用python拟合直方图

Question

我有直方图

H=hist(my_data,bins=my_bin,histtype='step',color='r')

我可以看到形状几乎是高斯的，但我想用高斯函数拟合这个直方图并打印我得到的均值和西格玛的值。你能救我吗？

Answer 1

这里有一个处理py2.6和py3.2的例子：

from scipy.stats import norm
import matplotlib.mlab as mlab
import matplotlib.pyplot as plt

# read data from a text file. One number per line
arch = "test/Log(2)_ACRatio.txt"
datos = []
for item in open(arch,'r'):
    item = item.strip()
    if item != '':
        try:
            datos.append(float(item))
        except ValueError:
            pass

# best fit of data
(mu, sigma) = norm.fit(datos)

# the histogram of the data
n, bins, patches = plt.hist(datos, 60, normed=1, facecolor='green', alpha=0.75)

# add a 'best fit' line
y = mlab.normpdf( bins, mu, sigma)
l = plt.plot(bins, y, 'r--', linewidth=2)

#plot
plt.xlabel('Smarts')
plt.ylabel('Probability')
plt.title(r'$\mathrm{Histogram\ of\ IQ:}\ \mu=%.3f,\ \sigma=%.3f$' %(mu, sigma))
plt.grid(True)

plt.show()

Answer 2

这是一个使用scipy.optimize拟合非线性函数（如高斯函数）的示例，即使数据位于不完全范围的直方图中，因此简单的均值估计也会失败。偏移常数也会导致简单的正常统计失败（只需删除普通高斯数据的p [3]和c [3]）。

from pylab import *
from numpy import loadtxt
from scipy.optimize import leastsq

fitfunc  = lambda p, x: p[0]*exp(-0.5*((x-p[1])/p[2])**2)+p[3]
errfunc  = lambda p, x, y: (y - fitfunc(p, x))

filename = "gaussdata.csv"
data     = loadtxt(filename,skiprows=1,delimiter=',')
xdata    = data[:,0]
ydata    = data[:,1]

init  = [1.0, 0.5, 0.5, 0.5]

out   = leastsq( errfunc, init, args=(xdata, ydata))
c = out[0]

print "A exp[-0.5((x-mu)/sigma)^2] + k "
print "Parent Coefficients:"
print "1.000, 0.200, 0.300, 0.625"
print "Fit Coefficients:"
print c[0],c[1],abs(c[2]),c[3]

plot(xdata, fitfunc(c, xdata))
plot(xdata, ydata)

title(r'$A = %.3f\  \mu = %.3f\  \sigma = %.3f\ k = %.3f $' %(c[0],c[1],abs(c[2]),c[3]));

show()

输出：

A exp[-0.5((x-mu)/sigma)^2] + k 
Parent Coefficients:
1.000, 0.200, 0.300, 0.625
Fit Coefficients:
0.961231625289 0.197254597618 0.293989275502 0.65370344131

Answer 3

以下是仅使用matplotlib.pyplot和numpy个包的另一种解决方案。 它仅适用于高斯拟合。它基于maximum likelihood estimation，已在此topic中提及。 这是相应的代码：

# Python version : 2.7.9
from __future__ import division
import numpy as np
from matplotlib import pyplot as plt

# For the explanation, I simulate the data :
N=1000
data = np.random.randn(N)
# But in reality, you would read data from file, for example with :
#data = np.loadtxt("data.txt")

# Empirical average and variance are computed
avg = np.mean(data)
var = np.var(data)
# From that, we know the shape of the fitted Gaussian.
pdf_x = np.linspace(np.min(data),np.max(data),100)
pdf_y = 1.0/np.sqrt(2*np.pi*var)*np.exp(-0.5*(pdf_x-avg)**2/var)

# Then we plot :
plt.figure()
plt.hist(data,30,normed=True)
plt.plot(pdf_x,pdf_y,'k--')
plt.legend(("Fit","Data"),"best")
plt.show()

和here是输出。

Answer 4

从Python 3.8开始，标准库提供NormalDist对象作为statistics模块的一部分。

可以使用NormalDist.from_samples方法从一组数据中构建NormalDist对象，并提供对其平均值（NormalDist.mean）和的访问。标准偏差（NormalDist.stdev）：

from statistics import NormalDist

# data = [0.7237248252340628, 0.6402731706462489, -1.0616113628912391, -1.7796451823371144, -0.1475852030122049, 0.5617952240065559, -0.6371760932160501, -0.7257277223562687, 1.699633029946764, 0.2155375969350495, -0.33371076371293323, 0.1905125348631894, -0.8175477853425216, -1.7549449090704003, -0.512427115804309, 0.9720486316086447, 0.6248742504909869, 0.7450655841312533, -0.1451632129830228, -1.0252663611514108]
norm = NormalDist.from_samples(data)
# NormalDist(mu=-0.12836704320073597, sigma=0.9240861018557649)
norm.mean
# -0.12836704320073597
norm.stdev
# 0.9240861018557649

Answer 5

我很困惑norm.fit显然只适用于扩展的采样值列表。我尝试给它提供两个数字列表或元组列表，但它似乎只会使所有内容变平并威胁输入作为单个样本。由于我已经有一个基于数百万个样本的直方图，因此如果不需要，我不想扩展它。幸运的是，正态分布的计算很简单，所以...

# histogram is [(val,count)]
from math import sqrt

def normfit(hist):
    n,s,ss = univar(hist)
    mu = s/n
    var = ss/n-mu*mu
    return (mu, sqrt(var))

def univar(hist):
    n = 0
    s = 0
    ss = 0
    for v,c in hist:
        n += c
        s += c*v
        ss += c*v*v
    return n, s, ss

我确定这必须由库提供，但是由于我在任何地方都找不到，因此我将其发布在这里。随意指出正确的做法，并否决我：-）