Mean vs Median

This script compares the errors based on mean and median. It shows that the errors based on mean are more sensitive to outliers than the errors based upon median.

import os
import sys
import numpy as np

import easy_mpl
from easy_mpl import plot

import SeqMetrics
from SeqMetrics import mse, med_seq_error, mae, median_abs_error, mape, mdape, me, mde

print('python version: ', sys.version)
print('OS Name: ', os.name)
print('numpy: ', np.__version__)
print('easy_mpl: ', easy_mpl.__version__)
print('SeqMetrics: ', SeqMetrics.__version__)

python version:  3.8.20 (default, Nov 25 2025, 20:14:35)
[GCC 9.4.0]
OS Name:  posix
numpy:  1.24.4
easy_mpl:  0.21.5
SeqMetrics:  2.0.1

Consider that we have two outputs from our model. One output is normal and the other has an outlier. We will compare the errors based on mean and median.

true = np.random.random(100)
pred = np.random.random(100)
pred_o = pred.copy()
pred_o[2] = 40

First see that how the array pred_o has just one outlier. The outlier is at index 2.

plot(true,'--.', show=False, label="True")
plot(pred,'--.', show=False, label="Predicted")
plot(pred_o, '--.', show=False, label="Predicted with outlier")

<Axes: >

See the change in mean and median errors. The change in mean error is very high but the change in median error is very low.

print("Simple Error")
print("Mean: ", round(me(true, pred), 2), '-->', round(me(true, pred_o), 2))
print("Median: ", round(mde(true, pred), 2), '-->', round(mde(true, pred_o), 2))

Simple Error
Mean:  0.01 --> -0.38
Median:  0.04 --> 0.04

Similarly the change in mean squared error is very large but the change in median squared error is very small.

print("Squared Error")
print(round(mse(true, pred), 2), '-->', round(mse(true, pred_o), 2))
print(round(med_seq_error(true, pred),2), '-->', round(med_seq_error(true, pred_o), 2))

Squared Error
0.17 --> 15.72
0.1 --> 0.1

print("Asbolute Error")
print("Mean: ", round(mae(true, pred), 2), '-->', round(mae(true, pred_o), 2))
print("Median: ", round(median_abs_error(true, pred), 2), '-->', round(median_abs_error(true, pred_o), 2))

Asbolute Error
Mean:  0.34 --> 0.73
Median:  0.32 --> 0.32

print("Abolute Percentage Error")
print("Mean:", round(mape(true, pred), 2), '-->', round(mape(true, pred_o), 2))
print("Median:", round(mdape(true, pred), 2), '-->', round(mdape(true, pred_o), 2))

Abolute Percentage Error
Mean: 165.63 --> 234.51
Median: 52.26 --> 53.15

So the erros based on mean are more sensitive to outliers than the errors based upon median. Therefore, if the data has outliers and we want to ignore them, we should use median based errors. On the other hand, if we want to focus on the outliers, we should use mean based errors.