The message “intermediate result is being cached” is just a blind guess in the canned message reported by %timeit. It may or may not be true, and you should not assume it is correct. In particular, one of the most common reasons for the first run being slowest is that the array is in the … Read more
Here’s what I put together. I tried to closely emulate your screenshot. Given import numpy as np import scipy as sp import scipy.stats as stats import matplotlib.pyplot as plt %matplotlib inline # Raw Data heights = np.array([50,52,53,54,58,60,62,64,66,67,68,70,72,74,76,55,50,45,65]) weights = np.array([25,50,55,75,80,85,50,65,85,55,45,45,50,75,95,65,50,40,45]) Two detailed options to plot confidence intervals: def plot_ci_manual(t, s_err, n, x, x2, y2, ax=None): … Read more
In the solution below I used python3.4 as binary, but it’s safe to use with any version or binary of python. it works fine on windows too (except the downloading pip with wget obviously but just save the file locally and run it with python). This is great if you have multiple versions of python … Read more
Straightforward way to do that is: import numpy as np a=np.array([[1,2,3],[3,4,5]]) b=np.array([[1,2,3],[1,2,3]]) np.sum(a*b, axis=1) which avoids the python loop and is faster in cases like: def npsumdot(x, y): return np.sum(x*y, axis=1) def loopdot(x, y): result = np.empty((x.shape[0])) for i in range(x.shape[0]): result[i] = np.dot(x[i], y[i]) return result timeit npsumdot(np.random.rand(500000,50),np.random.rand(500000,50)) # 1 loops, best of 3: … Read more
Sympy makes this straightforward. >>> from sympy import Matrix >>> A = [[2, 3, 5], [-4, 2, 3], [0, 0, 0]] >>> A = Matrix(A) >>> A * A.nullspace()[0] Matrix([ [0], [0], [0]]) >>> A.nullspace() [Matrix([ [-1/16], [-13/8], [ 1]])]
1. Is there any tutorial/example on the usage of InfogainLoss layer?: A nice example can be found here: using InfogainLoss to tackle class imbalance. 2. Should the input to this layer, the class probabilities, be the output of a Softmax layer? Historically, the answer used to be YES according to Yair’s answer. The old implementation … Read more
For short arrays, using sets is probably the clearest and most readable way to do it. Another way is to use numpy.intersect1d. You’ll have to trick it into treating the rows as a single value, though… This makes things a bit less readable… import numpy as np A = np.array([[1,4],[2,5],[3,6]]) B = np.array([[1,4],[3,6],[7,8]]) nrows, ncols … Read more