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Python how to control curvature when joining two points

I have a original curve. I am developing a model curve matching closely the original curve. Everything is working fine but not matching. How to control the curvature of my model curve? Below code is based on answer here.
My code:
def curve_line(point1, point2):
a = (point2[1] - point1[1])/(np.cosh(point2[0]) - np.cosh(point1[0]))
b = point1[1] - a*np.sinh(point1[0])
x = np.linspace(point1[0], point2[0],100).tolist()
y = (a*np.cosh(x) + b).tolist()
return x,y
###### A sample of my code is given below
point1 = [10,100]
point2 = [20,50]
x,y = curve_line(point1, point2)
plt.plot(point1[0], point1[1], 'o')
plt.plot(point2[0], point2[1], 'o')
plt.plot(x,y) ## len(x)
My present output:
I tried following function as well:
y = (50*np.exp(-x/10) +2.5)
The output is:

Instead of just guessing the right parameters of your model function, you can fit a model curve to your data using curve_fit.
import numpy as np
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
x = np.array([ 1.92, 14.35, 21.50, 25.27, 27.34, 30.32, 32.31, 34.09, 34.21])
y = np.array([8.30, 8.26, 8.13, 7.49, 6.66, 4.59, 2.66, 0.60, 0.06])
def fun(x, a, b, c):
return a * np.cosh(b * x) + c
coef,_ = curve_fit(fun, x, y)
plt.plot(x, y, label='Original curve')
plt.plot(x, fun(x, *coef), label=f'Model: %5.3f cosh(%4.2f x + %4.2f)' % tuple(coef) )
plt.legend()
plt.show()
If it is important that the start and end points are closely fitted, you can pass uncertainties to curve_fit, adjusting them to lower values towards the ends, e.g. by
s = np.ones(len(x))
s[1:-1] = s[1:-1] * 3
coef,_ = curve_fit(fun, x, y, sigma=s)
Your other approach a * np.exp(b * x) + c will also work and gives -0.006 exp(0.21 x + 8.49).
In some cases you'll have to provide an educated guess for the initial values of the coefficients to curve_fit (it uses 1 as default).

Curve fitting multiple outputs from a single function with scipy

OK, I have a function which uses a range of parameters to calculate the effect on two separate variables over time. These variables have already been curve-matched to some existing data to minimize the variation (shown below)
I want to be able to check the previous working, and match new data. I have been trying to use the scipy.optimize.curve_fit function, by stacking the x and y data resulting from my function (as suggested here: fit multiple parametric curves with scipy).
It may not be the right method, or I may just be misunderstanding, but my code keeps running into a type error TypeError: Improper input: N=3 must not exceed M=2
My simplified prototype code was initially taken from here: https://docs.scipy.org/doc/scipy/reference/generated/scipy.optimize.curve_fit.html
import numpy as np
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
def func(x, a, b, c):
result = ([],[])
for i in x:
#set up 2 example curves
result[0].append(a * np.exp(-b * i) + c)
result[1].append(a * np.exp(-b * i) + c**2)
return result #as a tuple containing 2 lists
#Define the data to be fit with some noise:
xdata = list(np.arange(0, 10, 1))
y = func(xdata, 2.5, 5, 0.5)[0]
y2 = func(xdata, 1, 1, 2)[1]
#Add some noise
y_noise = 0.1 * np.random.normal(size=len(xdata))
y2_noise = 0.1 * np.random.normal(size=len(xdata))
ydata=[]
ydata2=[]
for i in range(len(y)): #clunky
ydata.append(y[i] + y_noise[i])
ydata2.append(y2[i] + y2_noise[i])
plt.scatter(xdata, ydata, label='data')
plt.scatter(xdata, ydata2, label='data2')
#plt.plot(xdata, y, 'k-', label='data (original function)')
#plt.plot(xdata, y2, 'k-', label='data2 (original function)')
#stack the data
xdat = xdata+xdata
ydat = ydata+ydata2
popt, pcov = curve_fit(func, xdat, ydat)
plt.plot(xdata, func(xdata, *popt), 'r-',
label='fit: a=%5.3f, b=%5.3f, c=%5.3f' % tuple(popt))
plt.xlabel('x')
plt.ylabel('y')
plt.legend()
plt.show()
Any help much appreciated !

