate summary to compute only a subset of the stats/diagnostics as shown in the docstring examples.
I tried a time series forecast with Python using statsmodel's arima function and it gave me a different result from the r's arima function.
I used the same hyper-parameters.
R's version :
fit <- arima(data[1:9000,3], order = c(3,0,3), seasonal = list(order = c(0,0,0)))
predd = forecast(fit,h=1000)
pred = cbind(data[9001:10000,3], predd$mean)
Python's version :
series = df[0:9000].copy()
model = ARIMA(series, order=(3, 0, 3))
model_fitted = model.fit()
predictions = model_fitted.predict(start=len(series), end=len(df)-1)
Attached are the plots results Plots of the R's and Python's arima
What am I doing wrong?
Is there any other Python package/function arima that I can use other than statsmodel for a univariate time series?
Any insight or guidance would be greatly appreciated. Thank you so much in advance.
Summary: I do not know how you created the first image you showed as "R's version", but when I run the R code you gave and plot the results, they look identical to the Python results to me and do not look like the "R's version" graph you included. My best guess is that somehow you were plotting in-sample predictions when you created that image showing R's results.
See below for details.
Details:
I started by downloading the dataset "dataset.txt" from the link you gave, https://gist.github.com/DouddaS/5043a340ff7d7b35b255b4f8f74fc534
Now, if I run the following R code:
library(forecast)
y <- read.csv('dataset.txt')
fit <- arima(y[1:9000, 1], order = c(3,0,3), seasonal = list(order = c(0,0,0)))
predd = forecast(fit,h=1000)
pred = cbind(y[9001:10000,1], predd$mean)
autoplot(pred)
This gives the following plot:
And when I run the following Python code:
y = pd.read_csv('dataset.txt')
model = sm.tsa.arima.ARIMA(y.iloc[:9000, 0], order=(3, 0, 3))
model_fitted = model.fit()
pred = model_fitted.predict(start=len(series), end=len(y)-1)
predd = pd.concat([y.iloc[9000:, 0], pred], axis=1)
predd.plot()
Then I get the following plot:
These look basically identical to me, and R's version looks nothing like the image that was posted in the question.
I have fitted a logisitic regression model to some data, everything is working great. I need to calculate the wald statistic which is a function of the model result.
My problem is that I do not understand, from the documentation, what the wald test requires as input? Specifically what is the R matrix and how is it generated?
I tried simply inputting the data I used to train and test the model as the R matrix, but I do not think this is correct. The documentation suggest examining the examples however none give an example of this test. I also asked the same question on crossvalidated but got shot down.
Kind regards
http://statsmodels.sourceforge.net/0.6.0/generated/statsmodels.discrete.discrete_model.LogitResults.wald_test.html#statsmodels.discrete.discrete_model.LogitResults.wald_test
The Wald test is used to test if a predictor is significant or not, of the form:
W = (beta_hat - beta_0) / SE(beta_hat) ~ N(0,1)
So somehow you'll want to input the predictors into the test. Judging from the example of the t.test and f.test, it may be simpler to input a string or tuple to indicate what you are testing.
Here is their example using a string for the f.test:
from statsmodels.datasets import longley
from statsmodels.formula.api import ols
dta = longley.load_pandas().data
formula = 'TOTEMP ~ GNPDEFL + GNP + UNEMP + ARMED + POP + YEAR'
results = ols(formula, dta).fit()
hypotheses = '(GNPDEFL = GNP), (UNEMP = 2), (YEAR/1829 = 1)'
f_test = results.f_test(hypotheses)
print(f_test)
And here is their example using a tuple:
import numpy as np
import statsmodels.api as sm
data = sm.datasets.longley.load()
data.exog = sm.add_constant(data.exog)
results = sm.OLS(data.endog, data.exog).fit()
r = np.zeros_like(results.params)
r[5:] = [1,-1]
T_test = results.t_test(r)
If you're still struggling getting the wald test to work, include your code and I can try to help make it work.
I'm trying to convert the following code from R to Python using the Statsmodels module:
model <- glm(goals ~ att + def + home - (1), data=df, family=poisson, weights=weight)
I've got a similar dataframe (named df) using pandas, and currently have the following line in Python (version 3.4 if it makes a difference):
model = sm.Poisson.from_formula("goals ~ att + def + home - 1", df).fit()
Or, using GLM:
smf.glm("goals ~ att + def + home - 1", df, family=sm.families.Poisson()).fit()
However, I can't get the weighting terms to work. Each record in the dataframe has a date, and I want more recent records to be more valuable for fitting the model than older ones. I've not seen an example of it being used, but surely if it can be done in R, it can be done on Statsmodels... right?
freq_weights is now supported on GLM Poisson, but unfortunately not on sm.Poisson
To use it, pass freq_weights when creating the GLM:
import statsmodels.api as sm
import statsmodels.formula.api as smf
formula = "goals ~ att + def + home - 1"
smf.glm(formula, df, family=sm.families.Poisson(), freq_weights=df['freq_weight']).fit()
I've encountered the same issue.
there is a workaround that should lead to same results. add the weight in logarithm scale (np.log(weight)) you need as one of the explanatory variables with beta equal to 1 (offset option).
