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C++ SequenceTree::directed_branch方法代码示例

本文整理汇总了C++中SequenceTree::directed_branch方法的典型用法代码示例。如果您正苦于以下问题:C++ SequenceTree::directed_branch方法的具体用法?C++ SequenceTree::directed_branch怎么用?C++ SequenceTree::directed_branch使用的例子?那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在SequenceTree的用法示例。


在下文中一共展示了SequenceTree::directed_branch方法的7个代码示例,这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞,您的评价将有助于系统推荐出更棒的C++代码示例。

示例1: do_SPR

/// Do a SPR move on T1, moving the subtree behind b1_ to branch b2
double do_SPR(SequenceTree& T1, int b1_,int b2)
{
    const_branchview b1 = T1.directed_branch(b1_);

    SequenceTree T2 = T1;

    //------ Generate the new topology ------//
    if (T2.directed_branch(b2).target() == b1.target() or
            T2.directed_branch(b2).source() == b1.target())
        ;
    else
        SPR(T2,b1.reverse(),b2);

    //------ Find the two new branches ------//
    vector<const_branchview> connected1;
    append(T1.directed_branch(b1.source(),b1.target()).branches_after(),connected1);

    vector<const_branchview> connected2;
    append(T2.directed_branch(b1.source(),b1.target()).branches_after(),connected2);

    assert(connected1.size() == 2);
    assert(connected2.size() == 2);

    //------- Place the split randomly -------//
    double L1 = connected1[0].length() + connected1[1].length();
    double L2 = connected2[0].length() + connected2[1].length();

    T2.directed_branch(connected2[0]).set_length( myrandomf() * L2 );
    T2.directed_branch(connected2[1]).set_length( L2 - T2.directed_branch(connected2[0]).length() );

    T1 = T2;

    return L2/L1;
}
开发者ID:msuchard,项目名称:BAli-Phy,代码行数:35,代码来源:sample-topology-SPR.C

示例2: choose_SPR_target

int choose_SPR_target(SequenceTree& T1, int b1_)
{
    const_branchview b1 = T1.directed_branch(b1_);

    //----- Select the branch to move to ------//
    dynamic_bitset<> subtree_nodes = T1.partition(b1.reverse());
    subtree_nodes[b1.target()] = true;

    vector<int> branches;
    vector<double> lengths;

    for(int i=0; i<T1.n_branches(); i++)
    {
        const_branchview bi = T1.branch(i);

        // skip branch if its contained in the subtree
        if (subtree_nodes[bi.target()] and
                subtree_nodes[bi.source()])
            continue;

        double L = 1.0;

        // down-weight branch if it is one of the subtree's 2 neighbors
        if (subtree_nodes[bi.target()] or
                subtree_nodes[bi.source()])
            L = 0.5;

        branches.push_back(i);
        lengths.push_back(L);
    }

    int b2 = branches[ choose(lengths) ];

    return b2;
}
开发者ID:msuchard,项目名称:BAli-Phy,代码行数:35,代码来源:sample-topology-SPR.C

示例3: update_lengths

bool update_lengths(const SequenceTree& Q,const SequenceTree& T,
		    valarray<double>& branch_lengths, 
		    valarray<double>& branch_lengths_squared, 
		    valarray<double>& node_lengths)
{
  // map branches from Q -> T
  vector<int> branches_map = extends_map(T,Q);
  if (not branches_map.size())
    return false;

  // incorporate lengths of branches that map to Q
  for(int b=0;b<Q.n_branches();b++)
  {
    int b2 = branches_map[b];
    double L = T.directed_branch(b2).length();
    branch_lengths[b] += L;
    branch_lengths_squared[b] += L*L;
  }

  // map nodes from T -> Q
  vector<int> nodes_map = get_nodes_map(Q,T,branches_map);

