[英]scipy.optimize.minimize(method=‘SLSQP’) memory issues when outside the bounds
我正在使用scipy.optimize.minimize(method='SLSQP') ，函數和約束是通過scipy.interpolate.LinearNDInterpolator進行插值的。 起始值為邊界內的隨機數。
我正在與：
優化有時會運行並給出合理的結果，但有時優化器會開始請求高達20GB的大量內存，然后我的計算機停止工作。 這總是發生在邊界之外的值。
scipy.interpolate.LinearNDInterpolator是否有可能無法與scipy.optimize.minimize(method='SLSQP') ？ 在邊界之外，我沒有模擬數據，因此插值給出fill_Value = 0或fill_value = 1e10。
當我使用scipy.optimize.fmin_slsqp時，會發生相同的行為
不幸的是我的代碼很大，但是有了這個數據集，我總是遇到內存問題：
#########################################
###Memory Leak scipy.optimize.minimize###
#########################################
import numpy as np
from scipy.optimize import minimize
from scipy.interpolate import LinearNDInterpolator
def objfun(x):
print x
return x[1]
points = np.array([[ 0.00000000e+00, 0.00000000e+00],[ 0.00000000e+00, 1.00334000e+00],[ 0.00000000e+00, 2.00669000e+00],[ 7.07700000e+02, 0.00000000e+00],[ 7.07700000e+02, 1.00334000e+00],[ 7.07700000e+02, 2.00669000e+00],[ 1.56890000e+03, 0.00000000e+00],[ 1.56890000e+03, 1.00334000e+00],[ 1.56890000e+03, 2.00669000e+00],[ 2.50080000e+03, 0.00000000e+00],[ 2.50080000e+03, 1.00334000e+00],[ 2.50080000e+03, 2.00669000e+00],[ 3.47090000e+03, 0.00000000e+00],[ 3.47090000e+03, 1.00334000e+00],[ 3.47090000e+03, 2.00669000e+00],[ 4.46380000e+03, 0.00000000e+00],[ 4.46380000e+03, 1.00334000e+00],[ 4.46380000e+03, 2.00669000e+00],[ 5.47130000e+03, 0.00000000e+00],[ 5.47130000e+03, 1.00334000e+00],[ 5.47130000e+03, 2.00669000e+00],[ 6.48890000e+03, 0.00000000e+00],[ 6.48890000e+03, 1.00334000e+00],[ 6.48890000e+03, 2.00669000e+00],[ 7.51360000e+03, 0.00000000e+00],[ 7.51360000e+03, 1.00334000e+00],[ 7.51360000e+03, 2.00669000e+00],[ 8.54350000e+03, 0.00000000e+00],[ 8.54350000e+03, 1.00334000e+00],[ 8.54350000e+03, 2.00669000e+00],[ 9.57740000e+03, 0.00000000e+00],[ 9.57740000e+03, 1.00334000e+00],[ 9.57740000e+03, 2.00669000e+00],[ 1.06143000e+04, 0.00000000e+00],[ 1.06143000e+04, 1.00334000e+00],[ 1.06143000e+04, 2.00669000e+00],[ 1.16535000e+04, 0.00000000e+00],[ 1.16535000e+04, 1.00334000e+00],[ 1.16535000e+04, 2.00669000e+00],[ 1.26945000e+04, 0.00000000e+00],[ 1.26945000e+04, 1.00334000e+00],[ 1.26945000e+04, 2.00669000e+00],[ 1.37369000e+04, 0.00000000e+00],[ 1.37369000e+04, 1.00334000e+00],[ 1.37369000e+04, 2.00669000e+00],[ 1.47804000e+04, 0.00000000e+00],[ 1.47804000e+04, 1.00334000e+00],[ 1.47804000e+04, 2.00669000e+00],[ 1.58248000e+04, 0.00000000e+00],[ 1.58248000e+04, 1.00334000e+00],[ 1.58248000e+04, 2.00669000e+00],[ 1.68698000e+04, 0.00000000e+00],[ 1.68698000e+04, 1.00334000e+00],[ 1.68698000e+04, 2.00669000e+00],[ 1.79153000e+04, 0.00000000e+00],[ 1.79153000e+04, 1.00334000e+00],[ 1.79153000e+04, 2.00669000e+00],[ 1.89612000e+04, 0.00000000e+00],[ 1.89612000e+04, 1.00334000e+00],[ 1.89612000e+04, 2.00669000e+00],[ 2.00074000e+04, 0.00000000e+00],[ 2.00074000e+04, 1.00334000e+00],[ 2.00074000e+04, 2.00669000e+00]])
