作为公式获得的矩阵如下：

公式矩阵由参数A和B以及alpha（即xx2和yy2的系数）组成，并且它们的A，B和Alpha值未知。 并且只有参数的极限是已知的（“ A”，“ B”介于两个数字1和5之间，除了偶数也包含数字1和5，“ alpha”介于两个数字1.01和1.5之间，可以包括可以是1.01和1.5的数字

例如： uu = alpha *（xx2）+ A *（yy2）+ B * xx2

yy2和xx2的值如下：

yy2= array([0.,0.03846154,0.07692308,0.11538462,0.15384615,0.19230769,0.23076923,0.26923077,0.30769231,0.34615385,0.38461538,0.42307692,0.46153846,0.5,0.53846154,0.57692308,0.61538462,0.65384615,0.69230769,0.73076923,0.76923077,0.80769231,0.84615385,0.88461538,0.92307692,0.96153846,1.        ])

xx2 = array([0.,0.01470588,0.02941176,0.04411765,0.05882353,0.07352941,0.08823529,0.10294118,0.11764706,0.13235294,0.14705882,0.16176471,0.17647059,0.19117647,0.20588235,0.22058824,0.23529412,0.25,0.26470588,0.27941176,0.29411765,0.30882353,0.32352941,0.33823529,0.35294118,0.36764706,0.38235294,0.39705882,0.41176471,0.42647059,0.44117647,0.45588235,0.47058824,0.48529412,0.51470588,0.52941176,0.54411765,0.55882353,0.57352941,0.58823529,0.60294118,0.61764706,0.63235294,0.64705882,0.66176471,0.67647059,0.69117647,0.70588235,0.72058824,0.73529412,0.75,0.76470588,0.77941176,0.79411765,0.80882353,0.82352941,0.83823529,0.85294118,0.86764706,0.88235294,0.89705882,0.91176471,0.92647059,0.94117647,0.95588235,0.97058824,0.98529412,1.        ])

从Excel中获得的矩阵如下，其中主矩阵为“ df_nump”：

path = '100802_1.xlsx' 

df_main = pd.read_excel(path)


df_summarry = df_main.iloc[1:2,3:15]
df_summarry.columns = df_main.iloc[0,3:15]

df = df_main.iloc[8:,0:].reset_index()


del df['index']



df.columns = ['x_' + str(x) for x in range(np.size(df,1))]


df_y = normalize(df.iloc[1:,0])


'''x coordinate #experiment data'''
df_y1 = (df.iloc[1:,0]) 


df_x = normalize(df.iloc[0,1:])


''' y coordinate #experiment data ''' 
df_x1 = (df.iloc[0,1:]) 


''' the main matrix ''' 
df_nump = df.iloc[1:,1:].fillna(0)

和代码，用于在两个矩阵（“ uu”，“ df_nump”）中查找最接近的元素，而无需重复[重复]，并返回两个矩阵的索引：

list1=np.array(df_y1)
list2=xx2

index_dfy1=[]
index_xx2=[]
for i in range(len(list1)):

  if (len(list2)) > 1: #When there are elements in list2

    temp_result = abs(list1[i] - list2) #Matrix subtraction

    min_val = np.amin(temp_result) #Getting the minimum value to get closest element
    min_val_index = np.where(temp_result == min_val) #To find index of minimum value

    closest_element = list2[min_val_index] #actual value of closest element in list2

    list2 = list2[list2 != closest_element] #Remove closest element after found

    print(i,list1[i],min_val_index[0][0],closest_element[0]) #List1 Index,Element to find,List2 Index,Closest Element
    index_dfy1.append(i)
    index_xx2.append(min_val_index[0][0])

  else: #All elements are already found

    print(i,'No further closest unique closest elements found in list2')

idx_dfy1 = index_dfy1
idx_xx2 = index_xx2

print(idx_dfy1)
print(idx_xx2)

###################################################

list1=np.array(df_x1)
list2=yy2

index_dfx1=[]
index_yy2=[]
for i in range(len(list1)):

  if (len(list2)) > 1: #When there are elements in list2

    temp_result = abs(list1[i] - list2) #Matrix subtraction

    min_val = np.amin(temp_result) #Getting the minimum value to get closest element
    min_val_index = np.where(temp_result == min_val) #To find index of minimum value

    closest_element = list2[min_val_index] #actual value of closest element in list2

    list2 = list2[list2 != closest_element] #Remove closest element after found

    print(i,Closest Element
    index_dfx1.append(i)
    index_yy2.append(min_val_index[0][0])

  else: #All elements are already found

    print(i,'No further closest unique closest elements found in list2')

idx_dfx1 = index_dfx1
idx_yy2 = index_yy2

print(idx_dfx1)
print(idx_yy2)

现在，我想在两个“ uu”，“ df_nump”矩阵之间进行优化，以确定“ a”，“ b”，“ alpha”参数的最佳值，并根据平方根确定误差量（ R平方）方法