You can use a 2D convolution on the 1s with a 3×3 kernel of 1s to identify the centers of the 3×3 squares, then dilate them and restore the non 1 numbers from scipy.signal import convolve2d from scipy.ndimage import binary_dilation # get 1s (as boolean) m = data==1 kernel = np.ones((3, 3)) # get centers … Read more
I assume X and Y can’t begin with 0. [1-9]\d{0,2} matches a number from 1 to 3 digits that doesn’t begin with 0. So the regexp to extract X and Y should be: ^([1-9]\d{0,2})000([1-9]\d{0,2})000$ Then you can use re.sub() to remove the zeroes between X and Y. regex = re.compile(r’^([1-9]\d{0,2})000([1-9]\d{0,2})000$’); i = 14000010000 istr = … Read more
Is this what you’re looking for? In [2]: num_rows = 10 # should be divisible by 2 In [3]: np.repeat(np.eye(num_rows // 2), 2, axis=0) Out[3]: array([[1., 0., 0., 0., 0.], [1., 0., 0., 0., 0.], [0., 1., 0., 0., 0.], [0., 1., 0., 0., 0.], [0., 0., 1., 0., 0.], [0., 0., 1., 0., 0.], … Read more
This is working under the assumption that the number of rows between the text is always consistent. # Open file, split on newlines into a list with open(‘file.txt’) as f: data = f.read().splitlines() # Break list into multiple lists 7 items long # this covers the text and the 6 sets of numbers that follow … Read more
This is the shortest: from scipy.ndimage import median_filter values = [1,1,1,0,1,1,1,1,1,1,1,2,1,1,1,10,1,1,1,1,1,1,1,1,1,1,0,1] print median_filter(values, 7, mode=”mirror”) and it works correctly in the edges (or you can choose how it works at the edges). And any general running X is done like this (running standard deviation as an example): import numpy from scipy.ndimage.filters import generic_filter values = … Read more
I have to agree with Majid Shirazi regarding the proposed solution by Quang Hong. Let’s have a look at: idx = np.where(img == (0.4, 0.4, 0.4)) Then, idx is a 3-tuple containing each a ndarray for all x-coordinates, y-coordinates, and “channel”-coordinates. From NumPy’s indexing, I can’t see a possibility to properly access/manipulate the values in … Read more
I made a great experience with Cython (Another thing than CPython…). You can make plain C code out of your program, compile it to an extension and just run it from another Python file. See this question from me for more information on building an extension including numpy: How to create a .pyd file?. Sample: … Read more
Being NumPy tagged, here’s a NumPy solution – In [846]: import numpy as np In [847]: t = (1,5,2,3,4,5,6,7,3,2,2,4,3) In [848]: a = [1,2,3] In [849]: np.in1d(t,a).sum() Out[849]: 7 # Alternatively with np.count_nonzero for summing booleans In [850]: np.count_nonzero(np.in1d(t,a)) Out[850]: 7 Another NumPy one with np.bincount for the specific case of positive numbered elements in … Read more
I was wondering how you got the array of lists. That usually takes some trickery. In [2]: >>> a = np.array([“0,1”, “2,3”, “4,5”]) …: >>> b = np.core.defchararray.split(a, sep=’,’) …: In [4]: b Out[4]: array([list([‘0’, ‘1’]), list([‘2’, ‘3’]), list([‘4’, ‘5’])], dtype=object) Simply calling array again doesn’t change things: In [5]: np.array(b) Out[5]: array([list([‘0’, ‘1’]), list([‘2’, … Read more
You do it exactly the same way as with a curve. If you have a function with 3 parameters and fit it to three points, you’ll get an exact solution with the curve going through all 3 points (it boils down to solving 3 equations with 3 unknowns): import numpy as np from scipy.optimize import … Read more