values = np.array([ 0.00000000e+00, 0.00000000e+00, 0.00000000e+00,4.29730000e+01, 5.72947500e-01, -5.35464000e-01,9.11676000e+01, 1.31063500e+00, -1.05937500e+00,1.38660750e+02, 2.11484000e+00, -1.50850500e+00,1.84497000e+02, 2.96052000e+00, -1.88466000e+00,2.28622000e+02, 3.83846000e+00, -2.19702000e+00,2.71163000e+02, 4.74426500e+00, -2.45397000e+00,3.12274500e+02, 5.67547500e+00, -2.66222500e+00,3.52102000e+02, 6.63058000e+00, -2.82711000e+00,3.90774000e+02, 7.60858000e+00, -2.95286000e+00,4.28399500e+02, 8.60879000e+00, -3.04289000e+00,4.65074500e+02, 9.63071000e+00, -3.10001500e+00,5.00881500e+02, 1.06739850e+01, -3.12655500e+00,5.35893000e+02, 1.17383500e+01, -3.12444000e+00,5.70166500e+02, 1.28235000e+01, -3.09540500e+00,6.03760000e+02, 1.39293500e+01, -3.04082500e+00,6.36721500e+02, 1.50557500e+01, -2.96194500e+00,6.69093500e+02, 1.62026000e+01, -2.85982000e+00,7.00915000e+02, 1.73698000e+01, -2.73539500e+00,7.32222000e+02, 1.85573500e+01, -2.58950000e+00,7.63042500e+02, 1.97651000e+01, -2.42286000e+00])
S22_Adh1Ad_inter = LinearNDInterpolator(points,values,1e10)
def Fsigcon(x):
rf1_int = x[1]
rf_eval=[]
x_eval=[]
interval = np.linspace(0,x[0],x[0]/0.01)
if interval.size == 0:
interval=np.array([x[0]])
for xcoord in interval:
rf_eval.append(rf1_int)
x_eval.append(xcoord)
val_interp = S22_Adh1Ad_inter(rf_eval,x_eval) #'nearest' #'linear' #'cubic'
out = (val_interp.min()-39.45)
return out
points = np.array([[ 0.00000000e+00, 0.00000000e+00],[ 0.00000000e+00, 1.99997000e-01],[ 0.00000000e+00, 4.00002000e-01],[ 7.07700000e+02, 1.39999000e-01],[ 7.07700000e+02, 3.39996000e-01],[ 1.56890000e+03, 8.00020000e-02],[ 1.56890000e+03, 2.79999000e-01],[ 2.50080000e+03, 1.99970000e-02],[ 2.50080000e+03, 2.20001000e-01],[ 2.50080000e+03, 1.90000200e+00],[ 3.47090000e+03, 1.60004000e-01],[ 3.47090000e+03, 3.60001000e-01],[ 4.46380000e+03, 9.99980000e-02],[ 4.46380000e+03, 3.00003000e-01],[ 5.47130000e+03, 4.00010000e-02],[ 5.47130000e+03, 2.39998000e-01],[ 5.47130000e+03, 3.00000000e+00],[ 6.48890000e+03, 1.80000000e-01],[ 6.48890000e+03, 3.79997000e-01],[ 7.51360000e+03, 1.20003000e-01],[ 7.51360000e+03, 3.20000000e-01],[ 8.54350000e+03, 5.99980000e-02],[ 8.54350000e+03, 2.60002000e-01],[ 9.57740000e+03, 0.00000000e+00],[ 9.57740000e+03, 1.99997000e-01],[ 9.57740000e+03, 4.00002000e-01],[ 1.06143000e+04, 1.39999000e-01],[ 1.06143000e+04, 3.39996000e-01],[ 1.16535000e+04, 8.00020000e-02],[ 1.16535000e+04, 2.79999000e-01],[ 1.26945000e+04, 1.99970000e-02],[ 1.26945000e+04, 2.20001000e-01],[ 1.26945000e+04, 1.90000200e+00],[ 1.37369000e+04, 1.60004000e-01],[ 1.37369000e+04, 3.60001000e-01],[ 1.47804000e+04, 9.99980000e-02],[ 1.47804000e+04, 3.00003000e-01],[ 1.58248000e+04, 4.00010000e-02],[ 1.58248000e+04, 2.39998000e-01],[ 1.58248000e+04, 3.00000000e+00],[ 1.68698000e+04, 1.80000000e-01],[ 1.68698000e+04, 3.79997000e-01],[ 1.79153000e+04, 1.20003000e-01],[ 1.79153000e+04, 3.20000000e-01],[ 1.89612000e+04, 5.99980000e-02],[ 1.89612000e+04, 2.60002000e-01],[ 2.00074000e+04, 0.00000000e+00],[ 2.00074000e+04, 1.99997000e-01],[ 2.00074000e+04, 4.00002000e-01]])
