我的数据如下所示（变量zipid1-zipid13和变量hospid的范围是1-13：
zipid1 zipid2 zipid3 zipid4 zipid5 zipid6 zipid7 zipid8 zipid9 zipid10 zipid11 zipid12 zipid13 hospid local
1 0 0 0 0 1 0 0 0 0 0 0 0 0 5 0
2 0 0 1 0 1 0 0 0 0 0 0 0 0 5 0
3 0 0 0 0 0 0 1 0 0 0 0 0 0 5 0
4 0 0 1 0 0 0 0 0 0 0 0 0 0 5 0
5 0 0 1 0 1 0 0 0 0 0 0 0 0 5 0
6 0 0 0 0 1 0 0 0 0 0 0 0 0 5 0
当zipid1 ==1 & hospid =1, zipid2 == 1 & hospid == 2等时，如何创建 local 变量= 1。否则= 0（即zipid = hospid）？
我尝试了ifelse，但效果不佳。
for (i in 1:13) {
name = paste0("zipid", i)
local$local <- with(local, ifelse(name == 1 & hospid == i, 1, 0))
}
谢谢！
答案 0 :(得分：1)
这是一个想法：
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给予
df$local <- unlist(lapply(1:nrow(df), function(x)df[x, paste("zipid", df$hospid, sep = "")[x]]))
它们起作用的方式是，我在# zipid1 zipid2 zipid3 zipid4 zipid5 zipid6 zipid7 zipid8 zipid9 zipid10 zipid11 zipid12 zipid13 hospid local
# 1 0 0 0 0 1 0 0 0 0 0 0 0 0 5 1
# 2 0 0 1 0 1 0 0 0 0 0 0 0 0 5 1
# 3 0 0 0 0 0 0 1 0 0 0 0 0 0 5 0
# 4 0 0 1 0 0 0 0 0 0 0 0 0 0 5 0
# 5 0 0 1 0 1 0 0 0 0 0 0 0 0 5 1
# 6 0 0 0 0 1 0 0 0 0 0 0 0 0 5 1
的每一行中取值，然后将其与hospid粘贴以制作类似zipid的值。我在与特定行相对应的特定列中查找值，并检查其是否为zipid5。
如果数据框中存在1，则可以使用NA将其删除。例如，na.omit在运行上面的代码之前。
答案 1 :(得分：1)
问题在于列名zipid1，zipid2等传达的是有效载荷数据，即数字。
我的建议是将数据从宽到长整形，从列名中提取数字，与hospid匹配，由id聚合，然后将结果与原始宽合并格式。
使用toString()进行汇总，以便在出现多个匹配项时获得有效结果。
library(data.table)
# reshape from wide to long format
melt(setDT(DT), id.vars = c("id", "hospid"), variable.name = "zipid")[
# turn column names into integer
, zipid := as.integer(stringr::str_replace(zipid, "zipid", ""))][
# if value is 1 and zipid and hospid do match then store number
value == 1L & zipid == hospid, local := hospid][
# aggregate only mathcing entries by id
!is.na(local), .(local = toString(local)), by = id][
# right join with original data
DT, on = "id"][
# change column order to meet OP's expectation
, setcolorder(.SD, names(DT))]
id zipid1 zipid2 zipid3 zipid4 zipid5 zipid6 zipid7 zipid8 zipid9 zipid10 zipid11 zipid12 zipid13 hospid local 1: 1 0 0 0 0 1 0 0 0 0 0 0 0 0 5 5 2: 2 0 0 1 0 1 0 0 0 0 0 0 0 0 5 5 3: 3 0 0 0 0 0 0 1 0 0 0 0 0 0 5 <NA> 4: 4 0 0 1 0 0 0 0 0 0 0 0 0 0 5 <NA> 5: 5 0 0 1 0 1 0 0 0 0 0 0 0 0 5 5 6: 6 0 0 0 0 1 0 0 0 0 0 0 0 0 5 5
通过重塑，DT中的相关信息可以压缩为
melt(setDT(DT), id.vars = c("id", "hospid"), variable.name = "zipid")[
, zipid := as.integer(stringr::str_replace(zipid, "zipid", ""))][
value == 1L]
id hospid zipid value 1: 2 5 3 1 2: 4 5 3 1 3: 5 5 3 1 4: 1 5 5 1 5: 2 5 5 1 6: 5 5 5 1 7: 6 5 5 1 8: 3 5 7 1
结果由
给出melt(setDT(DT), id.vars = c("id", "hospid"), variable.name = "zipid")[
, zipid := as.integer(stringr::str_replace(zipid, "zipid", ""))][
value == 1L][
zipid == hospid]
id hospid zipid value 1: 1 5 5 1 2: 2 5 5 1 3: 5 5 5 1 4: 6 5 5 1
因此，要将其与原始数据对象结合起来，我们可以对join进行更新：
tmp <-
melt(setDT(DT), id.vars = c("id", "hospid"), variable.name = "zipid")[
, zipid := as.integer(stringr::str_replace(zipid, "zipid", ""))][
value == 1L & zipid == hospid]
DT[tmp, on = "id", local := value][]
id zipid1 zipid2 zipid3 zipid4 zipid5 zipid6 zipid7 zipid8 zipid9 zipid10 zipid11 zipid12 zipid13 hospid local 1: 1 0 0 0 0 1 0 0 0 0 0 0 0 0 5 1 2: 2 0 0 1 0 1 0 0 0 0 0 0 0 0 5 1 3: 3 0 0 0 0 0 0 1 0 0 0 0 0 0 5 NA 4: 4 0 0 1 0 0 0 0 0 0 0 0 0 0 5 NA 5: 5 0 0 1 0 1 0 0 0 0 0 0 0 0 5 1 6: 6 0 0 0 0 1 0 0 0 0 0 0 0 0 5 1
这给出了预期的输出。无需聚合。
library(data.table)
DT <- fread("id zipid1 zipid2 zipid3 zipid4 zipid5 zipid6 zipid7 zipid8 zipid9 zipid10 zipid11 zipid12 zipid13 hospid local
1 0 0 0 0 1 0 0 0 0 0 0 0 0 5 0
2 0 0 1 0 1 0 0 0 0 0 0 0 0 5 0
3 0 0 0 0 0 0 1 0 0 0 0 0 0 5 0
4 0 0 1 0 0 0 0 0 0 0 0 0 0 5 0
5 0 0 1 0 1 0 0 0 0 0 0 0 0 5 0
6 0 0 0 0 1 0 0 0 0 0 0 0 0 5 0", drop = "local")
答案 2 :(得分：-1)
name是字符串的向量，在这种情况下，它被解释为字符串，而不是变量，请尝试ifelse(get(name)==1 &...