Question

def lamb_0(x):

    return (1-x)/2

def lamb_1(x):

    return (1+x)/2

x = np.array([0.5773502691896257645091488,-0.5773502691896257645091488])

Right_item_2 = (1/2)*np.array([(-1/2)*[(C_n[0,j]*lamb_0(x[0])+C_n[1,j]*lamb_1(x[0]))*(R_n[0,j]*lamb_0(x[0])+R_n[1,j]*lamb_1(x[0]))+(C_n[0,j]*lamb_0(x[1])+C_n[1,j]*lamb_1(x[1]))*(R_n[0,j]*lamb_0(x[1])+R_n[1,j]*lamb_1(x[1]))],(1/2)*[(C_n[0,j]*lamb_0(x[0])+C_n[1,j]*lamb_1(x[0]))*(R_n[0,j]*lamb_0(x[0])+R_n[1,j]*lamb_1(x[0]))+(C_n[0,j]*lamb_0(x[1])+C_n[1,j]*lamb_1(x[1]))*(R_n[0,j]*lamb_0(x[1])+R_n[1,j]*lamb_1(x[1]))]],dtype=float)

TypeError: can't multiply sequence by non-int of type 'float'
#mark:C_n,R_n,S_n are arraies in 2 dimension

Answer 1

首先将计算分为几行。

detector.isOperational();

据我所知，两种计算方法是相同的，下面的Right_item_2 = (1/2)*np.array([ (-1/2)*[ # Here you multiply a list by (-1/2). Should it be ( not [? (C_n[0,j]*lamb_0(x[0])+C_n[1,j]*lamb_1(x[0]))* (R_n[0,j]*lamb_0(x[0])+R_n[1,j]*lamb_1(x[0]))+ (C_n[0,j]*lamb_0(x[1])+C_n[1,j]*lamb_1(x[1]))* (R_n[0,j]*lamb_0(x[1])+R_n[1,j]*lamb_1(x[1])) ], (1/2)*[ # Here you multiply a list by (-1/2) (C_n[0,j]*lamb_0(x[0])+C_n[1,j]*lamb_1(x[0]))* (R_n[0,j]*lamb_0(x[0])+R_n[1,j]*lamb_1(x[0]))+ (C_n[0,j]*lamb_0(x[1])+C_n[1,j]*lamb_1(x[1]))* (R_n[0,j]*lamb_0(x[1])+R_n[1,j]*lamb_1(x[1])) ] ],dtype=float)。

res

仍有使res = (C_n[0,j]*lamb_0(x[0])+C_n[1,j]*lamb_1(x[0]))* (R_n[0,j]*lamb_0(x[0])+R_n[1,j]*lamb_1(x[0]))+ (C_n[0,j]*lamb_0(x[1])+C_n[1,j]*lamb_1(x[1]))* (R_n[0,j]*lamb_0(x[1])+R_n[1,j]*lamb_1(x[1])) Right_itm_2 = np.array[-.5*res, .5*res]计算更易读，更不重复的范围

TypeError：无法将序列乘以'float'类型的非整数，非常感谢

1 个答案: