将hcluster生成的ndarray转换为用于ete2包的Newick字符串

Question

我有一个通过运行以下命令创建的向量列表：

import hcluster
import numpy as np
from ete2 import Tree

vecs = [np.array(i) for i in document_list] 

其中document_list是我正在分析的web文档的集合。然后，我执行分层聚类：

Z = hcluster.linkage(vecs, metric='cosine') 

这将生成ndarray，例如：

[[ 12.          19.           0.           1.        ]
[ 15.          21.           0.           3.        ]
[ 18.          22.           0.           4.        ]
[  3.          16.           0.           7.        ]
[  8.          23.           0.           6.        ]
[  5.          27.           0.           6.        ]
[  1.          28.           0.           7.        ]
[  0.          21.           0.           2.        ]
[  5.          29.           0.18350472   2.        ]
[  2.          10.           0.18350472   3.        ]
[ 47.          30.           0.29289577   9.        ]
[ 13.          28.           0.29289577  13.        ]
[ 73.          32.           0.29289577  18.        ]
[ 26.          12.           0.42264521   5.        ]
[  5.          33.           0.42264521  12.        ]
[ 14.          35.           0.42264521  12.        ]
[ 19.          35.           0.42264521  18.        ]
[  4.          20.           0.31174826   3.        ]
[ 34.          21.           0.5         19.        ]
[ 38.          29.           0.31174826  21.        ]]

是否可以将此ndarray转换为可传递给ete2树()构造函数的newick字符串，以便我可以使用ete2提供的工具绘制和操作newick树？

尝试这样做有意义吗?如果没有，有没有其他方法可以使用相同的数据和ete2来生成树/树图(我意识到还有其他包可以绘制树状图，比如树状图和hcluster本身，但还是更喜欢使用ete2 )？

谢谢!

Answer 1

我使用下面的方法来做几乎相同的事情：

from hcluster import linkage, to_tree
from ete2 import Tree

#hcluster part
Y = dist_matrix(items, dist_fn)
Z = linkage(Y, "single")
T = to_tree(Z)

#ete2 section
root = Tree()
root.dist = 0
root.name = "root"
item2node = {T: root}

to_visit = [T]
while to_visit:
    node = to_visit.pop()
    cl_dist = node.dist /2.0
    for ch_node in [node.left, node.right]:
        if ch_node:
            ch = Tree()
            ch.dist = cl_dist
            ch.name = str(ch_node.id)
            item2node[node].add_child(ch)
            item2node[ch_node] = ch
            to_visit.append(ch_node)

# This is your ETE tree structure
tree = root

Answer 2

更新：

from hcluster import linkage, to_tree
from ete2 import Tree

#hcluster part
Y = dist_matrix(items, dist_fn)
Z = linkage(Y, "single")

R,T       = to_tree( mat, rd=True )
#print "ROOT", R, "TREE", T
root      = Tree()
root.dist = 0
root.name = 'root'
item2node = {R.get_id(): root}
to_visit  = T

while to_visit:
    node = to_visit.pop()
    #print "NODE", node
    cl_dist = node.dist / 2.0

    for ch_node in [node.get_left(), node.get_right()]:
        if ch_node:
            ch_node_id         = ch_node.get_id()
            ch_node_name       = str(ch_node_id)
            ch                 = Tree()
            ch.dist            = cl_dist
            ch.name            = ch_node_name

            if nodeNames:
                if ch_node_id < len(nodeNames):
                    ch.name    = nodeNames[ ch_node_id ]

            item2node[ch_node_id] = ch
            item2node[ch_node_id].add_child(ch)
            to_visit.append(ch_node)