  // incorprate lengths of branches that map to nodes in Q
  for(int i=T.n_leafbranches();i<T.n_branches();i++) 
  {
    const_branchview b = T.branch(i);
    int n1 = nodes_map[b.source()];
    int n2 = nodes_map[b.target()];

    if (n1 == n2)
      node_lengths[n1] += T.branch(i).length();
  }

  return true;
}
开发者ID:msuchard,项目名称:BAli-Phy,代码行数:35,代码来源:tree-mean-lengths.C

示例4: n_mutations

B n_mutations(const alphabet& a, const vector<int>& letters, const SequenceTree& T,const ublas::matrix<B>& cost,
	      ublas::matrix<B>& n_muts, const vector<const_branchview>& branches)
{
  int root = T.directed_branch(0).target();

  peel_n_mutations(a,letters,T,cost,n_muts,branches);

  return row_min(n_muts,root);
}
开发者ID:sibonli,项目名称:BAli-Phy,代码行数:9,代码来源:parsimony.C

示例5: temp

vector<vector<int> > get_all_parsimony_letters(const alphabet& a, const vector<int>& letters, const SequenceTree& T,
					       const ublas::matrix<int>& cost)
{
  int root = T.directed_branch(0).target();

  ublas::matrix<int> n_muts(T.n_nodes(), a.size());
  peel_n_mutations(a,letters,T,cost,n_muts, branches_toward_node(T,root) );

  // get an order list of branches point away from the root;
  vector<const_branchview> branches = branches_from_node(T,root);
  std::reverse(branches.begin(),branches.end());
  
  // Allocate space to store the letters for each node
  vector<vector<int> > node_letters(T.n_nodes());

  const unsigned A = a.size();

  // choose the cheapest letters at the root
  {
    double m = row_min(n_muts,root);
    for(int l=0;l<A;l++)
      if (n_muts(root,l) <= m)
	node_letters[root].push_back(l);
  }

  vector<double> temp(A);

  for(int i=0;i<branches.size();i++) 
  {
    int s = branches[i].source();
    int t = branches[i].target();

    vector<double> best(node_letters[s].size());

    for(int j=0;j<node_letters[s].size();j++) 
    {
      for(int l=0;l<A;l++)
	temp[l] = n_muts(t,l)+cost(l,node_letters[s][j]);
      best[j] = min(temp);
    }
    
    for(int l=0;l<A;l++) 
    {
      bool is_best = false;
      for(int j=0;j<node_letters[s].size() and not is_best;j++) 
	if (n_muts(t,l)+cost(l,node_letters[s][j]) <= best[j])
	  is_best=true;
      if (is_best)
	node_letters[t].push_back(l);
    }

  }

  return node_letters;
}
开发者ID:sibonli,项目名称:BAli-Phy,代码行数:55,代码来源:parsimony.C

示例6: assert

// mark nodes in T according to what node of Q they map to
vector<int> get_nodes_map(const SequenceTree& Q,const SequenceTree& T,
			  const vector<int>& branches_map)
{
  assert(branches_map.size() == Q.n_branches() * 2);

  vector<int> nodes_map(T.n_nodes(),-1);

  // map nodes from T -> Q that are in both trees
  for(int b=0;b<Q.n_branches();b++)
  {
    int Q_source = Q.branch(b).source();
    int Q_target = Q.branch(b).target();

    int b2 = branches_map[b];

    int T_source = T.directed_branch(b2).source();
    int T_target = T.directed_branch(b2).target();

    if (nodes_map[T_source] == -1)
      nodes_map[T_source] = Q_source;
    else
      assert(nodes_map[T_source] == Q_source);

    if (nodes_map[T_target] == -1)
      nodes_map[T_target] = Q_target;
    else
      assert(nodes_map[T_target] == Q_target);
  }

  // map the rest of the nodes from T -> Q
  for(int i=Q.n_leaves();i<Q.n_nodes();i++) 
  {
    unsigned D = Q[i].degree();
    if (D <= 3) continue;

