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ovito軟件分析Voronoi
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使用ovito分析Voronoi，輸入?yún)?shù)后，怎么得到結(jié)果？不明白？有會的嗎？ 發(fā)自小木蟲Android客戶端 |
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valenhou001
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meizhixin: 金幣+10 2017-02-28 13:32:37
meizhixin: 金幣+10 2017-02-28 13:32:37
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采用python script來計算吧： https://www.ovito.org/manual/pyt ... ute-voronoi-indices |
2樓2017-02-28 12:52:46
3樓2017-02-28 13:31:52
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老師好，我把路徑改了，在ovito圖形界面，run scripts，但是ovito圖形界面只是閃了下，沒報錯，也沒結(jié)果，啥也沒有，是腳本的問題嗎？我用的是vasp的POSAR文件，下面是ovito計算voro的腳本，我直接復(fù)制的，不知道對不對 # Import OVITO modules. from ovito.io import * from ovito.modifiers import * # Import NumPy module. import numpy # Load a simulation snapshot of a Co-B metallic glass. node = import_file("E:/ovito/Co65B35/POSCAR"  # Set atomic radii (required for polydisperse Voronoi tessellation). atypes = node.source.particle_properties.particle_type.type_list atypes[0].radius = 1.25 # Co atomic radius (atom type 1 in input file) atypes[1].radius = 0.82 # B atomic radius (atom type 2 in input file) # Set up the Voronoi analysis modifier. voro = VoronoiAnalysisModifier( compute_indices = True, use_radii = True, edge_count = 6, # Length after which Voronoi index vectors are truncated edge_threshold = 0.1 ) node.modifiers.append(voro) # Let OVITO compute the results. node.compute() # Make sure we did not lose information due to truncated Voronoi index vectors. if voro.max_face_order > voro.edge_count: print("Warning: Maximum face order in Voronoi tessellation is {0}, " "but computed Voronoi indices are truncated after {1} entries. " "You should consider increasing the 'edge_count' parameter to {0}." .format(voro.max_face_order, voro.edge_count)) # Note that it would be possible to automatically increase the 'edge_count' # parameter to 'max_face_order' here and recompute the Voronoi tessellation: # voro.edge_count = voro.max_face_order # node.compute() # Access computed Voronoi indices as NumPy array. # This is an (N)x(edge_count) array. voro_indices = node.output.particle_properties['Voronoi Index'].array # This helper function takes a two-dimensional array and computes a frequency # histogram of the data rows using some NumPy magic. # It returns two arrays (of equal length): # 1. The list of unique data rows from the input array # 2. The number of occurences of each unique row # Both arrays are sorted in descending order such that the most frequent rows # are listed first. def row_histogram(a): ca = numpy.ascontiguousarray(a).view([('', a.dtype)] * a.shape[1]) unique, indices, inverse = numpy.unique(ca, return_index=True, return_inverse=True) counts = numpy.bincount(inverse) sort_indices = numpy.argsort(counts)[::-1] return (a[indices[sort_indices]], counts[sort_indices]) # Compute frequency histogram. unique_indices, counts = row_histogram(voro_indices) # Print the ten most frequent histogram entries. for i in range(10): print("%s\t%i\t(%.1f %%)" % (tuple(unique_indices), counts, 100.0*float(counts)/len(voro_indices))) |
4樓2017-03-09 14:30:58
tongaigao
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7樓2019-03-13 08:16:21
cuso4852cuoh
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8樓2019-04-17 19:13:01
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9樓2019-04-17 19:13:52
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