我有一组调查回复，受访者可以选择零个或多个选项来回答问题＆＃34;您喜欢什么类型的水果？＆＃34;。还有一个写入答案的空间。在结果电子表格中，每个人的回答都在一个单元格中，不同类型的水果用逗号分隔，如下所示：
(df <- data.frame(id = c("A", "B", "C", "D", "E"),
data = c("oranges, apples, peaches, cherries, pineapples, strawberries",
"oranges, peaches, pears",
"pears, nectarines, cherries (bing, rainier)",
"apples, peaches, nectarines",
""),
stringsAsFactors = FALSE))
# id data
# 1 A oranges, apples, peaches, cherries, pineapples, strawberries
# 2 B oranges, peaches, pears
# 3 C pears, nectarines, cherries (bing, rainier)
# 4 D apples, peaches, nectarines
# 5 E
我想要做的是将响应分成长格式表格，我使用底部的代码几乎完成了。但是，有些受访者在写入回复中包含了逗号，我不想将其答案分成逗号。我知道所有原始的多项选择是什么; 我怎么能只拆分这些答案，让写入（用逗号）保持原样？我想最终得到这样的数据框：
id data
1 A oranges
2 A apples
3 A peaches
4 A cherries, pineapples, strawberries
5 B oranges
6 B peaches
7 B pears
8 C pears
9 C nectarines
10 C cherries (bing, rainier)
11 D apples
12 D peaches
13 D nectarines
多项选择是：
mc_answers <- c("oranges", "plums", "apples", "peaches", "pears", "nectarines")
到目前为止，我所做的是：
# use strsplit to create a list of the types of fruit each person likes
datalist <- strsplit(df$data, ", ")
names(datalist) <- df$id
# remove zero-length list elements (person E doesn't like any fruit)
datalist <- Filter(length, datalist)
# convert list elements to data frames
datalist_dfs <- lapply(datalist, data.frame, stringsAsFactors = FALSE)
datalist_dfs <- lapply(datalist_dfs, setNames, "data") # name each column 'data'
# add id column to each data frame
data_long <- mapply(function(x, y) "[<-"(x, "id", value = y), datalist_dfs,
names(datalist_dfs), SIMPLIFY = FALSE)
# combine into one big data frame
(data_per_person <- do.call('rbind', data_long))
# data id
# A.1 oranges A
# A.2 apples A
# A.3 peaches A
# A.4 cherries A # should
# A.5 pineapples A # be one
# A.6 strawberries A # entry
# B.1 oranges B
# B.2 peaches B
# B.3 pears B
# C.1 pears C
# C.2 nectarines C
# C.3 cherries (bing C # should be
# C.4 rainier) C # one entry
# D.1 apples D
# D.2 peaches D
# D.3 nectarines D
没有关于一个人可以选择多少水果的规则，但如果有写入答案，它总是最后一次。
答案 0 :(得分：3)
这一行之后：
datalist <- Filter(length, datalist)
执行：
datalist <- lapply(datalist, function(x) {
if(any(!x %in% mc_answers))
c(x[x %in% mc_answers], paste(x[!x %in% mc_answers], collapse = ", "))
else
x[x %in% mc_answers]
})
然后按原样运行其余代码，最终得到：
> (data_per_person <- do.call('rbind', data_long))
data id
A.1 oranges A
A.2 apples A
A.3 peaches A
A.4 cherries, pineapples, strawberries A
B.1 oranges B
B.2 peaches B
B.3 pears B
C.1 pears C
C.2 nectarines C
C.3 cherries (bing, rainier) C
D.1 apples D
D.2 peaches D
D.3 nectarines D
答案 1 :(得分：3)
您也可以尝试：
library(data.table)
library(devtools)
source_gist(11380733) ##
df1 <- cSplit(df, "data", sep=", ", "long")
indx <- df1$data %in% mc_answers
res <- rbindlist(list(df1[indx,], df1[!indx,][, list(data=paste(data, collapse=", ")), by=id]))[order(id)]
res
# id data
#1: A oranges
#2: A apples
#3: A peaches
#4: A cherries, pineapples, strawberries
#5: B oranges
#6: B peaches
#7: B pears
#8: C pears
#9: C nectarines
#10: C cherries (bing, rainier)
#11: D apples
#12: D peaches
#13: D nectarines
答案 2 :(得分：2)
这样的事情
do.call(rbind, lapply(split(df, df$id), function(x) {
v<-unlist(strsplit(x$data, ",\\s?"))
v<-c(v[v %in% mc_answers], paste(v[!v %in% mc_answers], collapse=", "))
v<-v[nchar(v)>0]
if (length(v)>0) {
data.frame(id=x$id[1], data=v)
} else {
NULL
}
}))
这里我们分开处理每个组，然后进行字符串拆分。然后我们将折叠所有不在mc_answers向量中的条目。它返回
id data
A.1 A oranges
A.2 A apples
A.3 A peaches
A.4 A cherries, pineapples, strawberries
B.1 B oranges
B.2 B peaches
B.3 B pears
C.1 C pears
C.2 C nectarines
C.3 C cherries (bing, rainier)
D.1 D apples
D.2 D peaches
D.3 D nectarines