Intro:
Chapter 4. 넘파이 기본: 배열과 벡터 연산
Chapter 5. Pandas
기초 개념
1
2
3
4
5
6
7
8
import numpy as np
import pandas as pd
logx = np.logspace(0, 1, 100)
linx = np.linspace(0, 10, 100)
df = pd.DataFrame()
df['logspace'] = logx #컬럼명 logspace의 Series 데이터
df['linspace'] = linx # 컬럼명 linespace의 Series 데이터
Column명 변경
DataFrame의 컬럼명 Rename 3가지 방법
pd.DataFrame.columns = [ '칼럼1', '칼럼2', … , ]pd.DataFrame.rename( { 'OLD칼럼명' : 'NEW칼럼명', …, }, axis=1)pd.DataFrame.rename(columns={ 'OLD칼럼명' : 'NEW칼럼명', …, })
###
몬티홀 딜레마
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
import numpy as np
import pandas as pd
import matplotlib.pyplot as plt
%matplotlib inline
# 그림 그리는 환경
import random
def init():
doors = ['A', 'B', 'C']
choice = random.choice(doors)
prize = random.choice(doors)
return choice, prize
def MC(mychoice, prize):
candidates = ["A", "B", "C"]
if mychoice == prize:
candidates.remove(prize)
else:
candidates = [x for x in candidates if x != mychoice and x != prize]
return random.choice(candidates)
def no_change():
choice, prize = init()
opened_door = MC(choice, prize)
return 1 if choice == prize else 0
def change():
choice, prize = init()
opened_door = MC(choice, prize)
return 0 if choice == prize else 1
from tqdm.auto import tqdm, trange
tries = 1000
result_change = []
result_nochange = []
for _ in tqdm(range(tries), position=0):
win_nochange = 0
win_change = 0
for _ in range(tries):
win_nochange += no_change()
win_change += change()
result_nochange.append(win_nochange)
result_change.append(win_change)
plt.figure(figsize=(20, 20))
plt.plot(result_change, 'o:', label='Change')
plt.plot(result_nochange, 'o:', label='No Change')
plt.legend()
plt.title('Monty Hall Dilemma')
plt.xlabel('Trial')
plt.ylabel('? / 100')
