# feature scaling xgbRegressor

I read for example in this answer: Does the performance of GBM methods profit from feature scaling?

that scaling doesn´t affect the performance of any tree-based method, not for lightgbm,xgboost,catboost or even decision tree.

When i do feature scaling and compare the rmse of a xgboost model without and with minmax scaling, i got a better rmse value with feature scaling. Here is the code:

from sklearn.preprocessing import MinMaxScaler
from sklearn.model_selection import train_test_split
from sklearn.metrics import mean_squared_error as MSE
import math
from math import sqrt
import pandas as pd
import numpy as np
from xgboost import XGBRegressor
import xgboost as xgb

X = data.drop(['colA'], axis=1)
y = data['colA']

scaler = MinMaxScaler()
scaler.fit(X)
minmax_scaled_X = scaler.transform(X)
minmax_scaled_X
y = np.array(y).reshape(-1, 1)
scaler.fit(y)
minmax_scaled_y = scaler.transform(y)

from sklearn.model_selection import train_test_split
xtrain, xtest, ytrain, ytest = train_test_split(minmax_scaled_X, minmax_scaled_y, test_size =0.3, random_state=0, shuffle=True)

xg_reg = xgb.XGBRegressor(objective ='reg:squarederror', colsample_bytree = 0.7, learning_rate = 0.05,
max_depth = 8, min_child_weight = 4, n_estimators = 600, subsample = 0.7)

xg_reg.fit(xtrain,ytrain)
preds = xg_reg.predict(xtest)
rmse = sqrt(MSE(ytest, preds))
print(rmse)


the result with min max scaling is 0.003, while the rmse without is about 3.8. I did the same with simple decision tree and got always a better result with minmax scaling.

Where is my mistake? In other posts like the link above, answers are about that it is not good to scale when using trees. Can I say, that min max scaling does have a positive effect on the rmse on my data?

You're also scaling $$y$$, then of course you are getting lower error. That question was regarding scaling $$X$$.
The same model will have very different error metrics when units on $$y$$ are changed: if I multiply all $$y$$ values by 100, the error will be 100 times larger, if I divide all $$y$$ values by 100 the error will be divided by 100.