Here is graphing example code that fits two different equations with a single shared parameter, if this looks like what you need it can easily be adapted for your specific problem.
import numpy as np
import matplotlib
import matplotlib.pyplot as plt
from scipy.optimize import curve_fit
y1 = np.array([ 16.00, 18.42, 20.84, 23.26])
y2 = np.array([-20.00, -25.50, -31.00, -36.50, -42.00])
comboY = np.append(y1, y2)
x1 = np.array([5.0, 6.1, 7.2, 8.3])
x2 = np.array([15.0, 16.1, 17.2, 18.3, 19.4])
comboX = np.append(x1, x2)
if len(y1) != len(x1):
raise(Exception('Unequal x1 and y1 data length'))
if len(y2) != len(x2):
raise(Exception('Unequal x2 and y2 data length'))
def function1(data, a, b, c): # not all parameters are used here, c is shared
return a * data + c
def function2(data, a, b, c): # not all parameters are used here, c is shared
return b * data + c
def combinedFunction(comboData, a, b, c):
# single data reference passed in, extract separate data
extract1 = comboData[:len(x1)] # first data
extract2 = comboData[len(x1):] # second data
result1 = function1(extract1, a, b, c)
result2 = function2(extract2, a, b, c)
return np.append(result1, result2)
# some initial parameter values
initialParameters = np.array([1.0, 1.0, 1.0])
# curve fit the combined data to the combined function
fittedParameters, pcov = curve_fit(combinedFunction, comboX, comboY, initialParameters)
# values for display of fitted function
a, b, c = fittedParameters
y_fit_1 = function1(x1, a, b, c) # first data set, first equation
y_fit_2 = function2(x2, a, b, c) # second data set, second equation
plt.plot(comboX, comboY, 'D') # plot the raw data
plt.plot(x1, y_fit_1) # plot the equation using the fitted parameters
plt.plot(x2, y_fit_2) # plot the equation using the fitted parameters
plt.show()
print('a, b, c:', fittedParameters)

Integration of a Gaussian function to count the number of particle under area

I need to count the number of particle under the fitted Gaussian curve. The area of the fitted curve can be found by integrating the function within the limit (mean-3*sigma) to (mean+3*sigma). Would you please help me to solve this. Thanks for your kind consideration.
import pylab as py
import numpy as np
from scipy import optimize
from scipy.stats import stats
import matplotlib.pyplot as plt
import pandas as pd
BackPFT='T067.csv'
df_180 = pd.read_csv(BackPFT, error_bad_lines=False, header=1)
x_180=df_180.iloc[:,3]
y_180=df_180.iloc[:,4]
#want to plot the distribution of s calculated by the following equation
s=np.sqrt((((16*x_180**2*38.22**2)/((4*38.22**2-y_180**2)**2))+1))-1
#Shape of this distribution is Gaussian
#I need to fit this distribution by following parameter
mean=0.433
sigma=0.014
draw=s
#Definition of bin number
bi=np.linspace(0.01,8, 1000)
data = py.hist(draw.dropna(), bins = bi)
#Definition of Gaussian function
def f(x, a, b, c):
return (a * py.exp(-(x - mean)**2.0 / (2 *sigma**2)))
x = [0.5 * (data[1][i] + data[1][i+1]) for i in xrange(len(data[1])-1)]
y = data[0]
#Fitting the peak of the distribution
popt, pcov = optimize.curve_fit(f, x, y)
chi2, p = stats.chisquare(popt)
x_fit = py.linspace(x[0], x[-1], 80000)
y_fit = f(x_fit, *popt)
plot(x_fit, y_fit, lw=3, color="r",ls="--")
plt.xlim(0,2)
plt.tick_params(axis='both', which='major', labelsize=20)
plt.show()
The problem is how to integrate the defined function (f) and count the number under the area. Here I attach the file T067.csv. Thanks in advance for your kind consideration.

BackPFT='T061.csv'
df_180 = pd.read_csv(BackPFT, skip_blank_lines=True ,skiprows=1,header=None,skipfooter=None,engine='python')
x_180=df_180.iloc[:,3]
y_180=df_180.iloc[:,4]
b=42.4
E=109.8
LET=24.19
REL=127.32
mean=0.339; m1=0.259
sigma=0.012; s1=0.015
s=np.sqrt((((16*x_180**2*b**2)/((4*b**2-y_180**2)**2))+1))-1
draw=s
bi=np.linspace(0,8, 2000)
binwidth=0.004
#I want to plot the dsitribution of s. This distribution has three gaussian peaks
data = py.hist(draw.dropna(), bins = bi,color='gray',)
#first Gaussian function for the first peak (peaks counted from the right)
def f(x, a, b, c):
return (a * py.exp(-(x - mean)**2.0 / (2 *sigma**2)))
# fitting the function (Gaussian)
x = [0.5 * (data[1][i] + data[1][i+1]) for i in xrange(len(data[1])-1)]
y = data[0]
popt, pcov = optimize.curve_fit(f, x, y)
chi, p = stats.chisquare(popt)
x_fit = py.linspace(x[0], x[-1], 80000)
y_fit = f(x_fit, *popt)
plot(x_fit, y_fit, lw=5, color="r",ls="--")
#integration of first function f
gaussF = lambda x, a: f(x, a, sigma, mean)
bins=((6*sigma)/(binwidth))
delta = ((mean+3*sigma) - (mean-3*sigma))/bins
f1 = lambda x : f(x, popt[0], sigma, mean)
result = quad(f1,mean-3*sigma,mean+3*sigma)
area = result[0] # this give the area after integration of the gaussian
numPar = area / delta # this gives the number of particle under the integrated area
print"\n\tArea under curve = ", area, "\n\tNumber of particel= ", numPar
The file T061.csv here. Thanks Dr. I Putu Susila for his kind co-operation and interest.

