Question :
Can I extract the underlying decision-rules (or ‘decision paths’) from a trained tree in a decision tree as a textual list?
Something like:
if A>0.4 then if B<0.2 then if C>0.8 then class='X'
Thanks for your help.
Answer #1:
I believe that this answer is more correct than the other answers here:
from sklearn.tree import _tree
def tree_to_code(tree, feature_names):
tree_ = tree.tree_
feature_name = [
feature_names[i] if i != _tree.TREE_UNDEFINED else "undefined!"
for i in tree_.feature
]
print "def tree({}):".format(", ".join(feature_names))
def recurse(node, depth):
indent = " " * depth
if tree_.feature[node] != _tree.TREE_UNDEFINED:
name = feature_name[node]
threshold = tree_.threshold[node]
print "{}if {} <= {}:".format(indent, name, threshold)
recurse(tree_.children_left[node], depth + 1)
print "{}else: # if {} > {}".format(indent, name, threshold)
recurse(tree_.children_right[node], depth + 1)
else:
print "{}return {}".format(indent, tree_.value[node])
recurse(0, 1)
This prints out a valid Python function. Here’s an example output for a tree that is trying to return its input, a number between 0 and 10.
def tree(f0):
if f0 <= 6.0:
if f0 <= 1.5:
return [[ 0.]]
else: # if f0 > 1.5
if f0 <= 4.5:
if f0 <= 3.5:
return [[ 3.]]
else: # if f0 > 3.5
return [[ 4.]]
else: # if f0 > 4.5
return [[ 5.]]
else: # if f0 > 6.0
if f0 <= 8.5:
if f0 <= 7.5:
return [[ 7.]]
else: # if f0 > 7.5
return [[ 8.]]
else: # if f0 > 8.5
return [[ 9.]]
Here are some stumbling blocks that I see in other answers:
- Using
tree_.threshold == -2
to decide whether a node is a leaf isn’t a good idea. What if it’s a real decision node with a threshold of -2? Instead, you should look attree.feature
ortree.children_*
. - The line
features = [feature_names[i] for i in tree_.feature]
crashes with my version of sklearn, because some values oftree.tree_.feature
are -2 (specifically for leaf nodes). - There is no need to have multiple if statements in the recursive function, just one is fine.
Answer #2:
I created my own function to extract the rules from the decision trees created by sklearn:
import pandas as pd
import numpy as np
from sklearn.tree import DecisionTreeClassifier
# dummy data:
df = pd.DataFrame({'col1':[0,1,2,3],'col2':[3,4,5,6],'dv':[0,1,0,1]})
# create decision tree
dt = DecisionTreeClassifier(max_depth=5, min_samples_leaf=1)
dt.fit(df.ix[:,:2], df.dv)
This function first starts with the nodes (identified by -1 in the child arrays) and then recursively finds the parents. I call this a node’s ‘lineage’. Along the way, I grab the values I need to create if/then/else SAS logic:
def get_lineage(tree, feature_names):
left = tree.tree_.children_left
right = tree.tree_.children_right
threshold = tree.tree_.threshold
features = [feature_names[i] for i in tree.tree_.feature]
# get ids of child nodes
idx = np.argwhere(left == -1)[:,0]
def recurse(left, right, child, lineage=None):
if lineage is None:
lineage = [child]
if child in left:
parent = np.where(left == child)[0].item()
split = 'l'
else:
parent = np.where(right == child)[0].item()
split = 'r'
lineage.append((parent, split, threshold[parent], features[parent]))
if parent == 0:
lineage.reverse()
return lineage
else:
return recurse(left, right, parent, lineage)
for child in idx:
for node in recurse(left, right, child):
print node
The sets of tuples below contain everything I need to create SAS if/then/else statements. I do not like using do
blocks in SAS which is why I create logic describing a node’s entire path. The single integer after the tuples is the ID of the terminal node in a path. All of the preceding tuples combine to create that node.
In [1]: get_lineage(dt, df.columns)
(0, 'l', 0.5, 'col1')
1
(0, 'r', 0.5, 'col1')
(2, 'l', 4.5, 'col2')
3
(0, 'r', 0.5, 'col1')
(2, 'r', 4.5, 'col2')
(4, 'l', 2.5, 'col1')
5
(0, 'r', 0.5, 'col1')
(2, 'r', 4.5, 'col2')
(4, 'r', 2.5, 'col1')
6
Answer #3:
I modified the code submitted by Zelazny7 to print some pseudocode:
def get_code(tree, feature_names):
left = tree.tree_.children_left
right = tree.tree_.children_right
threshold = tree.tree_.threshold
features = [feature_names[i] for i in tree.tree_.feature]
value = tree.tree_.value
def recurse(left, right, threshold, features, node):
if (threshold[node] != -2):
print "if ( " + features[node] + " <= " + str(threshold[node]) + " ) {"
if left[node] != -1:
recurse (left, right, threshold, features,left[node])
print "} else {"
if right[node] != -1:
recurse (left, right, threshold, features,right[node])
print "}"
else:
print "return " + str(value[node])
recurse(left, right, threshold, features, 0)
if you call get_code(dt, df.columns)
on the same example you will obtain:
if ( col1 <= 0.5 ) {
return [[ 1. 0.]]
} else {
if ( col2 <= 4.5 ) {
return [[ 0. 1.]]
} else {
if ( col1 <= 2.5 ) {
return [[ 1. 0.]]
