bigdata_part03_01

# exam1) 주어진 캘리포니아 주택 데이터 첫번째 행부터 순서대로 80%까지의 데이터를 추출한후
# 'total_bedrooms'변수의 결측값(NA)을 'total_bedrooms'변수의 중앙값으로 대체하고
# 대체전의 'total_bedrooms' 변수의  표준편차값과 대체후에 표준편차의 차이를 구하시오 
library(dplyr)

## Warning: 패키지 'dplyr'는 R 버전 4.1.3에서 작성되었습니다

## 
## 다음의 패키지를 부착합니다: 'dplyr'

## The following objects are masked from 'package:stats':
## 
##     filter, lag

## The following objects are masked from 'package:base':
## 
##     intersect, setdiff, setequal, union

house<-read.csv("housing.csv")
# ① 주어진 데이터 첫번째 행부터 순서대로 80%추출
nrow(house)

## [1] 20640

rownum<-nrow(house)*0.8
rownum

## [1] 16512

house1<-house[1:rownum,]
# ② 데이터 구조 확인
house1 %>% glimpse

## Rows: 16,512
## Columns: 10
## $ longitude          <dbl> -122.23, -122.22, -122.24, -122.25, -122.25, -122.2~
## $ latitude           <dbl> 37.88, 37.86, 37.85, 37.85, 37.85, 37.85, 37.84, 37~
## $ housing_median_age <int> 41, 21, 52, 52, 52, 52, 52, 52, 42, 52, 52, 52, 52,~
## $ total_rooms        <int> 880, 7099, 1467, 1274, 1627, 919, 2535, 3104, 2555,~
## $ total_bedrooms     <int> 129, 1106, 190, 235, 280, 213, 489, 687, 665, 707, ~
## $ population         <int> 322, 2401, 496, 558, 565, 413, 1094, 1157, 1206, 15~
## $ households         <int> 126, 1138, 177, 219, 259, 193, 514, 647, 595, 714, ~
## $ median_income      <dbl> 8.3252, 8.3014, 7.2574, 5.6431, 3.8462, 4.0368, 3.6~
## $ median_house_value <int> 452600, 358500, 352100, 341300, 342200, 269700, 299~
## $ ocean_proximity    <chr> "NEAR BAY", "NEAR BAY", "NEAR BAY", "NEAR BAY", "NE~

# ③ 결측치(NA) 확인
colSums(is.na(house1))

##          longitude           latitude housing_median_age        total_rooms 
##                  0                  0                  0                  0 
##     total_bedrooms         population         households      median_income 
##                159                  0                  0                  0 
## median_house_value    ocean_proximity 
##                  0                  0

summary(is.na(house1))

##  longitude        latitude       housing_median_age total_rooms    
##  Mode :logical   Mode :logical   Mode :logical      Mode :logical  
##  FALSE:16512     FALSE:16512     FALSE:16512        FALSE:16512    
##                                                                    
##  total_bedrooms  population      households      median_income  
##  Mode :logical   Mode :logical   Mode :logical   Mode :logical  
##  FALSE:16353     FALSE:16512     FALSE:16512     FALSE:16512    
##  TRUE :159                                                      
##  median_house_value ocean_proximity
##  Mode :logical      Mode :logical  
##  FALSE:16512        FALSE:16512    
## 

# ④ 결측치 대체 전 표준편차 구하기
df1<-sd(house1$total_bedrooms,na.rm=TRUE)
df1

## [1] 435.9006

# ⑤ 결측치를 중앙값을 대체
df2<-median(house1$total_bedrooms,na.rm=TRUE)
df2

## [1] 436

house1$total_bedrooms<-ifelse(is.na(house1$total_bedrooms),df2,house1$total_bedrooms)
# 절대 혼동하지 마세요 house1$total_bedrooms==NA
# house1 %>% filter(!is.na(total_bedrooms))
colSums(is.na(house1))

##          longitude           latitude housing_median_age        total_rooms 
##                  0                  0                  0                  0 
##     total_bedrooms         population         households      median_income 
##                  0                  0                  0                  0 
## median_house_value    ocean_proximity 
##                  0                  0

# ⑥ 결측치 대체후 표준편차 구하기
df3<-sd(house1$total_bedrooms)
df3

## [1] 433.9254

df1-df3

## [1] 1.975147