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python - 温度传感器线性组合曲线拟合

转载 作者:行者123 更新时间:2023-12-01 01:36:21 24 4
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我正在尝试对不同的温度传感器进行线性组合,并使用应变传感器对其进行曲线化。

我所做的是可以安装一个温度传感器和一个应变传感器。

但我不知道如何在一个应变传感器上对不同温度传感器进行线性组合。

这是我的尝试:

def process_data_curve_fitting(temperature, strain):

#mean_T = (temperature[[i for i in temperature.columns.tolist() if str(i)[:2] == 'TW']].mean(axis=1))
print("process data")

T1 = temperature['T1'].tolist()
T2 = temperature['T2'].tolist()
T3 = temperature['T3'].tolist()
T4 = temperature['T4'].tolist()
T5 = temperature['T5'].tolist()
T6 = temperature['T6'].tolist()
T7 = temperature['T7'].tolist()
T8 = temperature['T8'].tolist()
T9 = temperature['T9'].tolist()
T10 = temperature['T10'].tolist()

df = pd.DataFrame(list(zip(T1, T2, T3, T4, T5, T6, T7, T8, T9, T10)))
mean_T = df.mean(axis = 1)

print(mean_T)
Sensor_Names = [ 'W_A1', 'W_A2', 'W_F1', 'W_F2', 'W_F4', 'W_S1', 'W_S2', 'W_S3', 'W_S4', 'W_KF1', 'W_KF2', 'W_KF3', 'W_KF4', 'W_DB1', 'W_DB2']
ys = []
for i in range(len(strain)):
cof = np.polyfit(mean_T, strain[i], 2)
poly = np.polyval(cof, mean_T)
ys.append(poly)
print (cof)
print (poly)

for i in range(len(strain)):
fig = plt.figure()
plt.scatter(mean_T, strain[i],s=0.1)
# fig.savefig(r'c:\\ahmed\\'+Sensor_Names[i]+'.png')
plt.plot(mean_T, ys[i], color='r')
fig.savefig(r'c:\\ahmed\\'+"Curve_fitting__" + Sensor_Names[i]+'.png',dpi=300)

plt.ylabel('strain' + Sensor_Names[i])
plt.xlabel('temperature')

请看一下等式 enter image description here

最佳答案

作为两个温度传感器的“概念验证”(这里既不添加噪音也不考虑实际参数):

import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import leastsq

def strain( t, a, b, c, d ):
return a * t**3 + b * t**2 + c * t + d

def residuals( params, x1Data, x2Data, yData ):
s1, s2, a, b, c, d = params
cxData = [ (s1**2 * x1 + s2**2 * x2) /( s1**2 + s2**2 ) for x1, x2 in zip( x1Data, x2Data) ]
diff = [ strain( x, a, b, c, d ) -y for x, y in zip( cxData, yData ) ]
return diff

timeList = np.linspace( 0, 25, 55 )
t1List = np.fromiter( ( 5 + 25. * (1 - np.exp( -t / 9. ) )for t in timeList ), np.float )
t2List = np.fromiter( (30. * (1 - np.exp( -t / 7. ) ) * ( 1 - np.exp( -t / 3. ) ) for t in timeList ), np.float )

combinedList = np.fromiter( ( (.7 * a + .2 * b)/.9 for a, b in zip( t1List, t2List ) ), np.float )
strainList = np.fromiter( ( strain( t, .01, -.1, .88, .2 ) for t in combinedList ), np.float )

fit, ier = leastsq( residuals, [.71,.22, 0,0, .1, .1 ], args=( t1List, t2List, strainList ), maxfev=5000 )
print fit
fittedT = [ (fit[0]**2 * x1 + fit[1]**2 *x2 ) /( fit[0]**2 + fit[1]**2 ) for x1, x2 in zip( t1List, t2List) ]
fittedS = [ strain( t, *(fit[2:]) ) for t in fittedT ]

fig = plt.figure()
ax = fig.add_subplot( 3, 1, 1 )
bx = fig.add_subplot( 3, 1, 2 )
cx = fig.add_subplot( 3, 1, 3 )
ax.plot( timeList, t1List )
ax.plot( timeList, t2List )
ax.plot( timeList, combinedList )
bx.plot( combinedList, strainList, linestyle='', marker='x' )
bx.plot( fittedT, fittedS )
cx.plot( timeList, fittedT ,'--')
cx.plot( timeList, combinedList,':' )
plt.show()

给予

[ 4.21350842e+03  2.25221499e+03  1.00000000e-02 -1.00000000e-01 8.80000000e-01  2.00000000e-01]

并显示:

fit

顶部:温度 1(蓝色)和 2(橙色)以及线性组合(绿色)中心:“模拟数据”(蓝色)和拟合(橙色)底部:拟合温度(蓝色)、真实温度(橙色)

根据实际数据,可能需要一些调整。

关于python - 温度传感器线性组合曲线拟合,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/52364406/

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