    // get a branch of Q pointing into the node
    const_branchview outside = *(Q[i].branches_in());
    // get a branch of T pointing into the node
    outside = T.directed_branch(branches_map[outside.name()]);

    list<const_branchview> branches;
    typedef list<const_branchview>::iterator list_iterator;
    append(outside.branches_after(),branches);
    for(list_iterator b = branches.begin() ; b != branches.end();)
    {
      int node = (*b).target();
      if (nodes_map[node] == -1)
	nodes_map[node] = i;

      if (nodes_map[node] == i) {
	append((*b).branches_after(),branches);
	b++;
      }
      else {
	list_iterator prev = b;
	b++;
	branches.erase(prev);
      }
    }
    assert(branches.size() == D-3);
  }

  for(int i=0;i<nodes_map.size();i++)
    assert(nodes_map[i] != -1);

  return nodes_map;
}
开发者ID:msuchard,项目名称:BAli-Phy,代码行数:68,代码来源:tree-mean-lengths.C

示例7: main

int main(int argc,char* argv[])
{ 
  try {
    //---------- Parse command line  -------//
    variables_map args = parse_cmd_line(argc,argv);

    //----------- Load alignment and tree ---------//
    alignment A;
    SequenceTree T;
    if (args.count("tree"))
      load_A_and_T(args,A,T,false);
    else
      A = load_A(args,false);

    const alphabet& a = A.get_alphabet();
    
    //------- Load groups and find branches -------//
    vector<sequence_group> groups;
    if (args.count("groups")) 
      groups = load_groups(A,args["groups"].as<string>());

    for(int i=0;i<groups.size();i++) {
      cerr<<groups[i].name<<": ";
      for(int j=0;j<groups[i].taxa.size();j++)
	cerr<<A.seq(groups[i].taxa[j]).name<<" ";
      cerr<<endl;
    }

    vector<int> group_branches;
    if (args.count("tree"))
    {
      for(int i=0;i<groups.size();i++)
      {
	dynamic_bitset<> p(T.n_leaves());
	for(int j=0;j<groups[i].taxa.size();j++)
	  p[groups[i].taxa[j]] = true;

	int found = -1;
	for(int b=0;b<2*T.n_branches() and found == -1;b++)
	  if (p == branch_partition(T,b))
	    found = b;
	if (found == -1)
	  throw myexception()<<"I can't find group "<<i+1<<" on the tree!";
	
	group_branches.push_back(found);
      }
    }

    vector<string> group_names;
    for(int i=0;i<groups.size();i++)
      group_names.push_back(groups[i].name);

    vector<Partition> splits;
    if (args.count("split")) 
    {
      vector<string> split = args["split"].as<vector<string> >();
      for(int i=0;i<split.size();i++) 
	splits.push_back(Partition(group_names,split[i]));
    }

    //-------------------------------------------//

    Matrix C(A.length(),A.n_sequences()+1);
    for(int i=0;i<C.size1();i++)
      for(int j=0;j<C.size2();j++)
	C(i,j) = 0;

    // yes but, how much more conservation THAN EXPECTED do we see?

    for(int c=0;c<C.size1();c++) 
    {
      vector<bool> interesting(groups.size(), true);

      //-------------------------------------------------------//
      vector<int> leaf_letters( T.n_leaves() );
      for(int j=0;j<leaf_letters.size();j++)
	leaf_letters[j] = A(c,j);
      vector<vector<int> > node_letters = get_all_parsimony_letters(a,leaf_letters,T,unit_cost_matrix(a));

      vector<vector<int> > initial_value(groups.size());
      for(int g=0;g<groups.size();g++) 
      {
	int n = T.directed_branch(group_branches[g]).target();
	initial_value[g] = node_letters[n];
      }

      //------------ find 'group conserved at' values ----------//
      vector<int> value(groups.size(),alphabet::gap);

      for(int g=0;g<groups.size();g++) 
      {
	vector<int> temp;
	for(int i=0;i<groups[g].taxa.size();i++)
	  temp.push_back(A(c,groups[g].taxa[i]));

	int best = most_common(temp);
	int count = number_of(temp,best);
	
	if (count >= groups[g].taxa.size()-1 and count >=3 and count> groups[g].taxa.size()/2)
	  value[g] = best;
//.........这里部分代码省略.........
开发者ID:,项目名称:,代码行数:101,代码来源:


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