[人工智能]q-rung Orthopair fuzzy set 运算规则python实现

模糊集运算规则合集之三，q-rung Orthopair fuzzy set 。
广义直觉模糊集合上的的运算规则，包括加法，乘法，数乘，幂乘，距离，得分函数精确函数，暂时我就用到这么多，之后有新研究会随时补充。

def scorenew1(df):#得分函数1
    score=[]
    for i in range(len(df)):
        score.append(0.5*(1-df[i][0])*(1+(1-df[i][0]-df[i][1])))
    return score

def scorenew2(df,q=2):#得分函数2
    score=[]
    for i in range(len(df)):
        score.append(0.5*(1+(df[i][0]**q)-(df[i][1]**q)))
    return score
def scorenew4(df,q=2):#得分函数3
    score=[]
    for i in range(len(df)):
        score.append(((df[i][0]**q)-(df[i][1]**q)))
    return score
def ltimes(lamda, coords2,q=2):  # 数乘运算
    d1 = []
    d2 = []
    d = []
    for x, y in coords2:
        d1.append((1 - (1 - x**q) ** lamda)**(1/q))
        d2.append(y ** lamda)
    t = list(zip(*[d1, d2]))
    for i in t:
        d.append(list(i))
    return d
def plus(coords1, coords2,q=2):  # 和运算
    d1=[]
    d2=[]
    d=[]
    for (x, y) in zip(coords1, coords2):
        d1.append(((x[0]**q)+(y[0]**q)-(x[0]**q)*(y[0]**q))**(1/q))
        d2.append(x[1]*y[1])
    t=list(zip(*[d1,d2]))
    for i in t:
        d.append(list(i))
    return d
def times(coords1, coords2,q=2):  # 积运算
    d1=[]
    d2=[]
    d=[]
    for (x, y) in zip(coords1, coords2):
        d1.append(x[0]*y[0])
        d2.append(((x[1]**q)+(y[1]**q)-(x[1]**q)*(y[1]**q))**(1/q))
    t=list(zip(*[d1,d2]))
    for i in t:
        d.append(list(i))
    return d
'''一些小例子'''
l1=[[0.6,0.5],[0.8,0.6]]
l2=[[0.6,0.2],[0.9,0.3]]
print(ltimes(1,l1))
print(times(l1,l2))
print(plus(l1,l2))
print(distance(l1[0],l2[1]))