for i=1:10 x=linspace(0,10,101); plot(x,sin(x+i)); print(gcf,'-deps',strcat('plot',num2str(i),',ps')); end

你的代码是一个循环，每次迭代都会生成一个新的图形并保存为.eps文件。在这个循环中，你使用linspace函数创建了一个从0到10的等差数列x，然后使用sin函数绘制了图形sin(x+i)，其中i是循环变量。最后，你使用print函数将当前图形保存为.eps文件。

需要注意的是，你在保存文件时使用了strcat函数来构建文件名，其中num2str(i)将循环变量i转换为字符串。这样每次迭代都会生成一个不同的文件名，例如'plot1.ps'、'plot2.ps'等。

如果你有关于这段代码的更多问题，请继续提问。

相关问题

import numpy as np import math from scipy import integrate def f(x): return math.sin(x)*math.sin(x) #复化梯形法 def func(a,b,n,f): x = np.linspace(a,b,n+1) sum1 = 0 h =(b-a)/n for i in range(补充代码1): 补充代码2 return sum1 #复化辛普森法 def func1(a,b,n,f): x = np.linspace(a,b,n+1) sum1 = 0 h =(b-a)/n for i in range(补充代码3): 补充代码4 return sum1 #复化科特斯法 def func2(a,b,n,f): x = np.linspace(a,b,n+1) sum1 = 0 h =(b-a)/n for i in range( 补充代码5 ): 补充代码6 return sum1 answer = func(0,1,100,f) answer1 = func1(0,1,100,f) answer2 = func2(0,1,100,f) print(answer,answer1,answer2) #integrate积分作为精确值 value, error = integrate.quad(f,0, 1) print(value,error) print("error: ", abs(answer-value), abs(answer1-value), abs(answer2-value))import numpy as np import matplotlib.pyplot as plt plt.rcParams['font.sans-serif']=['SimHei'] #用来正常显示中文 标签 plt.rcParams['axes.unicode_minus']=False #用来正常显示负号 f=lambda x:np.sin(x)*np.sin(x) #向前差商 def fo_df(x,h): plt.plot([x-h,x],[f(x-h),f(x)],'r--',label='前') result= 补充代码7 return result #向后差商 def back_df(x,h): plt.plot([x,x+h],[f(x),f(x+h)],'k--',label='后') result= 补充代码8 return result#中心差商 def cen_df(x,h): a=(f(x-h)+f(x+h))/2 plt.plot([x-h,x+h],[f(x-h)+f(x)-a,f(x+h)+f(x)-a],'g--',label='中') result= 补充代码9 return result xx=np.linspace(-0.5,1.5,20) yy=f(xx) plt.plot(xx,yy) print('前',fo_df(0.5,0.5)) print('后',back_df(0.5,0.5)) print('中',cen_df(0.5,0.5)) plt.legend(loc='best') plt.show() xx=np.linspace(-1,1,20) yy=f(xx) plt.plot(xx,yy) print('前',fo_df(0.5,0.25)) print('后',back_df(0.5,0.25)) print('中',cen_df(0.5,0.25)) plt.legend(loc='best') plt.show()

代码中有一些没有补充的部分，我来帮你补充一下。

复化梯形法：

def func(a,b,n,f):
    x = np.linspace(a,b,n+1)
    sum1 = 0
    h =(b-a)/n
    for i in range(n):
        sum1 += (f(x[i])+f(x[i+1]))*h/2
    return sum1

复化辛普森法：

def func1(a,b,n,f):
    x = np.linspace(a,b,n+1)
    sum1 = 0
    h =(b-a)/n
    for i in range(n):
        sum1 += h/6*(f(x[i])+4*f((x[i]+x[i+1])/2)+f(x[i+1]))
    return sum1

复化科特斯法：

def func2(a,b,n,f):
    x = np.linspace(a,b,n+1)
    sum1 = 0
    h =(b-a)/n
    for i in range(n//2):
        sum1 += h/3*(f(x[2*i])+4*f((x[2*i]+x[2*i+1])/2)+f(x[2*i+1]))
    return sum1

向前差商：

def fo_df(x,h):
    plt.plot([x-h,x],[f(x-h),f(x)],'r--',label='前')
    result= (f(x)-f(x-h))/h
    return result

向后差商：

def back_df(x,h):
    plt.plot([x,x+h],[f(x),f(x+h)],'k--',label='后')
    result= (f(x+h)-f(x))/h
    return result

中心差商：

def cen_df(x,h):
    a=(f(x-h)+f(x+h))/2
    plt.plot([x-h,x+h],[f(x-h)+f(x)-a,f(x+h)+f(x)-a],'g--',label='中')
    result= (f(x+h)-f(x-h))/(2*h)
    return result

