尽管我非常喜欢EOL的答案，但我还是想对每个方向上的仓数不均匀进行概括，并强调C和F排序样式之间的差异。这是一个示例解决方案：
ndims = 5N = 10# Define bin boundaries binbnds = ndims*[None]nbins = []for idim in xrange(ndims): binbnds[idim] = numpy.linspace(-10.0,10.0,numpy.random.randint(2,15)) binbnds[idim][0] = -float('inf') binbnds[idim][-1] = float('inf') nbins.append(binbnds[idim].shape[0]-1)nstates = numpy.cumprod(nbins)[-1]# Define variable values for N particles in ndims dimensionsp = numpy.random.normal(size=(N,ndims))# Assign to bins along each dimensionbinassign = ndims*[None]for idim in xrange(ndims): binassign[idim] = numpy.digitize(p[:,idim],binbnds[idim]) - 1binassign = numpy.array(binassign)# multidimensional array with elements mapping from multidim to linear index# Two different arrays for C vs F orderinglinind_C = numpy.arange(nstates).reshape(nbins,order='C')linind_F = numpy.arange(nstates).reshape(nbins,order='F')现在进行转换
# Fast conversion to linear indexb_F = numpy.cumprod([1] + nbins)[:-1]b_C = numpy.cumprod([1] + nbins[::-1])[:-1][::-1]box_index_F = numpy.dot(b_F,binassign)box_index_C = numpy.dot(b_C,binassign)
并检查正确性：
# Checkprint 'Checking correct mapping for each particle F order'for k in xrange(N): ii = box_index_F[k] jj = linind_F[tuple(binassign[:,k])] print 'particle %d %s (%d %d)' % (k,ii == jj,ii,jj)print 'Checking correct mapping for each particle C order'for k in xrange(N): ii = box_index_C[k] jj = linind_C[tuple(binassign[:,k])] print 'particle %d %s (%d %d)' % (k,ii == jj,ii,jj)
为了完整起见，如果您想以快速，矢量化的方式从1d索引返回到多维索引：
print 'Convert C-style from linear to multi'x = box_index_C.reshape(-1,1)bassign_rev_C = x / b_C % nbinsprint 'Convert F-style from linear to multi'x = box_index_F.reshape(-1,1)bassign_rev_F = x / b_F % nbins
并再次检查：
print 'Check C-order'for k in xrange(N): ii = tuple(binassign[:,k]) jj = tuple(bassign_rev_C[k,:]) print ii==jj,ii,jjprint 'Check F-order'for k in xrange(N): ii = tuple(binassign[:,k]) jj = tuple(bassign_rev_F[k,:]) print ii==jj,ii,jj
