Question

我有一个蒙版的1D数据数组，其中包含我已屏蔽的nan值，现在打印为 - 。我希望将此数组拟合为高斯数，并使用拟合的均值和标准差创建直方图。我已经尝试过辣.stats.fit，但那不起作用（卑鄙和std刚刚返回＆＃39; nan＆＃39;）。然后我追求辣.mstats，但它看起来没有合适的功能。

是否有一个模块可以将高斯拟合到蒙板数组并输出均值和标准？

编辑：这是我的代码所以

def createRmsMatrix( self ):

    '''
    Creates an array of RMS values for each profile in one file.
    '''

    # Initialize RMS table of zeroes
    rmsMatrix = np.zeros( ( self.nSub, self.nChan ), dtype = float )

    # Loop over the time and frequency indices
    for time in np.arange( self.nSub ):
        for frequency in np.arange( self.nChan ):

            # Create a mask along the bin space
            mask = utils.binMask( self.data[time][frequency], 0.55 )

            #print(mask)

            rmsMatrix[time][frequency] = mu.rootMeanSquare( self.data[time][frequency][mask == 0] )

    # Mask the nan values in the array
    rmsMatrix = np.ma.array( rmsMatrix, mask = np.isnan( rmsMatrix ) )

    print( "Root Mean Square matrix created..." )

    return rmsMatrix

我的主要功能部分称之为：

    # Return the array of RMS values for each profile
    self.rmsArray = self.createRmsMatrix()

    # Reshape RMS array to be linear and store in a new RMS array
    self.linearRmsArray = np.reshape( self.rmsArray, ( self.nChan * self.nSub ) )

    # Best fit of data using a Gaussian fit
    mu, sigma = norm.fit( self.linearRmsArray )

    # Creates the histogram
    n, bins, patches = self.histogramPlot( self.linearRmsArray, mu, sigma, 'Root Mean Squared', 'Frequency Density', True )

histogramPlot对我来说只是一个方便的matplotlib组织者，我也会发布：

def histogramPlot( self, data, mean, stdDev, xAxis='x-axis', yAxis='y-axis', showPlot = False ):

    '''
    Plots and returns a histogram of some linear data using matplotlib
    and fits a Gaussian centered around the mean with a spread of stdDev.
    Use this function to set the x and y axis names.
    Can also toggle showing of the histogram in this function.
    '''

    # Plot the histogram
    n, bins, patches = plt.hist( self.linearRmsArray, bins=self.nChan, normed=True )

    # Add a 'best fit' normal distribution line
    xPlot = np.linspace( ( mean - (4*stdDev) ), ( mean + (4*stdDev) ), 1000 )
    yPlot = mlab.normpdf( xPlot, mean, stdDev )
    l = plt.plot(xPlot, yPlot, 'r--', linewidth=2)

    # Format axes
    plt.ylabel( yAxis )
    plt.xlabel( xAxis )
    #plt.title(r'$\mathrm{Histogram\ of\ data:}\ \mu=%.3f,\ \sigma=%.3f$' %(mu, sigma))
    plt.title(r'$\mu=%.3f,\ \sigma=%.3f$' %(mean, stdDev))
    plt.grid(True)

    if showPlot == True:
        plt.show()

    return n, bins, patches

Answer 1

您试图使用scipy.norm.fit来为数据拟合正态分布，这意味着您的输入是应该是来自正态分布的随机样本的值集合。在这种情况下，均值和标准的最大似然估计。开发。只是数据的样本均值和样本标准差。对于包含nan的数据，您可以在调用nan之前删除scipy.norm.fit()，也可以直接使用numpy.nanmean和numpy.nanstd计算这些数据：

est_mean = np.nanmean(data)
est_stddev = np.nanstd(data)

例如，

In [18]: import numpy as np

In [19]: from scipy.stats import norm

In [20]: x = np.array([1, 4.5, np.nan, 3.3, 10.0, 4.1, 8.5, 17.1, np.nan])

In [21]: np.nanmean(x), np.nanstd(x)
Out[21]: (6.9285714285714288, 5.0366412520687653)

In [22]: norm.fit(x[np.isfinite(x)])
Out[22]: (6.9285714285714288, 5.0366412520687653)

请注意，x[np.isfinite(x)]是x中不是nan或inf的值数组。

如果你有一个蒙面数组，你可以使用mean和std方法：

In [36]: mx = np.ma.masked_array(x, np.isnan(x))

In [37]: mx
Out[37]: 
masked_array(data = [1.0 4.5 -- 3.3 10.0 4.1 8.5 17.1 --],
             mask = [False False  True False False False False False  True],
       fill_value = 1e+20)

In [38]: mx.mean(), mx.std()
Out[38]: (6.9285714285714288, 5.0366412520687653)

将高斯拟合到1D掩码数据数组

1 个答案: