In [82]: d Out[82]: A B C D 0 John Doe 45 True False 1 Jane Smith 32 False False 2 Alan Holmes 55 False True 3 Eric Lamar 29 True True Solution 1: In [83]: d.loc[d.C | d.D] Out[83]: A B C D 0 John Doe 45 True False 2 Alan Holmes 55 False … Read more
The numpy stores data in row major order. >>> a = np.array([[1,2,3,4], [5,6,7,8]]) >>> a.shape (2, 4) >>> a.shape = 4,2 >>> a array([[1, 2], [3, 4], [5, 6], [7, 8]]) If you change the shape, the order of data do not change. If you add a ‘F’, you can get what you want. >>> … Read more
You don’t want to get the intersections of the parallel lines; only the intersections of the vertical lines with those of the horizontal lines. Also, since you have vertical lines, calculating the slope will likely result in exploding or inf slopes, so you shouldn’t use the y = mx+b equations. You need to do two … Read more
np.where np.where(a > 0.5, 1, 0) # array([0, 0, 0, 1, 1, 1]) Boolean basking with astype (a > .5).astype(int) # array([0, 0, 0, 1, 1, 1]) np.select np.select([a <= .5, a>.5], [np.zeros_like(a), np.ones_like(a)]) # array([ 0., 0., 0., 1., 1., 1.]) Special case: np.round This is the best solution if your array values are … Read more
In scipy 0.14 or later, there is a new function scipy.interpolate.RegularGridInterpolator which closely resembles interp3. The MATLAB command Vi = interp3(x,y,z,V,xi,yi,zi) would translate to something like: from numpy import array from scipy.interpolate import RegularGridInterpolator as rgi my_interpolating_function = rgi((x,y,z), V) Vi = my_interpolating_function(array([xi,yi,zi]).T) Here is a full example demonstrating both; it will help you understand … Read more
Knowing that np.savetxt only works for 1D or 2D arrays, the general idea is: when fmt is a single formatting string it applies to all elements in the array (1D or 2D input array) when fmt is a sequence of formatting strings, it applies to each column of the 2D input array I’m presenting here … Read more
One other way to get a discrete distribution that looks like the normal distribution is to draw from a multinomial distribution where the probabilities are calculated from a normal distribution. import scipy.stats as ss import numpy as np import matplotlib.pyplot as plt x = np.arange(-10, 11) xU, xL = x + 0.5, x – 0.5 … Read more
reshape works for this a = np.arange(3) # a.shape = (3,) b = a.reshape((3,1)) # b.shape = (3,1) b2 = a.reshape((-1,1)) # b2.shape = (3,1) c = b.reshape((3,)) # c.shape = (3,) c2 = b.reshape((-1,)) # c2.shape = (3,) note also that reshape doesn’t copy the data unless it needs to for the new shape … Read more
numpy.maximum.accumulate works for me. >>> import numpy >>> numpy.maximum.accumulate(numpy.array([11,12,13,20,19,18,17,18,23,21])) array([11, 12, 13, 20, 20, 20, 20, 20, 23, 23])
My guess is that you want to estimate the shape parameter and the scale of the Weibull distribution while keeping the location fixed. Fixing loc assumes that the values of your data and of the distribution are positive with lower bound at zero. floc=0 keeps the location fixed at zero, f0=1 keeps the first shape … Read more