https://www.dataq.or.kr/www/sub/a_07.do
- 예시문제
- 기출문제
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Code and Output
Code : BAEPracticalQuiz1.r
# 출력을 원할 경우 print() 함수 활용 # 예시) print(df.head()) # setwd(), getwd() 등 작업 폴더 설정 불필요 # 파일 경로 상 내부 드라이브 경로(C: 등) 접근 불가 # 데이터 파일 읽기 예제 a <- read.csv("data/mtcars.csv", header=TRUE)
# 사용자 코딩 # print(head(a)) # need to wrap with print() # str(a) # summary(a) qsec <- a$qsec print(summary(qsec)) qsec_cvt <- (qsec - min(qsec)) / (max(qsec) - min(qsec)) print(summary(qsec_cvt)) qsec_over_median <-qsec_cvt[qsec_cvt>0.5] print(qsec_over_median) ans = length(qsec_over_median)
# 답안 제출 예시 # print(변수명) print(ans)
Output
Min. 1st Qu. Median Mean 3rd Qu. Max. 14.50 16.89 17.71 17.85 18.90 22.90
Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0000 0.2848 0.3821 0.3987 0.5238 1.0000
[1] 0.5880952 0.6809524 0.6547619 1.0000000 0.5238095 0.5916667 0.6428571 [8] 0.6559524 0.5238095
[1] 9
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Code and Output
Code : BAEPracticalQuiz2.r
# 출력을 원할 경우 print() 함수 활용 # 예시) print(df.head()) # setwd(), getwd() 등 작업 폴더 설정 불필요 # 파일 경로 상 내부 드라이브 경로(C: 등) 접근 불가 X_test = read.csv('data/X_test.csv') X_train = read.csv('data/X_train.csv') y_train = read.csv('data/y_train.csv')
# 사용자 코딩 library(rpart) library(caret)
# 1. Data Skimming # print(str(X_train)) # print(str(y_train)) # print(str(X_test)) # print(summary(X_train)) # print(summary(y_train)) # table(y_train[2])
# 2. Data Pre-processing # print(head(X_train[order(X_train$총구매액, decreasing=T),])) # don't remove outliers in 최대구매액 # print(head(X_train[is.na(X_train$환불금액),])) # remove or replace as 0 when 환불금액 == 0? X_train[is.na(X_train$환불금액),]$환불금액 <- 0 # replace as 0 # print(summary(X_train)) # print(help(sample)) n <- nrow(X_train) set.seed(230623) idx <- sample(n, n * 0.7, replace=F) X_train_new <- X_train[idx, -1] # remove cust_id y_train_new <- y_train[idx, 2] X_valid <- X_train[-idx, -1] y_valid <- y_train[-idx, 2] # print(str(X_train_new)) # print(summary(X_train_new)) # print(str(y_train_new))
# 3. Model Fitting # print(summary(train_new[, 1])) # idx doesn't start from 0! model <- rpart(y_train_new ~ ., data = X_train_new, method = "class", cp = 0.023) printcp(model)
# 4. Validation pred <- predict(model, newdata = X_valid, method = "class") pred_new <- pred[, 2] # print(summary(pred)) # print(summary(pred_new)) pred_new[pred_new>=0.5] <- 1 pred_new[pred_new<0.5] <- 0 print(summary(pred_new)) print(summary(y_valid)) print(cor(pred[, 2], y_valid)) # 0.2347942 print(cor(pred_new, y_valid)) # 0.2313989 # print(table(pred_new, y_valid)) # pred_new 0 1 # 0 463 176 # 1 204 207 # Seems not good print(confusionMatrix(table(pred_new, y_valid))) # Accuracy : 0.6381 # 95% CI : (0.6082, 0.6672) # No Information Rate : 0.6352 # P-Value [Acc > NIR] : 0.4374
# 5. Submission # 답안 제출 참고 # 아래 코드 변수명과 수험번호를 개인별로 변경하여 활용 # write.csv(변수명,'003000000.csv',row.names=F) pred2 <- predict(model, newdata = X_test, method = "class") ans <- data.frame(X_test[, 1], pred2[, 2]) colnames(ans) <- c("custid", "gender") print(head(ans)) # print(summary(ans)) write.csv(ans,'data/003000000.csv',row.names=F)
Output
Classification tree: rpart(formula = y_train_new ~ ., data = X_train_new, method = "class", cp = 0.023) Variables actually used in tree construction: [1] 내점일수 주구매상품 Root node error: 933/2450 = 0.38082 n= 2450 CP nsplit rel error xerror xstd 1 0.057342 0 1.00000 1.00000 0.025761 2 0.023000 2 0.88532 0.94212 0.025446 Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0000 0.0000 0.0000 0.3914 1.0000 1.0000 Min. 1st Qu. Median Mean 3rd Qu. Max. 0.0000 0.0000 0.0000 0.3648 1.0000 1.0000[1] 0.2347942 [1] 0.2313989
It seems not good
Confusion Matrix and Statistics y_valid pred_new 0 1 0 463 176 1 204 207 Accuracy : 0.6381 95% CI : (0.6082, 0.6672) No Information Rate : 0.6352 P-Value [Acc > NIR] : 0.4374 Kappa : 0.231 Mcnemar's Test P-Value : 0.1660 Sensitivity : 0.6942 Specificity : 0.5405 Pos Pred Value : 0.7246 Neg Pred Value : 0.5036 Prevalence : 0.6352 Detection Rate : 0.4410 Detection Prevalence : 0.6086 Balanced Accuracy : 0.6173 'Positive' Class : 0Not quite there yet, but better.
custid gender 1 3500 0.3144928 2 3501 0.2298137 3 3502 0.3144928 4 3503 0.5560209
Anyway, it meets the submission format.
