pdtimeout

Generate TimeOut errors with pandas.apply

• Free software: Apache Software License 2.0
• Documentation: https://pdtimeout.readthedocs.io.

Basic usage : Raise a TimeOut Error

• Define a pandas DataFrame

df = pd.DataFrame({'number': [1, 0.5, 0.2, 2, 0.3]}, index=[0, 2, 4, 6, 8])

Index	Number
0	1
2	0.5
4	0.2
6	2
8	0.3

• Define a function to apply on the DataFrame and set the timeout value

@timeout(4)
 def sleep_and_triple(row):
     number_ = row['number']

     time.sleep(number_)

     return number_ * 3

This function first sleeps for number seconds and then returns the triple of the input value. Since the highest number is 2, a timeout value of 4 should not trigger a TimeOut error.

• Apply function on DataFrame

df['result'] = df.apply(sleep_and_halve, axis=1)
print(df)

Index	Number	Result
0	1	3
2	0.5	1.5
4	0.2	0.6
6	2	6
8	0.3	0.9

• Change the timeout value to 1.7 seconds and re-apply function on DataFrame

@timeout(1.7)
def sleep_and_triple(row):
    number_ = row['number']

    time.sleep(number_)

    return number_ * 3

df['result'] = df.apply(sleep_and_halve, axis=1)
print(df)

The following TimeOut error is triggered:

>>> "TimeoutError: ('Time expired', 'occurred at index 6')"

The row index (pandes .loc) of the row triggering the TimeOut error is given in the error message. The row with index 6 sleeps for 2 seconds which is longer than the timeout value, thus the error is triggered.

Return a default value in case of timeout

@timeout(1, replace_value='TimeOut')
def sleep_and_triple(row):
    number_ = row['number']

    time.sleep(number_)

    return number_ * 3

df['result'] = df.apply(sleep_and_halve, axis=1)
print(df)

Index	Number	Result
0	1	TimeOut
2	0.5	1.5
4	0.2	0.6
6	2	TimeOut
8	0.3	0.9

Return the execution time for each row

The time_apply can be used to monitor the execution time of each row as follows:

@time_apply()
def sleep_and_triple(row):
    number_ = row['number']

    time.sleep(number_)

    return number_ * 3

df['result'] = df.apply(sleep_and_halve, axis=1)
print(df)

Index	Number	Result
0	1	1.000667
2	0.5	0.500193
4	0.2	0.205290
6	2	2.005164
8	0.3	0.301278

The returned value (number * 3) is replaced by the execution time for the associated row.

Credits

This package was created with Cookiecutter and the audreyr/cookiecutter-pypackage project template.