Matching line points with grid coordinates numerically
Dear Python users, I am having difficulty with numerically scaling to match line coordinates vs grid cell size coordinates. I want to calculate the following function: F = distance_of_crossed_line x intersected_cell_value The problem here is that when I calculate crossed_line_length in line coordinates that will unmatch vs grid coordinate, which is also another x,y with step e.g., dx=dy=2.5 each grid size. I want to do numerical calculation , say, F(distance, intersected_grid_value) function where, intersected_grid_value - values in intersected grid, distance - intersected_line_length (given below or can be found http://stackoverflow.com/questions/16377826/distance-for-each-intersected-po... ) import numpy as np import scipy as sp def distance_of_crossed_line(x0, x1, y0, y1): # slope m = (y1 - y0) / (x1 - x0) # Boundary of the selected points x_ceil = np.ceil(min(x0, x1)) x_floor = np.floor(max(x0, x1)) y_ceil = np.ceil(min(y0, y1)) y_floor = np.floor(max(y0, y1)) # calculate all intersected x coordinate x = np.arange(x_ceil, x_floor + 1) y = m * (x - x0) + y0 ax = zip(x, y) # calculate all intersected y coordinate y = np.arange(y_ceil, y_floor + 1) x = (y - y0) / m + x0 ax.extend(zip(x, y)) ax.append((x0, y0)) ax.append((x1, y1)) ax.sort() # Transpose ax = np.array(ax).T # Calculate difference of intersections in X dist_x = np.diff(ax[0]) # Calculate difference of intersections in Y dist_y = np.diff(ax[1]) return np.sqrt(dist_x**2 + dist_y**2) # PLEASE, note that line points are different from 2D array axis. they should be matched with each other. # 2D array. d_array = np.array[[4.5, 4.5, 4.5, 3.4, 2.5],[ 3.9, 4.5, 5.2, 4.5, 3.4],[3.9, 3.9, 2.5, 2.2, 1.9]] # Two sample points as a line x = np.array([ -80, -40 ]) y = np.array([ 60, 55 ]) # The problem: F = intersected_line_length * array_grid_values_where_line_crossed_area * It is not necessary for me to overlay lines onto grid cells properly, JUST, I need to calculate numerically accurate F function Thanks for the answer and guidance in advance, -- Bakhtiyor Zokhidov
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Bakhtiyor Zokhidov