values = np.array([ 0. , 0. , 0. , 0.010168, 0.010055, 0.046252,0.045731, 0.092687, 0.107056, 0.11196 , 0.19232 , 0.190859,0.29924 , 0.295611, 0.401297, 0.42018 , 0.450553, 0.564416,0.561699, 0.727387, 0.719631, 0.883825, 0.894486, 0. ,1.087256, 1.084631, 1.298136, 1.287209, 1.507127, 1.505308,1.424393, 1.740491, 1.839568, 1.993769, 1.981605, 2.251336,2.238475, 2.330676, 2.511822, 2.723058, 2.803453, 2.792818,3.104855, 3.08533 , 3.29549 , 3.393902, 0. , 3.721085,3.714504])
G_Adh1Ad_inter = LinearNDInterpolator(points,values,0)
def Gcon(x):
val_interp = G_Adh1Ad_inter(x[1],x[0])
out = (val_interp.min()-0.33)
return out
cons = (
{'type': 'ineq',
'fun' : Fsigcon},
{'type': 'ineq',
'fun' : Gcon}
)
amin = 0.0
amax = 3.0
bounds=[(amin,amax),(0.0, 20007.400000000001)]
a_start= 1.5343936873636999
rf1_start= 6824.9659188661817
res_int = minimize(objfun, [a_start,rf1_start],method='SLSQP',jac=None,bounds=bounds,constraints=cons,tol =1e-4,options={'iprint':2, 'disp': True , 'maxiter':1e2})
到目前為止，我自己的解決方案不是使用scipy.interpolate.LinearNDInterpolator，而是將其替換為多個一維插值。 對於3D問題和帶有scipy.interpolate.UnivariateSpline的三次插值，我需要在矩形網格上放置64個數據點。 在第一步中，我僅在坐標方向1上進行插值（16次），然后將問題簡化為具有16個點的2D模型。 在下一步中，我在坐標方向2上進行插值（4次），並進一步減少到剩下4個點的1D問題。 最后一步是在坐標方向3上進行一維插值。使用此過程，即使在邊界之外，scipy.optimize.minimize（method ='SLSQP'）也不會遇到內存問題。 插值代碼如下：
#!/usr/bin/python
# coding: utf8
# Filename: N1Dsplines.py
import numpy as np
from scipy.interpolate import UnivariateSpline
from scipy.interpolate import LinearNDInterpolator
def sort2lists(list1=[],list2=[]):
'''
Lists must be equal size
sorts after list1
'''
list1list2 = [(list1[i1],list2[i1]) for i1 in range(len(list1))]
list1list2 = sorted(list1list2, key=lambda x:x[0])
list1 = [i1[0] for i1 in list1list2]
list2 = [i1[1] for i1 in list1list2]
return list1,list2
def sortdictkeysandreturnvalues(dictionary):
keyvalues = [[i1,dictionary[i1]] for i1 in dictionary]
keyvalues = sorted(keyvalues, key= lambda x: x[0])
points = [i1[1] for i1 in keyvalues]
return points
def meshgrid2(*arrs):
if len(arrs)==1:
arrs = tuple(arrs[0])
# list with entries= length of the input arrays
lens = map(len, arrs)
# dim = number of arrays
dim = len(arrs)
# sz is the multiplication of the length of the *arrs
sz = 1
for s in lens:
sz*=s
ans = []
for i, arr in enumerate(arrs):
slc = [1]*dim
slc[i] = lens[i]
arr2 = np.asarray(arr).reshape(slc)
for j, sz in enumerate(lens):
if j!=i:
arr2 = arr2.repeat(sz, axis=j)
ans.append(arr2)
return tuple(ans)
def arr2points(*arrs):
'''
arrs = ([list1], [list2], ...)
