Python

Timing your Code

• Simply tracking your elapsed
• Using the timeit module
• Special %%timeit magic for Jupyter Notebook
• Timeit module used to calculate performance of functionality with available solutions.

Simply tracking your elapsed

Solution 1 implementation:

   def method_one(number):
      return [str(num) for num in range(number)]

Calling one time:

   import time
   #Current time before running code
   start_time = time.time()
   
   #run method
   method_one(100000)
   
   #Current time after running code
   end_time = time.time()
   
   #Calculate differences
   elapsed_time = end_time - start_time
   print(f'Elapsed Time = {elapsed_time}') # ==> Elapsed Time = 0.019519329071044922

Solution 2 implementation:

   def method_two(number):
      return list(map(str, range(number)))

Calling one time:

   import time
   #Current time before running code
   start_time = time.time()
   
   #run method
   method_two(100000)
   
   #Current time after running code
   end_time = time.time()
   
   #Calculate differences
   elapsed_time = end_time - start_time
   print(f'Elapsed Time = {elapsed_time}') # ==> Elapsed Time = 0.2002100944519043

Using the timeit module api

Solution1 implementation:

  #timeit api use to call same code multiple time to check performance multiple time.
  import timeit
  statement_1 = '''
  method_one(100)
  '''
  setup_1 = '''
  def method_one(number):
      return [str(num) for num in range(number)]
  '''

Calling multiple time:

  timeit.timeit(statement_1, setup_1, number = 1000000) # ==> 15.495843400000012

Solution2 implementation:

  #timeit api use to call same code multiple time to check performance multiple time.
  import timeit
  statement_2 = '''
  method_one(100)
  '''
  setup_2 = '''
  def method_two(number):
    return list(map(str, range(number)))
  '''

Calling multiple time:

  timeit.timeit(statement_2, setup_2, number = 1000000) # ==> 12.765108300000065

Using the timeit magic function of Jupyter Notebooks

Solution 1 implementation:

   def method_one(number):
      return [str(num) for num in range(number)]

Calling 1 time:

   method_one(10) # ==> ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

Calling multiple time:

Input:

   %%timeit
   method_one(100)

Output: ``‘python 16.3 µs ± 99.5 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

**Solution 2 implementation:**
   ```python
   def method_two(number):
       return list(map(str, range(number)))

Calling 1 time:

   method_two(10) # ==> ['0', '1', '2', '3', '4', '5', '6', '7', '8', '9']

Calling multiple time:

Input:

   %%timeit
   method_two(100)

Output:

   13.3 µs ± 263 ns per loop (mean ± std. dev. of 7 runs, 100000 loops each)

---

python-documentation is maintained by ravaan-techky.

---