Short answer: from scipy import interpolate def f(x): x_points = [ 0, 1, 2, 3, 4, 5] y_points = [12,14,22,39,58,77] tck = interpolate.splrep(x_points, y_points) return interpolate.splev(x, tck) print(f(1.25)) Long answer: scipy separates the steps involved in spline interpolation into two operations, most likely for computational efficiency. The coefficients describing the spline curve are computed, using … Read more
Why do you want to do so? what you are doing is already pretty direct. Yeah, if you want to fetch them for the plot, when the plot is already made, simply use the get_ydata() method. B = plt.boxplot(data) [item.get_ydata() for item in B[‘whiskers’]] It returns an array of the shape (2,) for each whiskers, … Read more
A direct conversion is not supported ATM. Contributions are welcome! Try this, should be ok on memory as the SpareSeries is much like a csc_matrix (for 1 column) and pretty space efficient In [37]: col = np.array([0,0,1,2,2,2]) In [38]: data = np.array([1,2,3,4,5,6],dtype=”float64″) In [39]: m = csc_matrix( (data,(row,col)), shape=(3,3) ) In [40]: m Out[40]: <3×3 … Read more
import numpy as np import scipy import scipy.ndimage as ndimage import scipy.ndimage.filters as filters import matplotlib.pyplot as plt fname=”/tmp/slice0000.png” neighborhood_size = 5 threshold = 1500 data = scipy.misc.imread(fname) data_max = filters.maximum_filter(data, neighborhood_size) maxima = (data == data_max) data_min = filters.minimum_filter(data, neighborhood_size) diff = ((data_max – data_min) > threshold) maxima[diff == 0] = 0 labeled, num_objects … Read more
If I get your question correctly, you could do something like this. >>> import matplotlib.pyplot as plt >>> testList =[(0, 6.0705199999997801e-08), (1, 2.1015700100300739e-08), (2, 7.6280656623374823e-09), (3, 5.7348209304555086e-09), (4, 3.6812203579604238e-09), (5, 4.1572516753310418e-09)] >>> from math import log >>> testList2 = [(elem1, log(elem2)) for elem1, elem2 in testList] >>> testList2 [(0, -16.617236475334405), (1, -17.67799605473062), (2, -18.691431541177973), (3, … Read more
You can also check if all the eigenvalues of matrix are positive. If so, the matrix is positive definite: import numpy as np def is_pos_def(x): return np.all(np.linalg.eigvals(x) > 0)
EDIT Changed aspect=”normal to aspect=”auto’ since that changed in more recent versions of matplotlib (thanks to @Luke19). Assuming : import matplotlib.pyplot as plt To make a figure without the frame : fig = plt.figure(frameon=False) fig.set_size_inches(w,h) To make the content fill the whole figure ax = plt.Axes(fig, [0., 0., 1., 1.]) ax.set_axis_off() fig.add_axes(ax) Then draw your … Read more
>>> import math >>> import numpy as np >>> import scipy >>> math.pi == np.pi == scipy.pi True So it doesn’t matter, they are all the same value. The only reason all three modules provide a pi value is so if you are using just one of the three modules, you can conveniently have access … Read more
As fuglede says, the issue here is that np.float64 can’t handle a number as large as exp(1234.1). Try using np.float128 instead: >>> cc = np.array([[0.120,0.34,-1234.1]], dtype=np.float128) >>> cc array([[ 0.12, 0.34, -1234.1]], dtype=float128) >>> 1 / (1 + np.exp(-cc)) array([[ 0.52996405, 0.58419052, 1.0893812e-536]], dtype=float128) Note however, that there are certain quirks with using extended precision. … Read more
NumPy’s transpose() effectively reverses the shape of an array. If the array is one-dimensional, this means it has no effect. In NumPy, the arrays array([1, 2, 3]) and array([1, 2, 3]) are actually the same – they only differ in whitespace. What you probably want are the corresponding two-dimensional arrays, for which transpose() would work … Read more