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| for j in range(Ytrain.shape[1]): | ||
| x=linear_model.Lasso(alpha=a).fit(Xtrain, Ytrain[:, j]) | ||
| mdl.append(x) | ||
| ypred = mdl[i].predict(Xtest) |
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I would move this code to the inner loop and predict with x instead - the reason is because if you evaluate with mdl[i], you're not evaluating all the models you just trained. Instead, if you're on the first alpha, you'll only train 1 model - mdl[0].
| x=linear_model.Lasso(alpha=a).fit(Xtrain, Ytrain[:, j]) | ||
| mdl.append(x) | ||
| ypred = mdl[i].predict(Xtest) | ||
| error.append(mean_squared_error(ypred,Ytest[:, j])) |
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Same thing here for the same exact reason.
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Then to save on memory, write some code to update your model list. The logic is the following:
- Evaluate models with first alpha
- On the next iteration, save your models and errors in another object
- Write code that will see if the MSE in the current alpha level is less than the MSE in the previous alpha level
- If it is, replace the model object in your original list, and update the alpha associated with this model
- Otherwise, don't update and continue
| print(error) | ||
| minmse_index = error.index(min(error)) | ||
| best_alpha = alphas[minmse_index] | ||
| print(best_alpha) |
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This is why all of your models have a single alpha value. But remember - each model corresponding to each target variable should have a unique "best" alpha. The comments above should resolve this.
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| GCP2MET_models = [] | ||
| for i in range(Ytrain.shape[1]): | ||
| mdl_G2M =linear_model.Lasso(alpha=best_alpha).fit(Xtrain, Ytrain[:, i]) |
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In the case described above, you will need to run this code with the arugment alpha=best_alpha[i] to get the correct values.
| # In[7]: | ||
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| kf = KFold(n_splits=3) |
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So I had you run hold out previously just to get the concept down. In practice, you don't have to run hold out. You can just report the results from k-fold CV,
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| r, p_value = pearsonr(Ypred, Ytest[:, i]) | ||
| # Compute P-value based on r-value | ||
| pval = norm.sf(abs(r))*2 |
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Ah, this may be where things are going wrong. I would run this code once you get a vector of r values out of the loop. I think this will return different values because your distribution will be different. I may be wrong, but it's worth a shot.
ScottCampit
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I just made comments inline for your linear regression and LASSO code. Take a look and let me know if you have any questions, or if you want to pitch additional ideas, let me know too.
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upload .py version of linear regression, Lasso, and met2gcp