Branches and dendropy

Dendropy contains many subclasses inside each object, then extract information might actually end up being a task of finding the correct one subclass. Here I present the call of those subclasses for getting some common metrics in branches

Terminal branch lengths

Given the following mock tree:

import dendropy

str_tree = '((A:1,B:2):3,(C:4,D:5):6);'
tree = dendropy.Tree.get_from_string(str_tree, 'newick')

print( tree.as_ascii_plot(plot_metric = 'length') )
        /-- A
/------+
|       \---- B
+              
|             /-------- C 
\------------+           
              \---------- D

We can get terminal branch length by iterating nodes in ‘postorder’:

df = {}
for nd in tree.postorder_edge_iter():
    if nd.length is None:
        continue
    taxn = nd.head_node.taxon
    if taxn:
        df[taxn.label] = nd.length

df
{'A': 1.0, 'B': 2.0, 'C': 4.0, 'D': 5.0}

Collapse nodes into a polytomy

Between two internal nodes theres is a edge (i.e., branch length) and when this edge is so short, you might want to collapse these two nodes, thus creating a polytomy.

Given mock the following tree:

str_tree = '((A:1,B:2):3,((C:0,D:0):0,E:7):8);'
tree = dendropy.Tree.get_from_string(str_tree, 'newick')

print( tree.as_string('newick') )
((A:1.0,B:2.0):3.0,((C:0.0,D:0.0):0.0,E:7.0):8.0);

We can see from the above string that ‘C’ and ‘D’ form a single clade and ‘E’ is the sister taxon of these ones. Now, lets try to collapse these three taxa into a single clade by using a threshold (i.e., min_len = 0):

min_len = 0

for nd in tree.postorder_edge_iter():
    if nd.length is None:
        continue
    if nd.is_internal() and nd.length <= min_len:
            nd.collapse()

print( tree.as_string('newick') )
((A:1.0,B:2.0):3.0,(C:0.0,D:0.0,E:7.0):8.0);

Notice that now ‘C’, ‘D’, and ‘E’ are inside the same parenthesis.

2020

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