@user1121588 answered most of this in a comment, but to answer fully… “Bitwise and” (&) behaves much the same as logical_and on boolean arrays, but it doesn’t convey the intent as well as using logical_and, and raises the possibility of getting misleading answers in non-trivial cases (packed or sparse arrays, maybe). To use logical_and on … Read more
You need to strip off the trailing ‘;’ from the lines. A possible workaround if you know you have 5 columns is: np.loadtxt(‘test.csv’, delimiter=”;”, usecols=range(5)) Or, use genfromtext instead which handles missing values np.genfromtxt(‘test.csv’, delimiter=”;”)[:,:-1]
numpy.argmax only returns the index of the first occurrence. You could apply argmax to a reversed view of the array: import numpy as np a = np.array([0,0,4,4,4,4,2,2,2,2]) b = a[::-1] i = len(b) – np.argmax(b) – 1 i # 5 a[i:] # array([4, 2, 2, 2, 2]) Note numpy doesn’t copy the array but instead … Read more
For the specific example you’ve given: dividing an (l,m,n) array by (m,) you can use np.newaxis: a = np.arange(1,61, dtype=float).reshape((3,4,5)) # Create a 3d array a.shape # (3,4,5) b = np.array([1.0, 2.0, 3.0, 4.0]) # Create a 1-d array b.shape # (4,) a / b # Gives a ValueError a / b[:, np.newaxis] # The … Read more
As you already guessed there is no need to create a figure. You basically need three steps. Normalize your data, apply the colormap, save the image. matplotlib provides all the necessary functionality: import numpy as np import matplotlib.pyplot as plt # some data (512×512) import scipy.misc data = scipy.misc.lena() # a colormap and a normalization … Read more
Given the title of your question, I will assume that the discretization size is constant. You can find this discretization size (or at least, strictly, n times that size as you may not have two adjacent samples in your data) np.diff(np.unique(data)).min() This finds the unique values in your data (np.unique), finds the differences between then … Read more
sns.jointplot returns a JointGrid object, which gives you access to the matplotlib axes and you can then manipulate from there. import seaborn as sns import numpy as np # example data X = np.random.randn(1000,) Y = 0.2 * np.random.randn(1000) + 0.5 h = sns.jointplot(X, Y) # JointGrid has a convenience function h.set_axis_labels(‘x’, ‘y’, fontsize=16) # … Read more
Each element of the tuple contains one of the indices for each nonzero value. Therefore, the length of each tuple element is the number of nonzeros in the array. From your example, the indices of the nonzeros are [0, 0], [1, 0], and [1, 1]. The first element of the tuple is the first index … Read more
As @Phoenix suggested, you might have an incomplete installation of numpy in your site-packages folder. Find your site-packages folder. SITE_PACKAGES_FOLDER=$(python3 -c “import sysconfig; print(sysconfig.get_paths()[‘purelib’])”) echo $SITE_PACKAGES_FOLDER Check for extraneous numpy packages from your site-packages folder. ls $SITE_PACKAGES_FOLDER/numpy* Trash extraneous packages. pip install trash-cli trash $SITE_PACKAGES_FOLDER/numpy* Reinstall numpy. pip install –upgrade numpy
I got the same error, but in my case I am subtracting dict.key from dict.value. I have fixed this by subtracting dict.value for corresponding key from other dict.value. cosine_sim = cosine_similarity(e_b-e_a, w-e_c) here I got error because e_b, e_a and e_c are embedding vector for word a,b,c respectively. I didn’t know that ‘w’ is string, … Read more