我正在尝试将this answer应用于我的代码,以显示scipy.optimize.differential_evolution方法的进度条。
我认为differential_evolution
将对func
(称为最小化的函数)进行popsize * maxiter
次评估,但显然并非如此。
下面的代码应显示一个进度条,最多增加到100%
:
[####################] 100%
但是实际上,随着DEdist()
函数的求值次数比popsize * maxiter
(我用作total
函数的updt()
参数)的次数要多得多
如何计算differential_evolution
执行的功能评估的总数?可以做到吗?
from scipy.optimize import differential_evolution as DE
import sys
popsize, maxiter = 10, 50
def updt(total, progress, extra=""):
"""
Displays or updates a console progress bar.
Original source: https://stackoverflow.com/a/15860757/1391441
"""
barLength, status = 20, ""
progress = float(progress) / float(total)
if progress >= 1.:
progress, status = 1, "\r\n"
block = int(round(barLength * progress))
text = "\r[{}] {:.0f}% {}{}".format(
"#" * block + "-" * (barLength - block),
round(progress * 100, 0), extra, status)
sys.stdout.write(text)
sys.stdout.flush()
def DEdist(model, info):
updt(popsize * maxiter, info['Nfeval'] + 1)
info['Nfeval'] += 1
res = (1. - model[0])**2 + 100.0 * (model[1] - model[0]**2)**2 + \
(1. - model[1])**2 + 100.0 * (model[2] - model[1]**2)**2
return res
bounds = [[0., 10.], [0., 10.], [0., 10.], [0., 10.]]
result = DE(
DEdist, bounds, popsize=popsize, maxiter=maxiter,
args=({'Nfeval': 0},))
答案 0 :(得分:1)
来自help(scipy.optimize.differential_evolution)
:
maxiter : int, optional
The maximum number of generations over which the entire population is
evolved. The maximum number of function evaluations (with no polishing)
is: ``(maxiter + 1) * popsize * len(x)``
默认情况下也为polish=True
:
polish : bool, optional
If True (default), then `scipy.optimize.minimize` with the `L-BFGS-B`
method is used to polish the best population member at the end, which
can improve the minimization slightly.
所以您需要更改两件事:
1在此处使用正确的薄膜:
updt(popsize * (maxiter + 1) * len(model), info['Nfeval'] + 1)
2通过polish=False
参数:
result = DE(
DEdist, bounds, popsize=popsize, maxiter=maxiter, polish=False,
args=({'Nfeval': 0},))
此后,您将看到进度条在达到100%时完全停止。