如何拟合温度/热分布数据? [英] How can fit the data on temperature/thermal profile?
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问题描述
我有一个包含某个温度曲线的数据集，我想在温度曲线上拟合或映射测量点，如下所示:
停留时间:30 分钟
斜坡时间:1 分钟
周期数:1000 个周期
测量点周期:16 分钟
测量点可以发生在高位 +150 或低位 -40
注意: T0(初始时间)不明确，所以时间参考不明确，例如.T0=0.
我已经在 Pandas DataFrame 中获取了数据:
将 numpy 导入为 np将熊猫导入为 pd从 scipy.optimize 导入 curve_fitdf = pd.read_csv('D:SOF.csv', header=None)数据 = {'A': A[:,0], 'B': B[:,0], '温度': 温度 [:,0],'S':S, 'C':C , 'Measurement_Points':MP}dff = pd.DataFrame(data, columns=['A','B','Temperature','S','C','MP'], index = id_set[:,0])# 温度范围是 [-40,+150]# MP 的范围是 [0-3000] 从第一个 MP 到最后一个MP = int(len(dff)/480) # 计算测量点数打印(MP)范围内循环(MP):j = 周期 * 480#使用来自 DataFrame 的温度列的每 480 个值的平均值或平均值来传递以适应热剖面Mean_temp = np.mean(df['Temperature'].iloc[j:j+480]) # 使用来自 numpy 的 Mean#Mean_temp = df.groupby('Temperature').mean() #by 使用 groupby到目前为止，我只是从 scipy.optimize 中找到了 curve_fit 基于这个
解决方案
这将是我的起点:
将 matplotlib.pyplot 导入为 plt将 numpy 导入为 np从 scipy.optimize 导入 curve_fit###生成测试数据def temp( t , 低, 高, 周期, 斜坡):tRed = t % 周期停留 = 周期/2. - 斜坡如果 tRed <住:出 = 高elif tRed <停留+斜坡:输出 = 高 - ( tRed - 停留 )/斜坡 * ( 高 - 低 )elif tRed <2 * 停留 + 斜坡:出 = 低elif tRed <= 句号:输出 = 低 + ( tRed - 2 * 停留 - 斜坡)/斜坡 * ( 高 - 低 )别的:断言 0返回 + np.random.normal()### 有点拟合数据的连续函数### 但最终会得到周期和水平.### 坡道的定义不太明确def fit_func( t, low, high, period, s, delta):回报(高+低)/2.+(高-低)/2.* np.tanh( s * np.sin( 2 * np.pi * ( t - delta )/period ) )time1List = np.arange( 300 ) * 16time2List = np.linspace( 0, 300 * 16, 7213 )tempList = np.fromiter( ( temp(t - 6.3 , 41, 155, 63.3, 2.05 ) for t in time1List ), np.float )funcList = np.fromiter( ( fit_func(t , 41, 155, 63.3, 10., 0 ) for t in time2List ), np.float )溶胶, 错误 = 曲线拟合 (fit_func, time1List, tempList, [ 40, 150, 63, 10, 0 ] )打印溶胶拟合低、拟合高、拟合周期、拟合S、拟合关闭 = solrealHigh = fit_func( fitPeriod/4., *sol)realLow = fit_func( 3/4. * fitedPeriod, *sol)打印高，低:"，[ realHigh, realLow ]打印apprx 斜坡:"，fitPeriod/(2 * np.pi * fitS ) * 2realAmp = realHigh - realLow斜坡X，斜坡Y = zip(* [ [ t，d ] for t，d in zip(time1List，tempList)如果((d realLow + 0.05 * realAmp)))])topX, topY = zip( *[ [ t, d ] for t, d in zip( time1List, tempList ) if ( ( ( d > realHigh - 0.05 * realAmp ) ) ] )botX, botY = zip( *[ [ t, d ] for t, d in zip( time1List, tempList ) if ( ( ( d 提供:
<代码>>>[155.0445024 40.7417905 63.29983807 13.07677546 -26.36945489]>>高、低:[155.04450237880076, 40.741790521444436]>>大约斜坡:1.540820542195840
有几点需要注意.如果斜坡与停留时间相比较小，我的拟合函数效果会更好.此外，您会在这里找到几篇讨论阶跃函数拟合的帖子.一般来说，由于拟合需要有意义的导数，离散函数是一个问题.至少有两种解决方案.a) 制作一个连续版本，拟合，并使结果根据您的喜好离散或 b) 提供离散函数和手动连续导数.
