我正在从另一个来源将数据导入R中（即，我无法轻易更改传入的格式/值）。
变量中包括一个或多个以下可能值的变量：
全部都在同一个“单元格”中，因此可能的数据如下所示：
示例输入数据帧（df）
df <- read.table(text =
"row lives.with.whom
1 'Mother (biological mother, foster mother, step mother, etc.), Father (biological father, foster father, step father, etc.), Grandparent(s) (biological, foster, step, etc.), Brother(s) older than 18, Sister(s) older than 18, Other adults (aunts, uncles, etc.)'
2 ''
3 'Mother (biological mother, foster mother, step mother, etc.), Sister(s) older than 18'
4 'Mother (biological mother, foster mother, step mother, etc.), Father (biological father, foster father, step father, etc.)'", header = T)
在R中，我如何有效地创建规则以将这些响应解析为单独的列，每种类型的家庭成员都划分为一列，以使输出看起来像这样：
示例输出数据帧
mother <- c(1,0,1,1)
father <- c(1,0,0,1)
adult.brother <- c(1,0,0,0)
adult.sister <- c(1,0,1,0)
grandparent <- c(1,0,0,0)
other.adult <- c(1,0,0,0)
output.df <- cbind(mother, father, adult.brother, adult.sister, grandparent, other.adult)
colnames(output.df) <- c("Mother", "Father", "Brother", "Sister", "Grandparent", "Other adult")
output.df
Mother Father Brother Sister Grandparent Other adult
[1,] 1 1 1 1 1 1
[2,] 0 0 0 0 0 0
[3,] 1 0 0 1 0 0
[4,] 1 1 0 0 0 0
TIA
答案 0 :(得分：1)
这是一个tidyverse选项，可以帮助您入门
library(tidyverse)
rel <- list("Mother", "Father", "Brother", "Sister", "Grandparent", "Other adult")
names(rel) <- unlist(rel)
bind_cols(df[, 1, drop = F], map(rel, ~+str_detect(tolower(df[, 2]), tolower(.x))))
# row Mother Father Brother Sister Grandparent Other adult
#1 1 1 1 1 1 1 1
#2 2 0 0 0 0 0 0
#3 3 1 0 0 1 0 0
#4 4 1 1 0 0 0 0
df <- read.table(text =
"row lives.with.whom
1 'Mother (biological mother, foster mother, step mother, etc.), Father (biological father, foster father, step father, etc.), Grandparent(s) (biological, foster, step, etc.), Brother(s) older than 18, Sister(s) older than 18, Other adults (aunts, uncles, etc.)'
2 ''
3 'Mother (biological mother, foster mother, step mother, etc.), Sister(s) older than 18'
4 'Mother (biological mother, foster mother, step mother, etc.), Father (biological father, foster father, step father, etc.)'", header = T)
答案 1 :(得分：1)
尝试一下：
rel<-list("Mother", "Father", "Brother", "Sister", "Grandparent", "Other adult")
for(i in 1:6){
df$i<-if_else(grepl(rel[[i]],df$lives.with.whom),1,0)
colnames(df)[i+2]<-rel[[i]]
}
答案 2 :(得分：0)
嘿，欢迎来到Stack Overflow！以下是一些有关如何在Stack Overflow上提出更好的问题的链接，以便人们轻松地帮助您（向前）。
谈到您的问题，我做了一些假设并试图解决它。正如莫里斯（Maurits）所提到的，您需要提供一个可复制的示例，以便有人可以给出具体的答案，在此之前，这是我能提出的最佳答案。
library(tidyr)
library(dplyr)
# create nested lists with names of mothers and fathers for two ppl
mother <- list(list("bio_1","step_1","foster_1"), list("bio_2", "stp_2", "foster_2"))
father <- list(list("bio_1", "foster_1", "other_1"), list("bio_2", "stp_2", "foster_2"))
# convert to data frame
test_object <- data_frame(person = c(1,2),mother,father)
# print
test_object
# A tibble: 2 x 3
person mother father
<dbl> <list> <list>
1 1 <list [3]> <list [3]>
2 2 <list [3]> <list [3]>
# first unnest the lists and get to the inner list
# then convert from wide to long form data
# do another unnnest to get the actual data in the long format
test_object %>%
unnest(.) %>%
gather(data = ., key = relationship, value = name, -person) %>%
unnest() -> test_object
test_object
# A tibble: 12 x 3
person relationship name
<dbl> <chr> <chr>
1 1 mother bio_1
2 1 mother step_1
3 1 mother foster_1
4 2 mother bio_2
5 2 mother stp_2
6 2 mother foster_2
7 1 father bio_1
8 1 father foster_1
9 1 father other_1
10 2 father bio_2
11 2 father stp_2
12 2 father foster_2
这里是指向tidyverse和data.table的链接，其中包含许多用于解决大多数数据仓库/争用问题的软件包和功能。