Profilen met cProfile en line_profiler

import cProfile

def slow():
    total = 0
    for i in range(1_000_000):
        total += i
    return total

cProfile.run("slow()")

# python -m cProfile -s cumtime my_script.py
#
# ncalls  tottime  percall  cumtime  percall filename:lineno(function)
#   1000    0.500    0.001    2.100    0.002 utils.py:10(process)

import cProfile, pstats, io

pr = cProfile.Profile()
pr.enable()

# ... code to profile ...

pr.disable()
stream = io.StringIO()
ps = pstats.Stats(pr, stream=stream).sort_stats("cumulative")
ps.print_stats(10)   # top 10
print(stream.getvalue())

import cProfile, pstats

cProfile.run("my_function()", "profile.out")
stats = pstats.Stats("profile.out")
stats.sort_stats("tottime")
stats.print_stats("mymodule")   # only mymodule functions

# pip install line_profiler
# Decorate target function:
from line_profiler import profile

@profile
def process(data):
    result = []
    for item in data:          # <- which line is slow?
        result.append(item*2)
    return result

# Run: kernprof -l -v script.py

# kernprof -l -v script.py
#
# Line #  Hits  Time  Per Hit  % Time  Line Contents
# ======================================================
#     4    1  2.0   2.0       1.0    result = []
#     5  1000  120.0  0.1     60.0    for item in data:
#     6  1000   80.0  0.1     39.0        result.append(item*2)

# In a Jupyter cell:
# %prun -s cumulative my_function(data)

# %load_ext line_profiler
# %lprun -f my_function my_function(data)

# cumtime = total time including callees (find root cause)
# tottime = time in function itself (find where work happens)
#
# Optimise the function with highest tottime first

# Before optimising:
# profile shows: process_row() 95% of time

# Speedup: vectorise with NumPy
import numpy as np
arr = np.array(data)
result = arr * 2   # 100x faster than Python loop

# pip install py-spy

# Profile for 30 s and show flamegraph:
# py-spy top --pid 12345
# py-spy record -o profile.svg --pid 12345 --duration 30

import timeit

result = timeit.timeit(
    "[x*2 for x in range(1000)]",
    number=10_000
)
print(f"{result:.3f} s for 10k runs")