编辑
所以这是我使用您新发布的数据集所做的工作:
将 matplotlib.pyplot 导入为 plt将 numpy 导入为 np从 scipy.optimize 导入 curve_fit，最小化def 分区( inList, n ):返回 zip( *[ iter( inList ) ] * n )def temp( t, 低, 高, 周期, 斜坡, 关闭):tRed = (t - off ) % 周期停留 = 周期/2. - 斜坡如果 tRed <住:出 = 高elif tRed <停留+斜坡:输出 = 高 - ( tRed - 停留 )/斜坡 * ( 高 - 低 )elif tRed <2 * 停留 + 斜坡:出 = 低elif tRed <= 句号:输出 = 低 + ( tRed - 2 * 停留 - 斜坡)/斜坡 * ( 高 - 低 )别的:断言 0回来def chi2( params, xData=None, yData=None, verbose=False ):低、高、周期、斜坡、关闭 = 参数th = np.fromiter( ( temp( t, low, high, period,ramp, off ) for t in xData ), np.float )差异 = ( th - yData )diff2 = diff**2out = np.sum( diff2 )如果冗长:打印 ' - - - - - -'打印打印差异打印差异2打印 ' - - - - - -'回来# ~ 返回def fit_func( t, low, high, period, s, delta):回报(高+低)/2.+(高-低)/2.* np.tanh( s * np.sin( 2 * np.pi * ( t - delta )/period ) )inData = np.loadtxt('SOF2.csv', skiprows=1, delimiter=',' )inData2 = inData[:, 2]xList = np.arange( len(inData2) )inData480 = 分区( inData2, 480 )xList480 = 分区( xList, 480 )inDataMean = np.fromiter( (np.mean( x ) for x in inData480 ), np.float )xMean = np.arange( len( inDataMean) ) * 16time1List = np.linspace( 0, 16 * len(inDataMean), 500)溶胶, 错误 = 曲线拟合 (fit_func, xMean, inDataMean, [ -40, 150, 60, 10, 10 ] )打印溶胶# ~ 打印 chi2([-49,155,62.5,1, 8.6], xMean, inDataMean )res = 最小化( chi2, [-44.12, 150.0, 62.0, 8.015, 12.3 ], args=( xMean, inDataMean ), method='nelder-mead' )# ~ 打印资源打印 res.x# ~ 打印 chi2( res.x, xMean, inDataMean, verbose=True )# ~ 打印 chi2( [-44.12, 150.0, 62.0, 8.015, 6.3], xMean, inDataMean, verbose=True )fig = plt.figure()ax = fig.add_subplot( 2, 1, 1 )bx = fig.add_subplot( 2, 1, 2 )对于 zip 中的 x,y(xList480, inData480):ax.plot( x, y, 标记='x', linestyle='', zorder=100 )bx.plot( xMean, inDataMean , marker='x', linestyle='' )bx.plot( time1List, fit_func( time1List, *sol ) )bx.plot( time1List, np.fromiter( ( temp( t , *res.x ) for t in time1List ), np.float) )bx.plot( time1List, np.fromiter( ( temp( t , -44.12, 150.0, 62.0, 8.015, 12.3 ) for t in time1List ), np.float) )plt.show()<代码>>>[-49.53569904 166.92138068 62.56131027 1.8547409 8.75673747]>>[-34.12188737 150.02194584 63.81464913 8.26491754 13.88344623]
如您所见，斜坡上的数据点不适合.那么，16 分钟的时间可能不是那么恒定?这将是一个问题，因为这不是局部 x 错误，而是累积效应.
I have a dataset consisting of a certain temperature profile and I wanna fit or map the measurement points on temperature profile which is following:
Dwell-time: 30 mins
Ramp-time: 1 min
Number of periods: 1000 cycles
Measure points period: 16 mins
Measure points can be happened in either in high regim +150 or low regim -40
Note: The T0 (initial time) is not clear so time reference is not clear eg. T0=0 .
