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from ete3 import ClusterTree
# Example of a minimalistic numerical matrix. It is encoded as a text
# string for convenience, but it usally be loaded from a text file.
matrix = """
#Names\tcol1\tcol2\tcol3\tcol4\tcol5\tcol6\tcol7
A\t-1.23\t-0.81\t1.79\t0.78\t-0.42\t-0.69\t0.58
B\t-1.76\t-0.94\t1.16\t0.36\t0.41\t-0.35\t1.12
C\t-2.19\t0.13\t0.65\t-0.51\t0.52\t1.04\t0.36
D\t-1.22\t-0.98\t0.79\t-0.76\t-0.29\t1.54\t0.93
E\t-1.47\t-0.83\t0.85\t0.07\t-0.81\t1.53\t0.65
F\t-1.04\t-1.11\t0.87\t-0.14\t-0.80\t1.74\t0.48
G\t-1.57\t-1.17\t1.29\t0.23\t-0.20\t1.17\t0.26
H\t-1.53\t-1.25\t0.59\t-0.30\t0.32\t1.41\t0.77
"""
print("Example numerical matrix")
print(matrix)
# #Names col1 col2 col3 col4 col5 col6 col7
# A -1.23 -0.81 1.79 0.78 -0.42 -0.69 0.58
# B -1.76 -0.94 1.16 0.36 0.41 -0.35 1.12
# C -2.19 0.13 0.65 -0.51 0.52 1.04 0.36
# D -1.22 -0.98 0.79 -0.76 -0.29 1.54 0.93
# E -1.47 -0.83 0.85 0.07 -0.81 1.53 0.65
# F -1.04 -1.11 0.87 -0.14 -0.80 1.74 0.48
# G -1.57 -1.17 1.29 0.23 -0.20 1.17 0.26
# H -1.53 -1.25 0.59 -0.30 0.32 1.41 0.77
#
#
# We load a tree structure whose leaf nodes correspond to rows in the
# numerical matrix. We use the text_array argument to link the tree
# with numerical matrix.
t = ClusterTree("(((A,B),(C,(D,E))),(F,(G,H)));", text_array=matrix)
print("Example tree", t)
# /-A
# /--------|
# | \-B
# /--------|
# | | /-C
# | \--------|
# | | /-D
#---------| \--------|
# | \-E
# |
# | /-F
# \--------|
# | /-G
# \--------|
# \-H
# Now we can ask the numerical profile associated to each node
A = t.search_nodes(name='A')[0]
print("A associated profile:\n", A.profile)
# [-1.23 -0.81 1.79 0.78 -0.42 -0.69 0.58]
#
# Or we can ask for the mean numerical profile of an internal
# partition, which is computed as the average of all vectors under the
# the given node.
cluster = t.get_common_ancestor("E", "A")
print("Internal cluster mean profile:\n", cluster.profile)
#[-1.574 -0.686 1.048 -0.012 -0.118 0.614 0.728]
#
# We can also obtain the std. deviation vector of the mean profile
print("Internal cluster std deviation profile:\n", cluster.deviation)
#[ 0.36565558 0.41301816 0.40676283 0.56211743 0.50704635 0.94949671
# 0.26753691]
# If would need to re-link the tree to a different matrix or use
# different matrix for different sub parts of the tree, we can use the
# link_to_arraytable method()
#
# Creates a matrix with all values = 1
matrix_ones = """
#Names\tcol1\tcol2\tcol3\tcol4\tcol5\tcol6\tcol7
A\t1\t1\t1\t1\t1\t1\t1
B\t1\t1\t1\t1\t1\t1\t1
C\t1\t1\t1\t1\t1\t1\t1
D\t1\t1\t1\t1\t1\t1\t1
E\t1\t1\t1\t1\t1\t1\t1
F\t1\t1\t1\t1\t1\t1\t1
G\t1\t1\t1\t1\t1\t1\t1
H\t1\t1\t1\t1\t1\t1\t1
"""
# Creates a matrix with all values = 0
matrix_zeros = """
#Names\tcol1\tcol2\tcol3\tcol4\tcol5\tcol6\tcol7
A\t0\t0\t0\t0\t0\t0\t0
B\t0\t0\t0\t0\t0\t0\t0
C\t0\t0\t0\t0\t0\t0\t0
D\t0\t0\t0\t0\t0\t0\t0
E\t0\t0\t0\t0\t0\t0\t0
F\t0\t0\t0\t0\t0\t0\t0
G\t0\t0\t0\t0\t0\t0\t0
H\t0\t0\t0\t0\t0\t0\t0
"""
# Re-associate left part of the tree with matrix of 1s, and right part
# with matrix of 0s. Note that rows without matches in both are
# obviated from association.
t.children[0].link_to_arraytable(matrix_ones)
t.children[1].link_to_arraytable(matrix_zeros)
print("A profile (using matrix with 1s", (t&"A").profile)
print("H profile (using matrix with 0s)", (t&"H").profile)
#A profile (using matrix with 1s [ 1. 1. 1. 1. 1. 1. 1.]
#H profile (using matrix with 0s) [ 0. 0. 0. 0. 0. 0. 0.]
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