以下解决方案仅需要numpy。它适用于凸多边形的顶点（在[行，列]坐标系中按顺时针顺序定义）。凹面多边形可以工作，但最终会切除凸出的点。
import numpy as npdef check(p1, p2, base_array): """ Uses the line defined by p1 and p2 to check array of input indices against interpolated value Returns boolean array, with True inside and False outside of shape """ idxs = np.indices(base_array.shape) # Create 3D array of indices p1 = p1.astype(float) p2 = p2.astype(float) # Calculate max column idx for each row idx based on interpolated line between two points max_col_idx = (idxs[0] - p1[0]) / (p2[0] - p1[0]) * (p2[1] - p1[1]) + p1[1] sign = np.sign(p2[0] - p1[0]) return idxs[1] * sign <= max_col_idx * signdef create_polygon(shape, vertices): """ Creates np.array with dimensions defined by shape Fills polygon defined by vertices with ones, all other values zero""" base_array = np.zeros(shape, dtype=float) # Initialize your array of zeros fill = np.ones(base_array.shape) * True # Initialize boolean array defining shape fill # Create check array for each edge segment, combine into fill array for k in range(vertices.shape[0]): fill = np.all([fill, check(vertices[k-1], vertices[k], base_array)], axis=0) # Set all values inside polygon to one base_array[fill] = 1 return base_array# (Row, Col) Vertices of Polygon (Defined Clockwise)vertices = np.array([ [5,12], [8,18], [13,14], [11,6], [4,6],])polygon_array = create_polygon([20,20], vertices)# This section prints numbers at each vertex for visual check, just comment out # to print an array of only zeros and onesfor n, vertex in enumerate(vertices): polygon_array[vertex[0],vertex[1]] = 10*(n+1)# Simple routine to print the final arrayfor row in polygon_array.tolist(): for c in row: print '{:4.1f}'.format(c), print ''