numpy polyfit passing through 0

Suppose I have x and y vectors with a weight vector wgt. I can fit a cubic curve (y = a x^3 + b x^2 + c x + d) by using np.polyfit as follows:
y_fit = np.polyfit(x, y, deg=3, w=wgt)
Now, suppose I want to do another fit, but this time, I want the fit to pass through 0 (i.e. y = a x^3 + b x^2 + c x, d = 0), how can I specify a particular coefficient (i.e. d in this case) to be zero?
Thanks

You can try something like the following:
Import curve_fit from scipy, i.e.
from scipy.optimize import curve_fit
import matplotlib.pyplot as plt
import numpy as np
Define the curve fitting function. In your case,
def fit_func(x, a, b, c):
# Curve fitting function
return a * x**3 + b * x**2 + c * x # d=0 is implied
Perform the curve fitting,
# Curve fitting
params = curve_fit(fit_func, x, y)
[a, b, c] = params[0]
x_fit = np.linspace(x[0], x[-1], 100)
y_fit = a * x_fit**3 + b * x_fit**2 + c * x_fit
Plot the results if you please,
plt.plot(x, y, '.r') # Data
plt.plot(x_fit, y_fit, 'k') # Fitted curve
It does not answer the question in the sense that it uses numpy's polyfit function to pass through the origin, but it solves the problem.
Hope someone finds it useful :)

You can use np.linalg.lstsq and construct your coefficient matrix manually. To start, I'll create the example data x and y, and the "exact fit" y0:
import numpy as np
import matplotlib.pyplot as plt
x = np.arange(100)
y0 = 0.07 * x ** 3 + 0.3 * x ** 2 + 1.1 * x
y = y0 + 1000 * np.random.randn(x.shape[0])
Now I'll create a full cubic polynomial 'training' or 'independent variable' matrix that includes the constant d column.
XX = np.vstack((x ** 3, x ** 2, x, np.ones_like(x))).T
Let's see what I get if I compute the fit with this dataset and compare it to polyfit:
p_all = np.linalg.lstsq(X_, y)[0]
pp = np.polyfit(x, y, 3)
print np.isclose(pp, p_all).all()
# Returns True
Where I've used np.isclose because the two algorithms do produce very small differences.
You're probably thinking 'that's nice, but I still haven't answered the question'. From here, forcing the fit to have a zero offset is the same as dropping the np.ones column from the array:
p_no_offset = np.linalg.lstsq(XX[:, :-1], y)[0] # use [0] to just grab the coefs
Ok, let's see what this fit looks like compared to our data:
y_fit = np.dot(p_no_offset, XX[:, :-1].T)
plt.plot(x, y0, 'k-', linewidth=3)
plt.plot(x, y_fit, 'y--', linewidth=2)
plt.plot(x, y, 'r.', ms=5)
This gives this figure,
WARNING: When using this method on data that does not actually pass through (x,y)=(0,0) you will bias your estimates of your output solution coefficients (p) because lstsq will be trying to compensate for that fact that there is an offset in your data. Sort of a 'square peg round hole' problem.
Furthermore, you could also fit your data to a cubic only by doing:
p_ = np.linalg.lstsq(X_[:1, :], y)[0]
Here again the warning above applies. If your data contains quadratic, linear or constant terms the estimate of the cubic coefficient will be biased. There can be times when - for numerical algorithms - this sort of thing is useful, but for statistical purposes my understanding is that it is important to include all of the lower terms. If tests turn out to show that the lower terms are not statistically different from zero that's fine, but for safety's sake you should probably leave them in when you estimate your cubic.
Best of luck!