I already fetched the data in Pandas DataFrame:
import numpy as np
import pandas as pd
from scipy.optimize import curve_fit
df = pd.read_csv('D:SOF.csv', header=None)
data = {'A': A[:,0], 'B': B[:,0], 'Temperature': Temperature[:,0],
'S':S, 'C':C , 'Measurement_Points':MP}
dff = pd.DataFrame(data, columns=['A','B','Temperature','S','C','MP'], index = id_set[:,0])
# Temperature's range is [-40,+150]
# MP's range is [0-3000] from 1st MP till last one
MP = int(len(dff)/480) # calculate number of measurement points
print(MP)
for cycle in range(MP):
j = cycle * 480
#use mean or average of each 480 values from temperature column of DataFrame to pass for fit on Thermal profile
Mean_temp = np.mean(df['Temperature'].iloc[j:j+480]) # by using Mean from numpy
#Mean_temp = df.groupby('Temperature').mean() #by using groupby
So far I just find curve_fit from scipy.optimize based on this answer and this post
but I am wondering how the fitting process could work right here in the other hand I would like Temperature values round only to the nearest either -40 or +150 .
I would be nice if someone can help me!
Update: Standard periodic Thermal profile pic is following:
expected result:
updated Data sample: data
解决方案
This would be my starting point:
import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import curve_fit
### to generate test data
def temp( t , low, high, period, ramp ):
tRed = t % period
dwell = period / 2. - ramp
if tRed < dwell:
out = high
elif tRed < dwell + ramp:
out = high - ( tRed - dwell ) / ramp * ( high - low )
elif tRed < 2 * dwell + ramp:
out = low
elif tRed <= period:
out = low + ( tRed - 2 * dwell - ramp)/ramp * ( high -low )
else:
assert 0
return out + np.random.normal()
### A continuous function that somewhat fits the data
### but definitively gets the period and levels.
### The ramp is less well defined
def fit_func( t, low, high, period, s, delta):
return ( high + low ) / 2. + ( high - low )/2. * np.tanh( s * np.sin( 2 * np.pi * ( t - delta ) / period ) )
time1List = np.arange( 300 ) * 16
time2List = np.linspace( 0, 300 * 16, 7213 )
tempList = np.fromiter( ( temp(t - 6.3 , 41, 155, 63.3, 2.05 ) for t in time1List ), np.float )
funcList = np.fromiter( ( fit_func(t , 41, 155, 63.3, 10., 0 ) for t in time2List ), np.float )
sol, err = curve_fit( fit_func, time1List, tempList, [ 40, 150, 63, 10, 0 ] )
print sol
fittedLow, fittedHigh, fittedPeriod, fittedS, fittedOff = sol
realHigh = fit_func( fittedPeriod / 4., *sol)
realLow = fit_func( 3 / 4. * fittedPeriod, *sol)
print "high, low : ", [ realHigh, realLow ]
print "apprx ramp: ", fittedPeriod/( 2 * np.pi * fittedS ) * 2
realAmp = realHigh - realLow
rampX, rampY = zip( *[ [ t, d ] for t, d in zip( time1List, tempList ) if ( ( d < realHigh - 0.05 * realAmp ) and ( d > realLow + 0.05 * realAmp ) ) ] )
topX, topY = zip( *[ [ t, d ] for t, d in zip( time1List, tempList ) if ( ( d > realHigh - 0.05 * realAmp ) ) ] )
botX, botY = zip( *[ [ t, d ] for t, d in zip( time1List, tempList ) if ( ( d < realLow + 0.05 * realAmp ) ) ] )
fig = plt.figure()
ax = fig.add_subplot( 2, 1, 1 )
bx = fig.add_subplot( 2, 1, 2 )
ax.plot( time1List, tempList, marker='x', linestyle='', zorder=100 )
ax.plot( time2List, fit_func( time2List, *sol ), zorder=0 )
bx.plot( time1List, tempList, marker='x', linestyle='' )
bx.plot( time2List, fit_func( time2List, *sol ) )
bx.plot( rampX, rampY, linestyle='', marker='o', markersize=10, fillstyle='none', color='r')
bx.plot( topX, topY, linestyle='', marker='o', markersize=10, fillstyle='none', color='#00FFAA')
bx.plot( botX, botY, linestyle='', marker='o', markersize=10, fillstyle='none', color='#80DD00')
bx.set_xlim( [ 0, 800 ] )
plt.show()
providing:
>> [155.0445024 40.7417905 63.29983807 13.07677546 -26.36945489]
>> high, low : [155.04450237880076, 40.741790521444436]
>> apprx ramp: 1.540820542195840
There is a few things to note. My fit function works better if the ramp is small compared to the dwell time. Moreover, one will find several posts here where the fitting of step functions is discussed. In general, as fitting requires a meaningful derivative, discrete functions are a problem. There are at least two solutions. a) make a continuous version, fit, and make the result discrete to your liking or b) provide a discrete function and a manual continuous derivative.