I can see there is an option for the exposure which doing the same as I explained above.
There are two solutions for setting up weights for Poisson regression. The first is to use freq_weigths in the GLM function as mentioned by MarkWPiper. The second is to just go with Poisson regression and pass the weights to exposure. As documented here: "Log(exposure) is added to the linear prediction with coefficient equal to 1." This does the same mathematical trick as mentioned by Yaron, although the parameter has a different original meaning. A sample code is as follows:
import statsmodels.api as sm
# or: from statsmodels.discrete.discrete_model import Poisson
fitted = sm.Poisson.from_formula("goals ~ att + def + home - 1", data=df, exposure=df['weight']).fit()
I am using LaasoCV from sklearn to select the best model is selected by cross-validation. I found that the cross validation gives different result if I use sklearn or matlab statistical toolbox.
I used matlab and replicate the example given in
http://www.mathworks.se/help/stats/lasso-and-elastic-net.html
to get a figure like this
Then I saved the matlab data, and tried to replicate the figure with laaso_path from sklearn, I got
Although there are some similarity between these two figures, there are also certain differences. As far as I understand parameter lambda in matlab and alpha in sklearn are same, however in this figure it seems that there are some differences. Can somebody point out which is the correct one or am I missing something? Further the coefficient obtained are also different (which is my main concern).
Matlab Code:
rng(3,'twister') % for reproducibility
X = zeros(200,5);
for ii = 1:5
X(:,ii) = exprnd(ii,200,1);
end
r = [0;2;0;-3;0];
Y = X*r + randn(200,1)*.1;
save randomData.mat % To be used in python code
[b fitinfo] = lasso(X,Y,'cv',10);
lassoPlot(b,fitinfo,'plottype','lambda','xscale','log');
disp('Lambda with min MSE')
fitinfo.LambdaMinMSE
disp('Lambda with 1SE')
fitinfo.Lambda1SE
disp('Quality of Fit')
lambdaindex = fitinfo.Index1SE;
fitinfo.MSE(lambdaindex)
disp('Number of non zero predictos')
fitinfo.DF(lambdaindex)
disp('Coefficient of fit at that lambda')
b(:,lambdaindex)
Python Code:
import scipy.io
import numpy as np
import pylab as pl
from sklearn.linear_model import lasso_path, LassoCV
data=scipy.io.loadmat('randomData.mat')
X=data['X']
Y=data['Y'].flatten()
model = LassoCV(cv=10,max_iter=1000).fit(X, Y)
print 'alpha', model.alpha_
print 'coef', model.coef_
eps = 1e-2 # the smaller it is the longer is the path
models = lasso_path(X, Y, eps=eps)
alphas_lasso = np.array([model.alpha for model in models])
coefs_lasso = np.array([model.coef_ for model in models])
pl.figure(1)
ax = pl.gca()
ax.set_color_cycle(2 * ['b', 'r', 'g', 'c', 'k'])
l1 = pl.semilogx(alphas_lasso,coefs_lasso)
pl.gca().invert_xaxis()
pl.xlabel('alpha')
pl.show()
I do not have matlab but be careful that the value obtained with the cross--validation can be unstable. This is because it influenced by the way you subdivide the samples.
Even if you run 2 times the cross-validation in python you can obtain 2 different results.
consider this example :
kf=sklearn.cross_validation.KFold(len(y),n_folds=10,shuffle=True)
cv=sklearn.linear_model.LassoCV(cv=kf,normalize=True).fit(x,y)
print cv.alpha_
kf=sklearn.cross_validation.KFold(len(y),n_folds=10,shuffle=True)
cv=sklearn.linear_model.LassoCV(cv=kf,normalize=True).fit(x,y)
print cv.alpha_
0.00645093258722
0.00691712356467
it's possible that alpha = lambda / n_samples
where n_samples = X.shape[0] in scikit-learn
another remark is that your path is not very piecewise linear as it could/should be. Consider reducing the tol and increasing max_iter.
hope this helps
I know this is an old thread, but:
I'm actually working on piping over to LassoCV from glmnet (in R), and I found that LassoCV doesn't do too well with normalizing the X matrix first (even if you specify the parameter normalize = True).
Try normalizing the X matrix first when using LassoCV.
If it is a pandas object,
(X - X.mean())/X.std()
It seems you also need to multiple alpha by 2
Though I am unable to figure out what is causing the problem, there is a logical direction in which to continue.
These are the facts:
Mathworks have selected an example and decided to include it in their documentation
Your matlab code produces exactly the result as the example.
The alternative does not match the result, and has provided inaccurate results in the past
This is my assumption:
The chance that mathworks have chosen to put an incorrect example in their documentation is neglectable compared to the chance that a reproduction of this example in an alternate way does not give the correct result.
The logical conclusion: Your matlab implementation of this example is reliable and the other is not.
This might be a problem in the code, or maybe in how you use it, but either way the only logical conclusion would be that you should continue with Matlab to select your model.