Python 3将元素识别为字符串
我定义了一个损失函数Python 3将元素识别为字符串,python,python-3.x,Python,Python 3.x,我定义了一个损失函数L（X，Y，c，b，d），其中X和Y应该是数组，c，b，d是浮点数。这是我的密码： def L(X,Y,c,b,d): error=0 a=Y[0] for i in range(len(X)): error = error + ((c*X[i]+(a/b)*np.sqrt(b**2-X[i]**2)+d)-Y[i])**2 return error 我要走了 "TypeError: unsupported operand t
L（X，Y，c，b，d）def L(X,Y,c,b,d):
error=0
a=Y[0]
for i in range(len(X)):
error = error + ((c*X[i]+(a/b)*np.sqrt(b**2-X[i]**2)+d)-Y[i])**2
return error
"TypeError: unsupported operand type(s) for ** or pow(): 'str' and 'int'",
def L(X,Y,c,b,d):
error=0
a=Y[0]
for i in range(len(X)):
error = error + ((c*X[i]+(a/b)*np.sqrt(b**2-X[i]**2)+d)-Y[i])**2
return error
def L_prime_c(X,Y,c,b,d):
p=0
for i in range(len(X)):
p=p+2*L(X,Y,c,b,d)*X[i]
return p
def L_prime_d(X,Y,c,b,d):
r=0
for i in range(len(X)):
r=r+2*L(X,Y,c,b,d)
return r
def fit(X,Y,learning_rate=1e-3,max_iter=20000,epsilon=1e-5):
c=1
b=1
d=Y[0]
for i in range(max_iter):
gradient_c = L_prime_c(X,Y,c,b,d)
c_new = c - learning_rate * gradient_c
gradient_b = L_prime_b(X,Y,c,b,d)
b_new = b - learning_rate * gradient_b
gradient_d = L_prime_d(X,Y,c,b,d)
d_new = d - learning_rate * gradient_d
if np.abs(c_new - c) < epsilon and np.abs(d_new - d) < epsilon and np.abs(b_new - b) < epsilon:
break
c=c_new
b=b_new
d=d_new
return np.array([c,b,d])
#getting X, Y
In = xlrd.open_workbook(r'C:\Users\gris_\Desktop\calculation4.xlsx')
isheet = In.sheet_by_name('Sheet1')
X=isheet.col_values(0)
Y=isheet.col_values(1)
a=Y[0]
#Calculation & Output
result=fit(X,Y,learning_rate=1e-3,max_iter=20000,epsilon=1e-5)
问题解决：我有6个不同长度的列，所以当从一列中读取数据时，python实际上添加了这些“以使较短的列与最长的列长度相同。当我删除其他列时，情况会很好。非常感谢您的帮助！：D看到您刚才添加的代码，尝试重写X，Y，如下所示：
X=list(map(float,isheet.col_values(0)))
Y=list(map(float,isheet.col_values(1)))
import numpy as np
# Another way of writing it
#def L(X,Y,c,b,d):
# error=0
# a=Y[0]
# for x,y in zip(X,Y):
# error += ((c*x+(a/b)*np.sqrt(b**2-x**2)+d)-y)**2
# return error
# Or even better:
def L(X,Y,c,b,d):
a = Y[0]
error = sum(((c*x+(a/b)*np.sqrt(b**2-x**2)+d)-y)**2 for x,y in zip(X,Y))
return error
X = [2.2,2.2,3.2]
Y = [2.0,2.1,3.1]
c = 1
b = 20
d = 1
print(L(X,Y,c,b,d))
29.148285465180138X = [2.2,2.2,3.2,'a']
Y = [2.0,2.1,3.1,'b']
返回：
TypeError:**或pow（）不支持的操作数类型：'str'和'int'XYX