我想得到的时间花在单元格执行除了原始的输出从单元格。
为此，我尝试了%%timeit -r1 -n1，但它没有公开在cell中定义的变量。
%%time适用于只包含1条语句的cell。
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
最好的方法是什么?
更新
我已经在nbeextension中使用执行时间相当长一段时间了。这是伟大的。
更新2021 - 03
到目前为止，这是正确的答案。从本质上讲，%%time和%%timeit现在都像预期的那样工作。
当前回答
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))
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
其他回答
当遇到麻烦时，什么意味着什么:
时间还是??时间
详情如下:
Usage, in line mode:
%timeit [-n<N> -r<R> [-t|-c] -q -p<P> -o] statement
or in cell mode:
%%timeit [-n<N> -r<R> [-t|-c] -q -p<P> -o] setup_code
code
code...
Time execution of a Python statement or expression using the timeit
module. This function can be used both as a line and cell magic:
- In line mode you can time a single-line statement (though multiple
ones can be chained with using semicolons).
- In cell mode, the statement in the first line is used as setup code
(executed but not timed) and the body of the cell is timed. The cell
body has access to any variables created in the setup code.
%time和%timeit现在成为ipython内置魔法命令的一部分
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
我只是在单元格的开头添加了%%time，就得到了时间。您可以在Jupyter Spark集群/虚拟环境中使用相同的方法。只需在单元格顶部添加%%time，就会得到输出。在使用Jupyter的星火集群上，我添加到单元格的顶部，我得到如下输出
[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
你可能还想查看python的剖析魔法命令% prunit给出类似-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
然后
%prun sum_of_lists(1000000)
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}
我发现它在处理大块代码时很有用。
