在 ipython notebook 中测量单元格执行时间的简单方法
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Simple way to measure cell execution time in ipython notebook
提问by colinfang
I would like to get the time spent on the cell execution in addition to the original output from cell.
除了单元格的原始输出之外，我还想获得单元格执行所花费的时间。
To this end, I tried %%timeit -r1 -n1but it doesn't expose the variable defined within cell.
为此，我尝试过，%%timeit -r1 -n1但它没有公开单元格中定义的变量。
%%timeworks for cell which only contains 1 statement.
%%time适用于仅包含 1 个语句的单元格。
In[1]: %%time
1
CPU times: user 4 μs, sys: 0 ns, total: 4 μs
Wall time: 5.96 μs
Out[1]: 1
In[2]: %%time
# Notice there is no out result in this case.
x = 1
x
CPU times: user 3 μs, sys: 0 ns, total: 3 μs
Wall time: 5.96 μs
What's the best way to do it?
最好的方法是什么？
Update
更新
I have been using Execute Time in Nbextensionfor quite some time now. It is great.
我在 Nbextension 中使用Execute Time已经有一段时间了。太好了。
采纳答案by Philipp Schwarz
Use cell magic and this project on github by Phillip Cloud:
使用 cell magic 和 Phillip Cloud 在 github 上的这个项目：
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:
通过将它放在笔记本顶部来加载它，或者如果您总是希望默认加载它，则将它放在您的配置文件中：
%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.
如果加载，后续单元格执行的每个输出都将包括执行它所花费的时间（以分钟和秒为单位）。
回答by ryanmc
%timeand %timeitnow come part of ipython's built-in magic commands
%time而%timeit现在来IPython中的一部分内置的魔法命令
回答by Salvador Dali
The only way I found to overcome this problem is by executing the last statement with print.
我发现解决这个问题的唯一方法是使用 print 执行最后一条语句。
Do not forget thatcell magic starts with %%and line magic starts with %.
不要忘记单元魔术以 开头，%%线魔术以 开头%。
%%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:
请注意，在单元格内执行的任何更改都不会在下一个单元格中考虑在内，这在有管道时是违反直觉的：
回答by eafit
This is not exactly beautiful but without extra software
这不是很漂亮，但没有额外的软件
class timeit():
from datetime import datetime
def __enter__(self):
self.tic = self.datetime.now()
def __exit__(self, *args, **kwargs):
print('runtime: {}'.format(self.datetime.now() - self.tic))
Then you can run it like:
然后你可以像这样运行它：
with timeit():
# your code, e.g.,
print(sum(range(int(1e7))))
% 49999995000000
% runtime: 0:00:00.338492
回答by blehman
Sometimes the formatting is different in a cell when using print(res), but jupyter/ipython comes with a display. See an example of the formatting difference using pandas below.
有时使用 时print(res)，单元格中的格式不同，但 jupyter/ipython 带有display. 请参阅下面使用 Pandas 的格式差异示例。
%%time
import pandas as pd
from IPython.display import display
df = pd.DataFrame({"col0":{"a":0,"b":0}
,"col1":{"a":1,"b":1}
,"col2":{"a":2,"b":2}
})
#compare the following
print(df)
display(df)
The displaystatement can preserve the formatting.
该display语句可以保留格式。
回答by Harry_pb
I simply added %%timeat the beginning of the cell and got the time. You may use the same on Jupyter Spark cluster/ Virtual environment using the same. Just add %%timeat the top of the cell and you will get the output. On spark cluster using Jupyter, I added to the top of the cell and I got output like below:-
我只是%%time在单元格的开头添加并获得了时间。您可以在 Jupyter Spark 集群/虚拟环境上使用相同的内容。只需%%time在单元格顶部添加，您将获得输出。在使用 Jupyter 的 Spark 集群上，我添加到单元格的顶部，得到如下输出：-
[1] %%time
import pandas as pd
from pyspark.ml import Pipeline
from pyspark.ml.classification import LogisticRegression
import numpy as np
.... code ....
Output :-
CPU times: user 59.8 s, sys: 4.97 s, total: 1min 4s
Wall time: 1min 18s
回答by vForce
An easier way is to use ExecuteTime plugin in jupyter_contrib_nbextensions package.
更简单的方法是使用 jupyter_contrib_nbextensions 包中的 ExecuteTime 插件。
pip install jupyter_contrib_nbextensions
jupyter contrib nbextension install --user
jupyter nbextension enable execute_time/ExecuteTime
回答by markroxor
you may also want to look in to python's profiling magic command %prunwhich gives something like -
您可能还想查看 python 的分析魔术命令%prun，它给出了类似的东西 -
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
然后
%prun sum_of_lists(1000000)
will return
将返回
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.
我发现它在处理大量代码时很有用。
回答by Mostafa Gazar
You can use timeitmagic function for that.
您可以timeit为此使用魔术功能。
%timeit CODE_LINE
Or on the cell
或者在细胞上
%%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
在https://nbviewer.jupyter.org/github/ipython/ipython/blob/1.x/examples/notebooks/Cell%20Magics.ipynb查看更多 IPython 魔法函数
回答by mina
import time
start = time.time()
"the code you want to test stays here"
end = time.time()
print(end - start)
