python - Solve equation with a set of points -

i have set of points. basically, have p = f(t).

i have, let's say, 50 measurements. 50 values of p, function of time. , these values follow established law.

what have find values of parameters in law, that's all. basically, have fit points best curve. here law:

p = v.t - ((v - w)(1 - exp(-k.t)) / k) 

what need find numeric value v, w , k. have t , p. have idea how that?

edit:

here screenshot of want obtain:

on picture:

• v vs
• w vi
• k k

and that's obtained reptilicus's help:

http://i.imgur.com/f59eo29.png

import numpy np scipy.optimize import curve_fit matplotlib.pyplot import * import xlrd  def myfunc(t, v, w, k):      y = v * t - ((v - w) * (1 - np.exp(-k * t)) / k)      return y  classeur = xlrd.open_workbook(path) names_sheets = classeur.sheet_names()  sheet = classeur.sheet_by_name(names_sheets[0])  row_start = 2  time = sheet.col_values(0, row_start) fluo = feuille.col_values(4, row_start)  time = [ index index in time if index ] fluo = [ index index in fluo if index ]  # generates fake data fit. youm read in  # data in csv or whatever you've x = np.array(time) y = np.array(fluo)  #fit data, return best fit parameters , covariance matrix #popt, pcov = curve_fit(myfunc, x, yn) popt, pcov = curve_fit(myfunc, x, y) print(popt) print(pcov)  #plot data clf() #matplotlib plot(x, y, "rs") #overplot best fit curve plot(x, myfunc(x, popt[0], popt[1], popt[2])) grid(true) show() 

not bad. managed extract data of excel workbook, , plotted it. can see, got linear regression, don't want. goal t reproduce fit got origin 8.

edit:

i have news. last guy did in team told me how did origin. in fact, use least square way well, find parameters chi 2 minimization. software iterations, , optimizes parameters.

edit 2:

because took me long figure out, share here results of researches. main problem facing fact values "too small". indeed, y values of order of 10^-7. explained here fitting curve: why small numbers better?, numbers of order of 1 better fit.

moreover, in case @ least, data of order, didn't need give initial parameters (by default, set 1). "normalized" values. example, transformed time values seconds hour, , multiplicated 10^7 y values, of order of 10^-7. then, transformed obtained parameters them in desired unities. here code:

import numpy np scipy.optimize import curve_fit, leastsq matplotlib.pyplot import *  def myfunc(t, vs, vi, k):      y = vs * t - ((vs - vi) * (1 - np.exp(-k * t)) / k)      return y  raw_x = some_input  raw_y = some_input   # scaling data time = [ index /3600 index in raw_x if index or index==0 ] fluo = [ index*10**7 index in raw_y if index or index==0 ]  x = np.array(temps) y = np.array(fluo)  popt, pcov = curve_fit(myfunc, x, y, maxfev=3000)  # unities popt2 = list() popt2 = [ popt[0] / 3600 * 10**-7, popt[1] / 3600 * 10**-7, popt[2] / 3600 ]  #plot data clf() #matplotlib plot(raw_x, raw_y, "rp") plot(raw_x, myfunc(raw_x, popt2[0], popt2[1], popt2[2]), 'b') grid(true) show() 

and here picture illustrating difference:

http://i.imgur.com/yxkjg5j.png

the blue plot fitting curve using parameters obtained units rescaling (and transformed in unities). green 1 curve obtained fitting in original unities.

thanks of help.

just use curve_fit in scipy.optimize:

import numpy np scipy.optimize import curve_fit pylab import *  def myfunc(t, v, w, k):     y = v * t - ((v - w) * (1 - np.exp(-k * t)) / k)     return y  # generates fake data fit. youm read in  # data in csv or whatever you've x = np.linspace(0,4,50) y = myfunc(x, 2.5, 1.3, 0.5) # add noise fake data make more realistic. . . yn = y + 0.2*np.random.normal(size=len(x))  #fit data, return best fit parameters , covariance matrix popt, pcov = curve_fit(myfunc, x, yn) print popt print pcov  #plot data clf() plot(x, yn, "rs") #overplot best fit curve plot(x, myfunc(x, popt[0], popt[1], popt[2])) grid(true) show() 

this gives plot below. red points (noisy) data, , blue line best fit curve, best fitting parameters particular data of:

[ 2.32751132, 1.27686053, 0.65986596] 

which pretty close expected parameters of 2.5, 1.3, 0.5. difference due noise added in fake data.