我有一个数据框,其中包含一连串混乱的家庭设施。我想将字符串分解为唯一的便利设施,为每个唯一的便利设施在数据框中创建一个新列,并在新列中记录字符串中各个便利设施的存在/不存在。使用嵌套的for
循环,我找到了一种完成任务的方法。但是,我想知道的是如何使用apply
系列函数或dplyr
方法之一来避免循环,从而获得相同的结果。
可复制数据:
df <- data.frame(
id = 1:4,
amenities = c('{"Wireless Internet","Wheelchair accessible",Kitchen,Elevator,"Buzzer/wireless intercom",Heating}',
'{TV,"Cable TV",Internet,"Wireless Internet","Air conditioning",Kitchen,"Smoking allowed","Pets allowed"}',
'{"Buzzer/wireless intercom",Heating,"Family/kid friendly","Smoke detector",Carbon monoxide}',
'{Washer,Dryer,Essentials,Shampoo,Hangers,"Laptop friendly workspace"}'))
到目前为止,我所做的是:
amenities_clean <- gsub('[{}"]', '', df$amenities) # remove unwanted stuff
amenities_split <- strsplit(amenities_clean, ",") # split rows into individual amenities
amenities_unique <- unique(unlist(strsplit(amenities_clean, ","))) # get a list of unique amenities
df[amenities_unique] <- NA # set up the columns for each amenity
要在新列中记录字符串中是否存在单个便利设施,我正在使用str_detect
以及嵌套的for
循环:
# record presence/absence of individual amenities in each new column:
library(stringr)
for(i in 1:ncol(df[amenities_unique])){
for(j in 1:nrow(df)){
df[amenities_unique][j,i] <-
ifelse(str_detect(amenities_split[j], names(df[amenities_unique][i])), 1, 0)
}
}
虽然这会产生警告,但它们似乎无害,因为结果看起来还不错:
df
id amenities Wireless Internet
1 1 {"Wireless Internet","Wheelchair accessible",Kitchen,Elevator,"Buzzer/wireless intercom",Heating} 1
2 2 {TV,"Cable TV",Internet,"Wireless Internet","Air conditioning",Kitchen,"Smoking allowed","Pets allowed"} 1
3 3 {"Buzzer/wireless intercom",Heating,"Family/kid friendly","Smoke detector",Carbon monoxide} 0
4 4 {Washer,Dryer,Essentials,Shampoo,Hangers,"Laptop friendly workspace"} 0
Wheelchair accessible Kitchen Elevator Buzzer/wireless intercom Heating TV Cable TV Internet Air conditioning Smoking allowed
1 1 1 1 1 1 0 0 1 0 0
2 0 1 0 0 0 1 1 1 1 1
3 0 0 0 1 1 0 0 0 0 0
4 0 0 0 0 0 0 0 0 0 0
Pets allowed Family/kid friendly Smoke detector Carbon monoxide Washer Dryer Essentials Shampoo Hangers Laptop friendly workspace
1 0 0 0 0 0 0 0 0 0 0
2 1 0 0 0 0 0 0 0 0 0
3 0 1 1 1 0 0 0 0 0 0
4 0 0 0 0 1 1 1 1 1 1
给出警告并考虑到嵌套循环的复杂性,如何使用apply
函数族中的函数或使用dplyr
来获得相同的结果?
一旦清洁便利设施,就可以使用cSplit_e
中的splitstackshape
。
df$amenities_clean <- gsub('[{}"]', '', df$amenities)
splitstackshape::cSplit_e(df, "amenities_clean", type = "character", fill = 0)