Finding the full width half maximum of a peak

You can use spline to fit the [blue curve - peak/2], and then find it's roots:

import numpy as np
from scipy.interpolate import UnivariateSpline

def make_norm_dist(x, mean, sd):
    return 1.0/(sd*np.sqrt(2*np.pi))*np.exp(-(x - mean)**2/(2*sd**2))

x = np.linspace(10, 110, 1000)
green = make_norm_dist(x, 50, 10)
pink = make_norm_dist(x, 60, 10)

blue = green + pink   

# create a spline of x and blue-np.max(blue)/2 
spline = UnivariateSpline(x, blue-np.max(blue)/2, s=0)
r1, r2 = spline.roots() # find the roots

import pylab as pl
pl.plot(x, blue)
pl.axvspan(r1, r2, facecolor='g', alpha=0.5)
pl.show()

Here is the result:

enter image description here

This worked for me in iPython (quick and dirty, can be reduced to 3 lines):

def FWHM(X,Y):
    half_max = max(Y) / 2.
    #find when function crosses line half_max (when sign of diff flips)
    #take the 'derivative' of signum(half_max - Y[])
    d = sign(half_max - array(Y[0:-1])) - sign(half_max - array(Y[1:]))
    #plot(X[0:len(d)],d) #if you are interested
    #find the left and right most indexes
    left_idx = find(d > 0)[0]
    right_idx = find(d < 0)[-1]
    return X[right_idx] - X[left_idx] #return the difference (full width)

Some additions can be made to make the resolution more accurate, but in the limit that there are many samples along the X axis and the data is not too noisy, this works great.

Even when the data are not Gaussian and a little noisy, it worked for me (I just take the first and last time half max crosses the data).

If your data has noise (and it always does in the real world), a more robust solution would be to fit a Gaussian to the data and extract FWHM from that:

import numpy as np
import scipy.optimize as opt

def gauss(x, p): # p[0]==mean, p[1]==stdev
    return 1.0/(p[1]*np.sqrt(2*np.pi))*np.exp(-(x-p[0])**2/(2*p[1]**2))

# Create some sample data
known_param = np.array([2.0, .7])
xmin,xmax = -1.0, 5.0
N = 1000
X = np.linspace(xmin,xmax,N)
Y = gauss(X, known_param)

# Add some noise
Y += .10*np.random.random(N)

# Renormalize to a proper PDF
Y /= ((xmax-xmin)/N)*Y.sum()

# Fit a guassian
p0 = [0,1] # Inital guess is a normal distribution
errfunc = lambda p, x, y: gauss(x, p) - y # Distance to the target function
p1, success = opt.leastsq(errfunc, p0[:], args=(X, Y))

fit_mu, fit_stdev = p1

FWHM = 2*np.sqrt(2*np.log(2))*fit_stdev
print "FWHM", FWHM

enter image description here

The plotted image can be generated by:

from pylab import *
plot(X,Y)
plot(X, gauss(X,p1),lw=3,alpha=.5, color='r')
axvspan(fit_mu-FWHM/2, fit_mu+FWHM/2, facecolor='g', alpha=0.5)
show()

An even better approximation would filter out the noisy data below a given threshold before the fit.

Here is a nice little function using the spline approach.

from scipy.interpolate import splrep, sproot, splev

class MultiplePeaks(Exception): pass
class NoPeaksFound(Exception): pass

def fwhm(x, y, k=10):
    """
    Determine full-with-half-maximum of a peaked set of points, x and y.

    Assumes that there is only one peak present in the datasset.  The function
    uses a spline interpolation of order k.
    """

    half_max = amax(y)/2.0
    s = splrep(x, y - half_max, k=k)
    roots = sproot(s)

    if len(roots) > 2:
        raise MultiplePeaks("The dataset appears to have multiple peaks, and "
                "thus the FWHM can't be determined.")
    elif len(roots) < 2:
        raise NoPeaksFound("No proper peaks were found in the data set; likely "
                "the dataset is flat (e.g. all zeros).")
    else:
        return abs(roots[1] - roots[0])

You should use scipy to solve it: first find_peaks and then peak_widths. With default value in rel_height(0.5) you're measuring the width at half maximum of the peak.

Finding the full width half maximum of a peak

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