} else {
return [[ 0. 1.]]
}
}
}
Answer #4:
Scikit learn introduced a delicious new method called export_text
in version 0.21 (May 2019) to extract the rules from a tree. Documentation here. It’s no longer necessary to create a custom function.
Once you’ve fit your model, you just need two lines of code. First, import export_text
:
from sklearn.tree import export_text
Second, create an object that will contain your rules. To make the rules look more readable, use the feature_names
argument and pass a list of your feature names. For example, if your model is called model
and your features are named in a dataframe called X_train
, you could create an object called tree_rules
:
tree_rules = export_text(model, feature_names=list(X_train.columns))
Then just print or save tree_rules
. Your output will look like this:
|--- Age <= 0.63
| |--- EstimatedSalary <= 0.61
| | |--- Age <= -0.16
| | | |--- class: 0
| | |--- Age > -0.16
| | | |--- EstimatedSalary <= -0.06
| | | | |--- class: 0
| | | |--- EstimatedSalary > -0.06
| | | | |--- EstimatedSalary <= 0.40
| | | | | |--- EstimatedSalary <= 0.03
| | | | | | |--- class: 1
Answer #5:
There is a new DecisionTreeClassifier
method, decision_path
, in the 0.18.0 release. The developers provide an extensive (well-documented) walkthrough.
The first section of code in the walkthrough that prints the tree structure seems to be OK. However, I modified the code in the second section to interrogate one sample. My changes denoted with # <--
Edit The changes marked by # <--
in the code below have since been updated in walkthrough link after the errors were pointed out in pull requests #8653 and #10951. It’s much easier to follow along now.
sample_id = 0
node_index = node_indicator.indices[node_indicator.indptr[sample_id]:
node_indicator.indptr[sample_id + 1]]
print('Rules used to predict sample %s: ' % sample_id)
for node_id in node_index:
if leave_id[sample_id] == node_id: # <-- changed != to ==
#continue # <-- comment out
print("leaf node {} reached, no decision here".format(leave_id[sample_id])) # <--
else: # < -- added else to iterate through decision nodes
if (X_test[sample_id, feature[node_id]] <= threshold[node_id]):
threshold_sign = "<="
else:
threshold_sign = ">"
print("decision id node %s : (X[%s, %s] (= %s) %s %s)"
% (node_id,
sample_id,
feature[node_id],
X_test[sample_id, feature[node_id]], # <-- changed i to sample_id
threshold_sign,
threshold[node_id]))
Rules used to predict sample 0:
decision id node 0 : (X[0, 3] (= 2.4) > 0.800000011921)
decision id node 2 : (X[0, 2] (= 5.1) > 4.94999980927)
leaf node 4 reached, no decision here
Change the sample_id
to see the decision paths for other samples. I haven’t asked the developers about these changes, just seemed more intuitive when working through the example.
Answer #6:
from StringIO import StringIO
out = StringIO()
out = tree.export_graphviz(clf, out_file=out)
print out.getvalue()
You can see a digraph Tree. Then, clf.tree_.feature
and clf.tree_.value
are array of nodes splitting feature and array of nodes values respectively. You can refer to more details from this github source.
Answer #7:
Just because everyone was so helpful I’ll just add a modification to Zelazny7 and Daniele’s beautiful solutions. This one is for python 2.7, with tabs to make it more readable:
def get_code(tree, feature_names, tabdepth=0):
left = tree.tree_.children_left
right = tree.tree_.children_right
threshold = tree.tree_.threshold
features = [feature_names[i] for i in tree.tree_.feature]
value = tree.tree_.value
def recurse(left, right, threshold, features, node, tabdepth=0):
if (threshold[node] != -2):
print 't' * tabdepth,
print "if ( " + features[node] + " <= " + str(threshold[node]) + " ) {"
if left[node] != -1:
recurse (left, right, threshold, features,left[node], tabdepth+1)
print 't' * tabdepth,
print "} else {"
if right[node] != -1:
recurse (left, right, threshold, features,right[node], tabdepth+1)
print 't' * tabdepth,
print "}"
else:
print 't' * tabdepth,
print "return " + str(value[node])
recurse(left, right, threshold, features, 0)
Answer #8:
I’ve been going through this, but i needed the rules to be written in this format
if A>0.4 then if B<0.2 then if C>0.8 then class='X'
So I adapted the answer of @paulkernfeld (thanks) that you can customize to your need
def tree_to_code(tree, feature_names, Y):
tree_ = tree.tree_
feature_name = [
feature_names[i] if i != _tree.TREE_UNDEFINED else "undefined!"
for i in tree_.feature
]
pathto=dict()
global k
k = 0
def recurse(node, depth, parent):
global k
indent = " " * depth
if tree_.feature[node] != _tree.TREE_UNDEFINED:
name = feature_name[node]
threshold = tree_.threshold[node]
s= "{} <= {} ".format( name, threshold, node )
if node == 0:
pathto[node]=s
else:
pathto[node]=pathto[parent]+' & ' +s
recurse(tree_.children_left[node], depth + 1, node)
s="{} > {}".format( name, threshold)
if node == 0:
pathto[node]=s
else:
pathto[node]=pathto[parent]+' & ' +s
recurse(tree_.children_right[node], depth + 1, node)
else:
k=k+1
print(k,')',pathto[parent], tree_.value[node])
recurse(0, 1, 0)