ans = [[x1,y1,z1...],[x1,y1,z2,...],...,[x1,y2,z1,...],...[x2,y1,z1]
'''
if len(arrs)==1:
arrs = tuple(arrs[0])
# list with entries= length of the input arrays
lens = map(len, arrs)
# dim = number of arrays
dim = len(arrs)
# sz is the multiplication of the length of the *arrs
sz = 1
for s in lens:
sz*=s
ans = [[] for i1 in range(sz)]
for i1, arr in enumerate(arrs):
count = 0
sz2 = 1
for s in lens[i1+1:]:
sz2*=s
for i4 in range(sz/(lens[i1]*sz2)):
for i2 in arr:
for i3 in range(sz2):
ans[count].append(i2)
count+=1
return ans
#
def N1Dsplines(pointsvaluesdict={}, point =(), scheme = (1,0), k=3):
'''
Geht nur für reguläre Grids nicht für scattered data points
pointsvaluesdict = {(x,y,z):value , (x,y,z):value ....}
point = (x,y,z)
k = interpolation order
scheme = (2,1,0) bedeutet erst wird in Richtung z=2 dann in Richtung y=1 und als letztes in Richtung x=0 interpoliert
'''
#Find the closest Datapoints to the point coordinates
closest = {i1:[] for i1 in scheme}
for i1 in scheme:
for count in range(k+1):
points = [i2 for i2 in pointsvaluesdict.keys() if i2[i1] not in closest[i1]]
closest[i1].append(min(points, key=lambda x:abs(x[i1]-point[i1]))[i1])
#Make pointgrid from closest points
liste = sortdictkeysandreturnvalues(closest)
pointsgrid = arr2points(liste)
pointsgrid = [tuple(i1) for i1 in pointsgrid]
valuesgrid = [pointsvaluesdict[i1] for i1 in pointsgrid]
griddict = {pointsgrid[i1]:valuesgrid[i1] for i1 in range(len(pointsgrid))}
#Get 1D Splines by letting scheme[>0]=const and vary scheme[0]
for i1 in scheme:
pointsinter = []
valuesinter = []
temp = closest.copy()
del temp[i1]
points = sortdictkeysandreturnvalues(temp)
points = arr2points(points)
for coords in points:
points2inter=[]
for i2 in pointsgrid:
i2red = list(i2)
del i2red[i1]
if coords == i2red:
points2inter.append(i2)
values2inter = [griddict[i2] for i2 in points2inter]
points2inter = [i2[i1] for i2 in points2inter]
points2inter,values2inter = sort2lists(list1=points2inter,list2=values2inter)
coords.insert(i1,point[i1])
pointsinter.append(coords)
s = UnivariateSpline(points2inter, values2inter,k=k, s=0)
value = s(point[i1])
valuesinter.append(value)
closest[i1]=[point[i1]]
pointsgrid = pointsinter
pointsgrid = [tuple(i1) for i1 in pointsgrid]
valuesgrid = valuesinter
griddict = {pointsgrid[i1]:valuesgrid[i1] for i1 in range(len(pointsgrid))}
return griddict
#
x = np.array([1, 2, 3, 4, 5, 6, 7 ,8 ,9, 10])
y = x
z= x
w = x
xx, yy, zz, ww = meshgrid2(x, y,z,w)
v= xx**2 + yy**2 +zz**2 +ww**2
xx = xx.flatten()
yy = yy.flatten()
zz = zz.flatten()
ww = ww.flatten()
v = v.flatten()
pointsvalues = [((xx[i1],yy[i1],zz[i1], ww[i1]),v[i1]) for i1 in range(len(v))]
pointsvaluesdict = {key: value for (key, value) in pointsvalues}
pointsvaluesdict.keys()
pointsvaluesdict.values()
point = (5.5,4.5,3.5,2.5)
5.5**2 + 4.5**2 + 3.5**2 +2.5**2== 69
N1Dsplines(pointsvaluesdict=pointsvaluesdict, point =(5.5,4.5,3.5,2.5), scheme = (0,1,2,3), k=3)
#Outside Bounds
N1Dsplines(pointsvaluesdict=pointsvaluesdict, point =(0,0,0,0), scheme = (0,1,2,3), k=3)
但是，為什么插值始終超出變量的定義范圍？
好像您為變量定義了max和min范圍。
我知道，根據我使用的scipy的版本，有時最小化函數將不接受我的界限，除非它們以浮點數給出（這是一個非常可怕的錯誤）。
您可以嘗試將邊界重新定義為
bnds = np.array( [ [min_var1, max_var1], [min_var2, max_var2] ], dtype = float )
這通常對我有用。
sudo -H pip install --upgrade scipy
sudo -H pip install --upgrade numpy
徹底解決了問題
Scipy.optimize.minimize method ='SLSQP'忽略約束
[英]Scipy.optimize.minimize method='SLSQP' ignores constraint
具有線性約束的Scipy.optimize.minimize SLSQP失敗
[英]Scipy.optimize.minimize SLSQP with linear constraints fails
scipy.optimize.minimize('SLSQP') 在給定 2000 暗變量時太慢
[英]scipy.optimize.minimize('SLSQP') too slow when given 2000 dim variable
python scipy.optimize.minimize “SLSQP求解器”在xo之間添加約束
[英]python scipy.optimize.minimize “SLSQP solver” adding constraint between xo
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