How to traverse a decision tree with a test sample in Python?

Question

Python

Hello.
There is a code for constructing a decision tree on a training sample.
Now we need to go through the tree with the test sample and get the classification result.
Thank you.

The code

Import math
import pandas as pd
from functools import reduce

data = {
    "Кредитная_история":["Плохое","Неизвестно","Неизвестно","Неизвестно","Неизвестно","Неизвестно","Плохая","Плохая","Хорошая","Хорошая","Хорошая","Хорошая","Хорошая","Плохая"],
    "Долг":["Высокий","Высокий","Низкий","Низкий","Низкий","Низкий","Низкий","Низкий","Низкий","Высокий","Высокий","Высокий","Высокий","Низкий"], 
    "Поручитель":["Нет","Нет","Нет","Нет","Нет","Адекватный","Нет","Адекватный","Нет","Адекватный","Нет","Нет","Нет","Нет"],
    "Доход":["0-15000","15000-30000","15000-30000","0-15000","35000+","35000+","0-15000","35000+","35000+","35000+","0-15000","15000-30000","35000+","15000-30000"],
    "Риск":["Высокий","Высокий","Средний","Высокий","Низкий","Низкий","Высокий","Средний","Низкий","Низкий","Высокий","Средний","Низкий","Высокий"],
}

df = pd.DataFrame(data)

# массив содержащий строку вида: кортеж(key, value)
convert_str = lambda string:[key+":"+str(value) for key, value in sorted(string.value_counts().items())]

Tree = {
    "name": df.columns[-1]+" "+str(cstr(df.iloc[:,-1])),
    "df": df,
    "edges":[],
}

def log_(x):
    i = len(df.iloc[:,-1].unique())
    return math.log(x)/math.log(6)

X = [Tree]
entropy = lambda s:-reduce(lambda x,y:x+y,map(lambda x:(x/len(s))*math.log2(x/len(s)),s.value_counts()))

def solution():
    while(len(X)!=0):
        n = X.pop(0)
        df_n = n["df"]
        if 0==entropy(df_n.iloc[:,-1]):
            continue
        attrs = {}
        for attr in df_n.columns[:-1]:
            attrs[attr] = {"entropy":0,"dfs":[],"values":[]}
            for value in sorted(set(df_n[attr])):
                df_m = df_n.query(attr+"=='"+value+"'")
                attrs[attr]["entropy"] += entropy(df_m.iloc[:,-1])*df_m.shape[0]/df_n.shape[0]
                attrs[attr]["dfs"] += [df_m]
                attrs[attr]["values"] += [value]
                pass
            pass
        if len(attrs)==0:
            continue
        attr = min(attrs,key=lambda x:attrs[x]["entropy"])
        for d,v in zip(attrs[attr]["dfs"],attrs[attr]["values"]):
            m = {"name":attr+"="+v,"edges":[],"df":d.drop(columns=attr)}
            n["edges"].append(m)
            X.append(m)
        pass

def tstr(tree,indent=""):
    s = indent+tree["name"]+str(cstr(tree["df"].iloc[:,-1]) if len(tree["edges"])==0 else "")+"\n"
    for e in tree["edges"]:
        s += tstr(e,indent+"  ")
        pass
    return s

solution()
print(tstr(Tree))

test set

data = {
    "Кредитная_история":["Плохое","Неизвестно","Неизвестно","Неизвестно","Неизвестно","Неизвестно","Плохая","Плохая","Хорошая","Хорошая","Хорошая","Хорошая","Хорошая","Плохая"],
    "Долг":["Высокий","Высокий","Низкий","Низкий","Низкий","Низкий","Низкий","Низкий","Низкий","Высокий","Высокий","Высокий","Высокий","Низкий"], 
    "Поручитель":["Нет","Нет","Нет","Нет","Нет","Адекватный","Нет","Адекватный","Нет","Адекватный","Нет","Нет","Нет","Нет"],
    "Доход":["0-15000","15000-30000","15000-30000","0-15000","35000+","35000+","0-15000","35000+","35000+","35000+","0-15000","15000-30000","35000+","15000-30000"],
}

Answer 1

Just thought of this before.

for i in range(len(data["Долг"])):
    temp = list()
    for key,_ in data.items():
        temp.append(key+"="+data[key][i])
        pass
    new_data.append(temp.copy())

print(new_data)
print('--------------------')


def bypass(tree, data_x, data_y, data_z):
    name_tree = tree["name"]
    target_tree = str(cstr(tree["df"].iloc[:,-1]) if len(tree["edges"])==0 else "")
    for i in [data_x, data_y, data_z]:
        if i == name_tree:
            for e in tree["edges"]:
                name_tree += bypass(e, data_x, data_y, data_z)
                pass
            print(i, name_tree)
    for e in tree["edges"]:
        name_tree += bypass(e, data_x, data_y, data_z)
        pass
    return name_tree
  
for i in range(len(new_data)):
    bypass(Tree, new_data[i][3], new_data[i][0], new_data[i][1])