EDIT
So here is what I get working with your newly posted data set:
import matplotlib.pyplot as plt
import numpy as np
from scipy.optimize import curve_fit, minimize
def partition( inList, n ):
return zip( *[ iter( inList ) ] * n )
def temp( t, low, high, period, ramp, off ):
tRed = (t - off ) % period
dwell = period / 2. - ramp
if tRed < dwell:
out = high
elif tRed < dwell + ramp:
out = high - ( tRed - dwell ) / ramp * ( high - low )
elif tRed < 2 * dwell + ramp:
out = low
elif tRed <= period:
out = low + ( tRed - 2 * dwell - ramp)/ramp * ( high -low )
else:
assert 0
return out
def chi2( params, xData=None, yData=None, verbose=False ):
low, high, period, ramp, off = params
th = np.fromiter( ( temp( t, low, high, period, ramp, off ) for t in xData ), np.float )
diff = ( th - yData )
diff2 = diff**2
out = np.sum( diff2 )
if verbose:
print '-----------'
print th
print diff
print diff2
print '-----------'
return out
# ~ return th
def fit_func( t, low, high, period, s, delta):
return ( high + low ) / 2. + ( high - low )/2. * np.tanh( s * np.sin( 2 * np.pi * ( t - delta ) / period ) )
inData = np.loadtxt('SOF2.csv', skiprows=1, delimiter=',' )
inData2 = inData[ :, 2 ]
xList = np.arange( len(inData2) )
inData480 = partition( inData2, 480 )
xList480 = partition( xList, 480 )
inDataMean = np.fromiter( (np.mean( x ) for x in inData480 ), np.float )
xMean = np.arange( len( inDataMean) ) * 16
time1List = np.linspace( 0, 16 * len(inDataMean), 500 )
sol, err = curve_fit( fit_func, xMean, inDataMean, [ -40, 150, 60, 10, 10 ] )
print sol
# ~ print chi2([-49,155,62.5,1 , 8.6], xMean, inDataMean )
res = minimize( chi2, [-44.12, 150.0, 62.0, 8.015, 12.3 ], args=( xMean, inDataMean ), method='nelder-mead' )
# ~ print res
print res.x
# ~ print chi2( res.x, xMean, inDataMean, verbose=True )
# ~ print chi2( [-44.12, 150.0, 62.0, 8.015, 6.3], xMean, inDataMean, verbose=True )
fig = plt.figure()
ax = fig.add_subplot( 2, 1, 1 )
bx = fig.add_subplot( 2, 1, 2 )
for x,y in zip( xList480, inData480):
ax.plot( x, y, marker='x', linestyle='', zorder=100 )
bx.plot( xMean, inDataMean , marker='x', linestyle='' )
bx.plot( time1List, fit_func( time1List, *sol ) )
bx.plot( time1List, np.fromiter( ( temp( t , *res.x ) for t in time1List ), np.float) )
bx.plot( time1List, np.fromiter( ( temp( t , -44.12, 150.0, 62.0, 8.015, 12.3 ) for t in time1List ), np.float) )
plt.show()
>> [-49.53569904 166.92138068 62.56131027 1.8547409 8.75673747]
>> [-34.12188737 150.02194584 63.81464913 8.26491754 13.88344623]
As you can see, the data point on the ramp does not fit in. So, it might be that the 16 min time is not that constant? That would be a problem as this is not a local x-error but an accumulating effect.
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