[STBDA2023] 03wk-012: 취업, 로지스틱

Author

김보람

Published

September 28, 2023

해당 자료는 전북대학교 최규빈 교수님 2023학년도 2학기 빅데이터분석특강 자료임

03wk-012: 취업, 로지스틱

최규빈
2023-09-21

1. 강의영상

https://youtu.be/playlist?list=PLQqh36zP38-z03THS4jG11HPcozk3ZfVS&si=Ry49nDAOI3PSu0Ja

2. Imports

import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
import sklearn.linear_model

3. 데이터 불러오기

df = pd.read_csv('https://raw.githubusercontent.com/guebin/MP2023/main/posts/employment.csv')
df
toeic gpa employment
0 135 0.051535 0
1 935 0.355496 0
2 485 2.228435 0
3 65 1.179701 0
4 445 3.962356 1
... ... ... ...
495 280 4.288465 1
496 310 2.601212 1
497 225 0.042323 0
498 320 1.041416 0
499 375 3.626883 1

500 rows × 3 columns

plt.plot(df.toeic,df.gpa,'o')
df_filtered = df[df.employment==1]
plt.plot(df_filtered.toeic,df_filtered.gpa,'o')

  • 주황색이 합격

4. 분석

A. 데이터 정리

X = pd.get_dummies(df[['toeic','gpa']])
y = df[['employment']]
X
toeic gpa
0 135 0.051535
1 935 0.355496
2 485 2.228435
3 65 1.179701
4 445 3.962356
... ... ...
495 280 4.288465
496 310 2.601212
497 225 0.042323
498 320 1.041416
499 375 3.626883

500 rows × 2 columns

B. Predictor (\(\star\star\star\))

- 여기가 중요함. \(y\)가 연속형이 아니라 범주형(취업/미취업)으로 이루어진 경우는 sklearn.linear_model.LogisticRegression() 이용하여 predictor를 만들 것

predictr = sklearn.linear_model.LogisticRegression()
predictr
LogisticRegression()
In a Jupyter environment, please rerun this cell to show the HTML representation or trust the notebook.
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C. 학습

predictr.fit(X,y)
/home/coco/anaconda3/envs/py38/lib/python3.8/site-packages/sklearn/utils/validation.py:1143: DataConversionWarning: A column-vector y was passed when a 1d array was expected. Please change the shape of y to (n_samples, ), for example using ravel().
  y = column_or_1d(y, warn=True)
LogisticRegression()
In a Jupyter environment, please rerun this cell to show the HTML representation or trust the notebook.
On GitHub, the HTML representation is unable to render, please try loading this page with nbviewer.org.

D. 예측

predictr.predict(X) 
array([0, 0, 0, 0, 1, 0, 0, 0, 1, 0, 0, 1, 1, 0, 1, 0, 0, 1, 1, 0, 1, 1,
       1, 1, 1, 1, 1, 0, 0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 1, 1,
       0, 0, 1, 0, 1, 0, 1, 1, 1, 0, 0, 0, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0,
       0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 1, 1,
       1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 1, 1, 0, 0, 0, 0, 1, 0, 1, 0,
       1, 0, 1, 0, 1, 1, 1, 0, 1, 1, 0, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0,
       1, 0, 0, 1, 1, 1, 1, 1, 1, 1, 1, 1, 0, 1, 1, 0, 1, 1, 0, 1, 1, 1,
       0, 0, 0, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0, 1, 0, 1, 1, 0, 1,
       1, 1, 1, 1, 0, 1, 1, 0, 1, 0, 0, 1, 0, 1, 0, 0, 1, 1, 0, 0, 1, 1,
       1, 1, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 1,
       0, 0, 1, 0, 1, 0, 0, 0, 1, 1, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 1, 0,
       1, 1, 1, 0, 0, 0, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 1, 1, 0, 0, 0, 0,
       0, 0, 1, 1, 0, 1, 1, 1, 0, 0, 1, 1, 1, 0, 1, 1, 0, 1, 0, 1, 1, 0,
       0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 0, 1, 0, 0, 0, 0, 1, 1, 0, 0, 1, 1,
       0, 0, 0, 0, 0, 0, 1, 0, 1, 1, 1, 1, 0, 0, 1, 1, 0, 0, 0, 1, 1, 0,
       0, 1, 1, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0, 0, 1, 1, 1,
       0, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0, 1, 1, 1, 0, 1, 1, 1, 1, 1, 0, 1,
       0, 0, 0, 0, 1, 1, 1, 0, 0, 0, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 1, 1,
       0, 1, 1, 0, 0, 0, 1, 0, 1, 0, 1, 1, 0, 0, 1, 1, 1, 1, 0, 1, 0, 0,
       0, 0, 1, 1, 0, 0, 1, 0, 0, 1, 0, 0, 0, 1, 1, 1, 0, 1, 0, 1, 0, 1,
       1, 1, 0, 1, 1, 1, 1, 1, 0, 1, 1, 1, 0, 1, 1, 1, 0, 1, 0, 0, 0, 1,
       0, 1, 0, 1, 1, 0, 1, 0, 1, 0, 1, 0, 1, 0, 0, 1, 0, 1, 0, 1, 1, 0,
       0, 0, 1, 1, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 1])

E. 평가

predictr.score(X,y)
0.882
(predictr.predict(X) == y.employment).mean()
0.882
plt.plot(df.toeic,df.gpa,'o')
df_filtered = df[predictr.predict(X)==1]
plt.plot(df_filtered.toeic,df_filtered.gpa,'o')

  • 이 정도면 합리적임