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Code and Output
Code : BAEPracticalQuiz2_2.r
# 출력을 원할 경우 print() 함수 활용 # 예시) print(df.head()) # setwd(), getwd() 등 작업 폴더 설정 불필요 # 파일 경로 상 내부 드라이브 경로(C: 등) 접근 불가 train = read.csv("data/customer_train.csv") # no file in the local environment test = read.csv("data/customer_test.csv")
# 0. Import Libraries library(randomForest) library(caret)
# 1. Data Skimming str(train[,-11]) # summary(train[,-11]) str(test) # summary(test)
# 2. Data Pre-processing data <- merge(train, test, all=TRUE) # all=TRUE : Outer Join # str(data) # summary(data) data$주구매상품 <- as.numeric(data$주구매상품) data$주구매지점 <- as.numeric(data$주구매지점) data <- subset(data, select = -환불금액) # do not need '' data$성별 <- as.factor(data$성별) # Y should be a factor # str(data) # summary(data) train <- data[1:3500,] test <- data[3501:5982,-10] # str(train) # str(test) # help(sample) idx <- sample(0:3500, 3500 * 0.7) # 0:3500 should be a sequence # str(idx) # summary(idx) train_x <- train[idx, -10] train_y <- train[idx, 10] valid_x <- train[-idx, -10] valid_y <- train[-idx, 10] # str(train_x) # str(train_y) # str(valid_x) # str(valid_y)
# 3. Model Fitting # help(randomForest) mf <- randomForest(train_x, train_y, mtry=2, ntree=150) print(mf)
# 4. Validation # help(predict.randomForest) pred <- predict(mf, valid_x) # not predict.randomForest # str(pred) # summary(pred) cm <- table(pred, valid_y) print(confusionMatrix(cm)) # from caret
# 5. Submission ans <- predict(mf, test) ans <- data.frame(pred=ans) # print(head(ans)) # str(ans) # summary(ans) # 답안 제출 참고 # 아래 코드는 예시이며 변수명 등 개인별로 변경하여 활용 write.csv(ans, "result.csv", row.names = FALSE) # 5.1 답안 확인 result = read.csv("result.csv") head(result)
Output
Call: randomForest(x = train_x, y = train_y, ntree = 150, mtry = 2) Type of random forest: classification Number of trees: 150 No. of variables tried at each split: 2 OOB estimate of error rate: 38.38% Confusion matrix: 0 1 class.error 0 1241 310 0.1998711 1 630 268 0.7015590valid_y pred 0 1 0 528 294 1 105 124 Accuracy : 0.6204 95% CI : (0.5902, 0.6498) No Information Rate : 0.6023 P-Value [Acc > NIR] : 0.1217 Kappa : 0.1417 Mcnemar's Test P-Value : <2e-16 Sensitivity : 0.8341 Specificity : 0.2967 Pos Pred Value : 0.6423 Neg Pred Value : 0.5415 Prevalence : 0.6023 Detection Rate : 0.5024 Detection Prevalence : 0.7821 Balanced Accuracy : 0.5654 'Positive' Class : 0It seems (still) not good
pred 1 1 2 0 3 0 4 0 5 1 6 1
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Code and Output
Code : BAEPracticalQuiz3.r
# 출력을 원할 경우 print() 함수 활용 # 예시) print(df.head()) # setwd(), getwd() 등 작업 폴더 설정 불필요 # 파일 경로 상 내부 드라이브 경로(C: 등) 접근 불가 # 데이터 파일 읽기 예제 a <- read.csv("data/blood_pressure.csv", header=TRUE)
# 사용자 코딩 # str(a) # summary(a) attach(a) # (a) sample_mean = mean(bp_after - bp_before) ans_a = round(sample_mean, 2) # (b) result = t.test(bp_after - bp_before, mu = 0, var.equal = TRUE) # print(result) ans_b = round(as.numeric(result[1]), 2) # (c) ans_c1 = round(as.numeric(result[3]), 4) if (ans_c1 < 0.05) { ans_c2 = "기각" } else { ans_c2 = "채택" }
# 답안 제출 예시 # print(변수명) print(ans_a) # -5.09 print(ans_b) # t-statistic = -3.34 print(ans_c1) # p-value = 0.0011 < 0.05 print(ans_c2) # 기각 detach(a)
Output
[1] -5.09 [1] -3.34 [1] 0.0011 [1] "기각"
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가답안 전체 공유 ☞ 5회 실기(2022.12.03) 필답형 가답안 공유 (Kaggle Discussion)
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문제 : 표준정규분포의 (-a, a) 구간은 99.7%? (☞ Kaggle Discussion Comment)
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Code and Output
Code : BAE5Q9.r
pnorm(2.75, 0, 1) # 0.9970202 pnorm(-2.75, 0, 1) # 0.002979763 pnorm(3, 0, 1) # 0.9986501 pnorm(-3, 0, 1) # 0.001349898 pnorm(2.75, 0, 1) - pnorm(-2.75, 0, 1) # 0.9940405 pnorm(3, 0, 1) - pnorm(-3, 0, 1) # 0.9973002
Output
- 정답 : 3 (:sob::sob::sob:)