Simple way to measure cell execution time in ipython notebook

Question

I would like to get the time spent on the cell execution in addition to the original output from cell.

To this end, I tried %%timeit -r1 -n1 but it does

Answer 1

you may also want to look in to python's profiling magic command %prunwhich gives something like -

def sum_of_lists(N):
    total = 0
    for i in range(5):
        L = [j ^ (j >> i) for j in range(N)]
        total += sum(L)
    return total

then

%prun sum_of_lists(1000000)

will return

14 function calls in 0.714 seconds  

Ordered by: internal time      

ncalls  tottime  percall  cumtime  percall filename:lineno(function)
    5    0.599    0.120    0.599    0.120 <ipython-input-19>:4(<listcomp>)
    5    0.064    0.013    0.064    0.013 {built-in method sum}
    1    0.036    0.036    0.699    0.699 <ipython-input-19>:1(sum_of_lists)
    1    0.014    0.014    0.714    0.714 <string>:1(<module>)
    1    0.000    0.000    0.714    0.714 {built-in method exec}

I find it useful when working with large chunks of code.

Answer 2

Use cell magic and this project on github by Phillip Cloud:

Load it by putting this at the top of your notebook or put it in your config file if you always want to load it by default:

%install_ext https://raw.github.com/cpcloud/ipython-autotime/master/autotime.py
%load_ext autotime

If loaded, every output of subsequent cell execution will include the time in min and sec it took to execute it.

Answer 3

If you want to print wall cell execution time here is a trick, use

%%time
<--code goes here-->

but here make sure that, the %%time is a magic function, so put it at first line in your code.

if you put it after some line of your code it's going to give you usage error and not gonna work.

Answer 4

import time
start = time.time()
"the code you want to test stays here"
end = time.time()
print(end - start)

Answer 5

You can use timeit magic function for that.

%timeit CODE_LINE

Or on the cell

%%timeit 

SOME_CELL_CODE

Check more IPython magic functions at https://nbviewer.jupyter.org/github/ipython/ipython/blob/1.x/examples/notebooks/Cell%20Magics.ipynb

Answer 6

The only way I found to overcome this problem is by executing the last statement with print.

Do not forget that cell magic starts with %% and line magic starts with %.

%%time
clf = tree.DecisionTreeRegressor().fit(X_train, y_train)
res = clf.predict(X_test)
print(res)

Notice that any changes performed inside the cell are not taken into consideration in the next cells, something that is counter intuitive when there is a pipeline: