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我工作的建設有一個skfuzzy模糊推理系統循環，我需要找到一種方法，以加快我的代碼：有效的方式來通過大numpy的陣列
import skfuzzy as fuzz
from skfuzzy import control as ctrl
import numpy as np
def FIS(s, r):
#Generate universe variables
a = ctrl.Antecedent(np.arange(0, 70.1, 0.1), 'a')
b = ctrl.Antecedent(np.arange(0, 6.01, 0.01), 'b')
c = ctrl.Consequent(np.arange(0, 12.01, 0.01), 'c')
#Generate fuzzy membership functions
#a
a['l'] = fuzz.trapmf(a.universe, [0.0, 0.0, 3.0, 6.0])
a['m'] = fuzz.trapmf(a.universe,[3.0, 6.0, 16.0, 24.0])
a['h'] = fuzz.trapmf(a.universe, [16.0, 24.0, 30.0, 45.0])
a['e'] = fuzz.trapmf(a.universe, [30.0, 45.0, 70.0, 70.0])
#b
b['l'] = fuzz.trapmf(b.universe, [0, 0, 0.01, 0.02])
b['m'] = fuzz.trapmf(b.universe,[0.01, 0.02, 0.03, 0.05])
b['h'] = fuzz.trapmf(b.universe, [0.03, 0.05, 0.10, 0.12])
b['e'] = fuzz.trapmf(b.universe, [0.10, 0.12, 6.00, 6.00])
#c
c['l'] = fuzz.trapmf(c.universe, [0, 0, 0.01, 0.02])
c['m'] = fuzz.trapmf(c.universe,[0.01, 0.02, 0.04, 0.05])
c['h'] = fuzz.trapmf(c.universe, [0.04, 0.05, 0.10, 0.20])
c['e'] = fuzz.trapmf(c.universe, [0.10, 0.20, 12.00, 12.00])
#FUZZY RULES
rule1 = ctrl.Rule(a['l'] & b['l'], c['l'])
rule2 = ctrl.Rule(a['l'] & b['m'], c['m'])
rule3 = ctrl.Rule(a['l'] & b['h'], c['h'])
rule4 = ctrl.Rule(a['m'] & b['l'], c['m'])
rule5 = ctrl.Rule(a['m'] & b['m'], c['m'])
rule6 = ctrl.Rule(a['m'] & b['h'], c['h'])
rule7 = ctrl.Rule(a['h'] & b['l'], c['h'])
rule8 = ctrl.Rule(a['h'] & b['m'], c['h'])
rule9 = ctrl.Rule(a['h'] & b['h'], c['e'])
rule10 = ctrl.Rule(a['e'], c['e'])
rule11 = ctrl.Rule(b['e'], c['e'])
#CONTROL SYSTEM
c_ctrl = ctrl.ControlSystem([rule1, rule2, rule3, rule4, rule5, rule6,
rule7, rule8, rule9, rule10, rule11])
c_simulation = ctrl.ControlSystemSimulation(c_ctrl)
c_simulation.input['a'] = s
c_simulation.input['b'] = r
c_simulation.compute()
value = c_simulation.output['c']
return value
#Fake data
s_data = np.random.RandomState(1234567890)
s_data = s_data.randint(0, 70, size=600000)
r_data = np.random.random_sample(600000)
vec1 = s_data.flatten().astype('float')
vec2 = r_data.flatten().astype('float')
#pre allocate output array
cert = np.zeros(np.shape(vec1))*np.nan
#Find index of all finite elements of the array
ind = np.where(np.isfinite(vec1))[0]
# classify
for k in xrange(len(ind)):
cert[ind[k]] = FIS(vec1[ind[k]], vec2[ind[k]])
我的計算，錄取比10 +小時完成。如何在沒有for循環的情況下執行這些計算？理想情況下，我會尋找一個numpy解決方案，但我願意接受其他解決方案。
所以你打電話給'FIS'幾千次？什麼花費在'FIS'上的時間？該功能的2/3設置可以只執行一次。我不知道'simulation.compute'的功能。無論如何，我的猜測是你應該擔心在每次FIS調用中花費的時間，而不是擔心循環機制。 – hpaulj
它當然看起來像FIS中'c_simulation.input ['a'] = s'的所有內容都不應該在那個循環中......你重新計算所有'len（ind）'時間。 – TemporalWolf
好點。我拿出你在for循環之外所建議的內容，但仍然有同樣的問題。問題是有600,000個計算要做。 – brainier7