it depends what sort of t-test you want to do (one sided or two sided dependent or independent) but it should be as simple as: from scipy.stats import ttest_ind cat1 = my_data[my_data[‘Category’]==’cat1′] cat2 = my_data[my_data[‘Category’]==’cat2′] ttest_ind(cat1[‘values’], cat2[‘values’]) >>> (1.4927289925706944, 0.16970867501294376) it returns a tuple with the t-statistic & the p-value see here for other t-tests … Read more
As mg007 suggested, some of the scipy.optimize routines allow for a callback function (unfortunately leastsq does not permit this at the moment). Below is an example using the “fmin_bfgs” routine where I use a callback function to display the current value of the arguments and the value of the objective function at each iteration. import … Read more
Months later, here’s a small class PCA, and a picture: #!/usr/bin/env python “”” a small class for Principal Component Analysis Usage: p = PCA( A, fraction=0.90 ) In: A: an array of e.g. 1000 observations x 20 variables, 1000 rows x 20 columns fraction: use principal components that account for e.g. 90 % of the … Read more
Here you go. You just need to use None (or alternatively np.newaxis) combined with broadcasting: In [6]: data – vector[:,None] Out[6]: array([[0, 0, 0], [0, 0, 0], [0, 0, 0]]) In [7]: data / vector[:,None] Out[7]: array([[1, 1, 1], [1, 1, 1], [1, 1, 1]])
Look at the Python code for the scipy sparse product. Notice that it calls the compiled code in 2 passes. It looks like the mkl code does the same thing https://software.intel.com/en-us/node/468640 If request=1, the routine computes only values of the array ic of length m + 1, the memory for this array must be allocated … Read more
Correct functions: import numpy as np from scipy import fftpack as scipy_fftpack from scipy import fft as scipy # FFTPACK RFFT 2D def fftpack_rfft2d(matrix): fftRows = scipy_fftpack.fft(matrix, axis=1) fftCols = scipy_fftpack.fft(fftRows, axis=0) return fftCols # FFTPACK IRFFT 2D def fftpack_irfft2d(matrix): ifftRows = scipy_fftpack.ifft(matrix, axis=1) ifftCols = scipy_fftpack.ifft(ifftRows, axis=0) return ifftCols.real You calculated the 2D FFT … Read more
Least squares (scipy.linalg.lstsq) is guaranteed to converge. In fact, there is a closed form analytical solution (given by (A^T A)^-1 A^Tb (where ^T is matrix transpose and ^-1 is matrix inversion) The standard optimization problem, however, is not generally solvable – we are not guaranteed to find a minimizing value. However, for the given equation, … Read more
So after obsessing a lot about my question, and much research, i finally have my answer. Everything is available in scipy , and i’m putting my code here so hopefully someone else can find this useful. The function takes in an array of N-d points, a curve degree, a periodic state (opened or closed) and … Read more
It looks like you could use numpy.bincount here: import numpy as np def radial_profile(data, center): y, x = np.indices((data.shape)) r = np.sqrt((x – center[0])**2 + (y – center[1])**2) r = r.astype(np.int) tbin = np.bincount(r.ravel(), data.ravel()) nr = np.bincount(r.ravel()) radialprofile = tbin / nr return radialprofile
The answer by DaveP is almost right… but can cause problems for very sparse matrices: if the last column(s) or row(s) are empty, they are dropped. So to be sure that everything works, the “shape” attribute must be stored too. This is the code I regularly use: import tables as tb from numpy import array … Read more