###function approximation f(x)=sin(x) ###2018.08.14 ###激活函数用的是sigmoid import numpy as np import math import matplotlib.pyplot as plt x = np.linspace(-3, 3, 600) # print(x) # print(x[1]) x_size = x.size y = np.zeros((x_size, 1)) # print(y.size) for i in range(x_size): y[i] = math.sin(2*math.pi*0.4*x[i])+ math.sin(2*math.pi*0.1*x[i]) + math.sin(2*math.pi*0.9*x[i]) # print(y) hidesize = 10 W1 = np.random.random((hidesize, 1)) # 输入层与隐层之间的权重 B1 = np.random.random((hidesize, 1)) # 隐含层神经元的阈值 W2 = np.random.random((1, hidesize)) # 隐含层与输出层之间的权重 B2 = np.random.random((1, 1)) # 输出层神经元的阈值 threshold = 0.005 max_steps = 1001 def sigmoid(x_): y_ = 1 / (1 + math.exp(-x_)) return y_ E = np.zeros((max_steps, 1)) # 误差随迭代次数的变化 Y = np.zeros((x_size, 1)) # 模型的输出结果 for k in range(max_steps): temp = 0 for i in range(x_size): hide_in = np.dot(x[i], W1) - B1 # 隐含层输入数据 # print(x[i]) hide_out = np.zeros((hidesize, 1)) # 隐含层的输出数据 for j in range(hidesize): # print("第{}个的值是{}".format(j,hide_in[j])) # print(j,sigmoid(j)) hide_out[j] = sigmoid(hide_in[j]) # print("第{}个的值是{}".format(j, hide_out[j])) # print(hide_out[3]) y_out = np.dot(W2, hide_out) - B2 # 模型输出 # print(y_out) Y[i] = y_out # print(i,Y[i]) e = y_out - y[i] # 模型输出减去实际结果。得出误差 ##反馈，修改参数 dB2 = -1 * threshold * e dW2 = e * threshold * np.transpose(hide_out) dB1 = np.zeros((hidesize, 1)) for j in range(hidesize): dB1[j] = np.dot(np.dot(W2[0][j], sigmoid(hide_in[j])), (1 - sigmoid(hide_in[j])) * (-1) * e * threshold) dW1 = np.zeros((hidesize, 1)) for j in range(hidesize): dW1[j] = np.dot(np.dot(W2[0][j], sigmoid(hide_in[j])), (1 - sigmoid(hide_in[j])) * x[i] * e * threshold) W1 = W1 - dW1 B1 = B1 - dB1 W2 = W2 - dW2 B2 = B2 - dB2 temp = temp + abs(e) E[k] = temp if k % 100 == 0: print(k) plt.figure() plt.plot(x, Y) plt.plot(x, Y, color='red', linestyle='--') plt.show()这个程序如何每迭代100次就输出一次图片

要每迭代100次输出一次图片，你需要在代码中添加相应的逻辑。以下是修改后的代码示例：

import numpy as np
import math
import matplotlib.pyplot as plt

x = np.linspace(-3, 3, 600)

x_size = x.size
y = np.zeros((x_size, 1))
for i in range(x_size):
    y[i] = math.sin(2*math.pi*0.4*x[i]) + math.sin(2*math.pi*0.1*x[i]) + math.sin(2*math.pi*0.9*x[i])

hidesize = 10
W1 = np.random.random((hidesize, 1))
B1 = np.random.random((hidesize, 1))
W2 = np.random.random((1, hidesize))
B2 = np.random.random((1, 1))
threshold = 0.005
max_steps = 1001

def sigmoid(x_):
    y_ = 1 / (1 + math.exp(-x_))
    return y_

E = np.zeros((max_steps, 1))
Y = np.zeros((x_size, 1))

for k in range(max_steps):
    temp = 0
    for i in range(x_size):
        hide_in = np.dot(x[i], W1) - B1
        hide_out = np.zeros((hidesize, 1))
        for j in range(hidesize):
            hide_out[j] = sigmoid(hide_in[j])

        y_out = np.dot(W2, hide_out) - B2
        Y[i] = y_out

        e = y_out - y[i]

        dB2 = -1 * threshold * e
        dW2 = e * threshold * np.transpose(hide_out)
        dB1 = np.zeros((hidesize, 1))
        for j in range(hidesize):
            dB1[j] = np.dot(np.dot(W2[0][j], sigmoid(hide_in[j])), (1 - sigmoid(hide_in[j])) * (-1) * e * threshold)

        dW1 = np.zeros((hidesize, 1))
        for j in range(hidesize):
            dW1[j] = np.dot(np.dot(W2[0][j], sigmoid(hide_in[j])), (1 - sigmoid(hide_in[j])) * x[i] * e * threshold)

        W1 = W1 - dW1
        B1 = B1 - dB1
        W2 = W2 - dW2
        B2 = B2 - dB2
        temp = temp + abs(e)

    E[k] = temp

    if k % 100 == 0:
        plt.figure()
        plt.plot(x, Y)
        plt.plot(x, Y, color='red', linestyle='--')
        plt.savefig(f'iteration_{k}.png')
        plt.close()

plt.figure()
plt.plot(x, Y)
plt.plot(x, Y, color='red', linestyle='--')
plt.show()

在上述示例中，我添加了一个条件判断语句`if k % 100 == 0`来判断是否达到每100次迭代的条件。如果满足条件，则在该迭代结束后绘制并保存图片，文件名为`iteration_k.png`，其中`k`表示当前迭代次数。

请根据你的具体需求和编程环境进行相应的调整。