If I know the time of a given validation with set values, can I estimate the time GridSearchCV will take for n values I want to cross-validate?


3 Answers 3


You could fit your model/pipeline (with default parameters) to your data once and see how long it takes to train. Then you would multiply that by how many times you want to train the model through grid search.

E.g. suppose you want to use a grid search to select the hyperparameters a, b and c of your pipeline.

params = {'a': [1, 2, 3, 4, 5],
          'b': [1, 2, 3, 4],
          'c': [1, 2, 3]}

cv = GridSearchCV(pipeline, params)

By default this should run a search for a grid of $5 \cdot 4 \cdot 3 = 60$ different parameter combinations. The default cross-validation is a 3-fold cv so the above code should train your model $60 \cdot 3 = 180$ times. By default GridSearch runs parallel on your processors, so depending on your hardware you should divide the number of iterations by the number of processing units available. Let's say for example I have 4 processors available, each processor should fit the model $ 180 / 4 = 45$ times. Now, if on average my model takes $10 sec$ to train, I'm estimating around $45 \cdot 10 / 60 = 7.5min$ training time. In practice it should be closer to $8min$ due to overhead.

Finally, because some parameters heavily affect the training time of that algorithm, I would suggest using the max_iter argument whenever available so that your estimation doesn't fall far off.

Please note : As of July 2021, the default folds is 5.

From sklearn documentation : Changed in version 0.22: cv default value if None changed from 3-fold to 5-fold.

  • $\begingroup$ thanks ! I'll try that out and see how long it takes. I think the average time was 0.4s that I had to run some thousands of time so it did take quite some time. I'll be able to estimate it better :) $\endgroup$ Mar 25, 2018 at 7:27
  • $\begingroup$ By default number of jobs (n_jobs) that GridSearchCV runs is 1. In case you want to use more one CPU at a time you should set n_jobs=-1 or n_jobs=<num_parallel_jobs_desired>. This may need extra memory as per documentation if the dataset is big and you may have to use pre_dispatch parameter. $\endgroup$
    – smishra
    Aug 19, 2018 at 14:12
  • $\begingroup$ I have 3 parameters with 10 levels to scan and the time for a run is about 19 seconds. Hence, 10*3*19=570/60=~10 minutes. But I definitely have to wait about 35-45 minutes. What am I doing wrong (either in time calculation or in scanning params?) $\endgroup$
    – Ben
    Oct 11, 2019 at 14:34
  • 1
    $\begingroup$ Probably some combination of parameters severely slows down your computation. You can try setting verbose=1 so that it prints your progress. $\endgroup$
    – Djib2011
    Oct 11, 2019 at 23:33

Let the search complete and then you can use cv_results_ attribute to compute the elapsed time as given below.

mean_fit_time= search_cv.cv_results_['mean_fit_time']
mean_score_time= search_cv.cv_results_['mean_score_time']
n_splits  = search_cv.n_splits_ #number of splits of training data
n_iter = pd.DataFrame(search_cv.cv_results_).shape[0] #Iterations per split

print(np.mean(mean_fit_time + mean_score_time) * n_splits * n_iter)
  • $\begingroup$ This solution is elegant and intuitive and leverages built in return attributes of GridSearchCV. $\endgroup$
    – Smitty
    Aug 1, 2021 at 18:09

Here is a little function I made to estimate the GridSearchCV time, though I think it could be improved by @Naveen Vuppula's comment.

import timeit

def esitmate_gridsearch_time(model, param_grid:dict, cv:int=5, processors:int=6):
    times = []
    for _ in range(5):
        start = timeit.default_timer()
        model.fit(X_train, y_train)
        model.score(X_train, y_train)
        times.append(timeit.default_timer() - start)

    single_train_time = np.array(times).mean() # seconds

    combos = 1
    for vals in param_grid.values():
        combos *= len(vals)

    num_models = combos * cv / processors
    seconds = num_models * single_train_time
    minutes = seconds / 60
    hours = minutes / 60

    print(hours, minutes, seconds)

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