77 lines
2.9 KiB
Python
77 lines
2.9 KiB
Python
import numpy as np
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from test import test
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def crossnew(pcross, lenchrom, chrom, sizepop, bound,final_partitions):
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# 本函数完成交叉操作
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# pcorss input : 交叉概率
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# lenchrom input : 染色体的长度
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# chrom input : 染色体群
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# sizepop input : 种群规模
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# ret output : 交叉后的染色体
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#chrom=individuals.chrom
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chrom = chrom.reshape(sizepop,-1)
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for i in range(sizepop):
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# 随机选择1个染色体进行交叉
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# 交叉概率决定是否进行交叉
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pick = np.random.rand()
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while pick == 0:
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pick = np.random.rand()
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if pick > pcross:
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continue
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flag = 0
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m = 0
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while flag == 0:
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pick = np.random.rand()
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while np.prod(pick) == 0:
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pick = np.random.rand()
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index = np.ceil(pick * sizepop)-1
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adr = np.load('adr.npy')
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m += 1
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# 随机选择交叉位置
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lo = []
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for d in range(len(adr)):
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for e in range(len(adr)):
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if adr[d, e] != 0:
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adr[d, e] = chrom[int(index), e]
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if len(np.unique(adr[d, :])) > 2:
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lo.append(d)
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a = round(np.random.rand() * len(lo))-1
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while a == -1:
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a = round(np.random.rand() * len(lo))-1
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pick = lo[a] # 交叉位置
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# 随机选择进行交叉的位置，即选择第几个变量进行交叉，注意：两个染色体交叉的位置相同
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exchangel = chrom[int(index), pick] # 交叉位置所属区域
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ek = np.unique(adr[pick, :]) # 交叉开始
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ek = ek[ek != exchangel]
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ek = ek[ek != 0]
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a = round(np.random.rand() * len(ek))-1
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while a == -1:
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a = round(np.random.rand() * len(ek))-1
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chrom[int(index), pick] = ek[a]
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t = ek[a]
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m = np.array(np.where(chrom[int(index), :] == ek[a]))
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m = m[m != pick]
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a = round(np.random.rand() * len(m))-1
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while a == -1:
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a = round(np.random.rand() * len(m))-1
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chrom[int(index), m[a]] = exchangel
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# 交叉结束
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# test实现了对染色体编码值的可行性检验
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# 具体来说，它通过检查每个变量的取值
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flag1 = test(lenchrom, bound, chrom[int(index), :],final_partitions)
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# Check the feasibility of chromosome 2
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# flag2 = test(lenchrom, bound, chrom[index[1], :])
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# If flag1 is 0
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if flag1 == 0:
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flag = 0
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chrom[int(index), pick] = exchangel
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chrom[int(index), m[a]] = t
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# chrom[index[1], pos] = v2
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else:
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flag = 1
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return chrom |