Python - calculating trendlines with errors

So I've got some data stored as two lists, and plotted them using
plot(datasetx, datasety)
Then I set a trendline
trend = polyfit(datasetx, datasety)
trendx = []
trendy = []
for a in range(datasetx[0], (datasetx[-1]+1)):
trendx.append(a)
trendy.append(trend[0]*a**2 + trend[1]*a + trend[2])
plot(trendx, trendy)
But I have a third list of data, which is the error in the original datasety. I'm fine with plotting the errorbars, but what I don't know is using this, how to find the error in the coefficients of the polynomial trendline.
So say my trendline came out to be 5x^2 + 3x + 4 = y, there needs to be some sort of error on the 5, 3 and 4 values.
Is there a tool using NumPy that will calculate this for me?

I think you can use the function curve_fit of scipy.optimize (documentation). A basic example of the usage:
import numpy as np
from scipy.optimize import curve_fit
def func(x, a, b, c):
return a*x**2 + b*x + c
x = np.linspace(0,4,50)
y = func(x, 5, 3, 4)
yn = y + 0.2*np.random.normal(size=len(x))
popt, pcov = curve_fit(func, x, yn)
Following the documentation, pcov gives:
The estimated covariance of popt. The diagonals provide the variance
of the parameter estimate.
So in this way you can calculate an error estimate on the coefficients. To have the standard deviation you can take the square root of the variance.
Now you have an error on the coefficients, but it is only based on the deviation between the ydata and the fit. In case you also want to account for an error on the ydata itself, the curve_fit function provides the sigma argument:
sigma : None or N-length sequence
If not None, it represents the standard-deviation of ydata. This
vector, if given, will be used as weights in the least-squares
problem.
A complete example:
import numpy as np
from scipy.optimize import curve_fit
def func(x, a, b, c):
return a*x**2 + b*x + c
x = np.linspace(0,4,20)
y = func(x, 5, 3, 4)
# generate noisy ydata
yn = y + 0.2 * y * np.random.normal(size=len(x))
# generate error on ydata
y_sigma = 0.2 * y * np.random.normal(size=len(x))
popt, pcov = curve_fit(func, x, yn, sigma = y_sigma)
# plot
import matplotlib.pyplot as plt
fig = plt.figure()
ax = fig.add_subplot(111)
ax.errorbar(x, yn, yerr = y_sigma, fmt = 'o')
ax.plot(x, np.polyval(popt, x), '-')
ax.text(0.5, 100, r"a = {0:.3f} +/- {1:.3f}".format(popt[0], pcov[0,0]**0.5))
ax.text(0.5, 90, r"b = {0:.3f} +/- {1:.3f}".format(popt[1], pcov[1,1]**0.5))
ax.text(0.5, 80, r"c = {0:.3f} +/- {1:.3f}".format(popt[2], pcov[2,2]**0.5))
ax.grid()
plt.show()
Then something else, about using numpy arrays. One of the main advantages of using numpy is that you can avoid for loops because operations on arrays apply elementwise. So the for-loop in your example can also be done as following:
trendx = arange(datasetx[0], (datasetx[-1]+1))
trendy = trend[0]*trendx**2 + trend[1]*trendx + trend[2]
Where I use arange instead of range as it returns a numpy array instead of a list.
In this case you can also use the numpy function polyval:
trendy = polyval(trend, trendx)

I have not been able to find any way of getting the errors in the coefficients that is built in to numpy or python. I have a simple tool that I wrote based on Section 8.5 and 8.6 of John Taylor's An Introduction to Error Analysis. Maybe this will be sufficient for your task (note the default return is the variance, not the standard deviation). You can get large errors (as in the provided example) because of significant covariance.
def leastSquares(xMat, yMat):
'''
Purpose
-------
Perform least squares using the procedure outlined in 8.5 and 8.6 of Taylor, solving
matrix equation X a = Y
Examples
--------
>>> from scipy import matrix
>>> xMat = matrix([[ 1, 5, 25],
[ 1, 7, 49],
[ 1, 9, 81],
[ 1, 11, 121]])
>>> # matrix has rows of format [constant, x, x^2]
>>> yMat = matrix([[142],
[168],
[211],
[251]])
>>> a, varCoef, yRes = leastSquares(xMat, yMat)
>>> # a is a column matrix, holding the three coefficients a, b, c, corresponding to
>>> # the equation a + b*x + c*x^2
Returns
-------
a: matrix
best fit coefficients
varCoef: matrix
variance of derived coefficents
yRes: matrix
y-residuals of fit
'''
xMatSize = xMat.shape
numMeas = xMatSize[0]
numVars = xMatSize[1]
xxMat = xMat.T * xMat
xyMat = xMat.T * yMat
xxMatI = xxMat.I
aMat = xxMatI * xyMat
yAvgMat = xMat * aMat
yRes = yMat - yAvgMat
var = (yRes.T * yRes) / (numMeas - numVars)
varCoef = xxMatI.diagonal() * var[0, 0]
return aMat, varCoef, yRes