Question

我试图在R中创建一堆带有循环的向量，如下所示：

    dlist <- c('"lnincome", "retprice", "age15to24"',
           '"retprice", "age15to24", "beer"',
           '"retprice", "age15to24"',
           '"retprice", "beer"',
           '"lnincome", "beer"')

    for (i in 1:length(dlist)) {
output_[[i]]<-dataprep(foo = smoking, 
                     predictors = c(cat(dlist[i])),
                     predictors.op = c("mean"),
                     special.predictors = list(
                       list("cigsale", 1988, "mean"),
                       list("cigsale", 1980, "mean"),
                       list("cigsale", 1975, "mean")
                     ),
                     dependent = c("cigsale"),
                     unit.variable = "statenum",
                     time.variable = c("year"),
                     treatment.identifier = 3,
                     controls.identifier = c(1:2, 4:39),
                     time.predictors.prior = c(1980:1988),
                     time.optimize.ssr = c(1975:1988),
                     time.plot = c(1975:2000),
                     unit.names.variable = c("state")
              )

#assign( paste("test_", eval(i), sep="") , output)
synth.out_[[i]]<-synth(output_i)
#assign( paste("synth_", eval(i), sep="") , synth.out)

path.plot(dataprep.res = output_i, synth.res = snyth.out_i)
    rm(synth.out, test)

    }

然而，当循环运行时，我收到的错误是＆＃34; object＆＃39; test _＆＃39;未找到。如何创建对象test_1是在循环的第一次迭代，然后是对象test_2，直到test_n（n = 5，这里）。

然后，如何在path.plot函数中调用output_i和synth_i的创建值？

提前致谢！

Answer 1

您正在寻找的是函数get()

如

get(paste("output_", i, sep=""))

我通常会建议你不要制作这么多变量，而是将它们放在一个命名列表中。为此，您可以更新以下内容。

output_ <- list()
synth.out_ <- list()
   for (item in dlist) {
output_[paste(item)]<-dataprep(foo = smoking, 
                 predictors = c(cat(dlist[i])),
                 predictors.op = c("mean"),
                 special.predictors = list(
                   list("cigsale", 1988, "mean"),
                   list("cigsale", 1980, "mean"),
                   list("cigsale", 1975, "mean")
                 ),
                 dependent = c("cigsale"),
                 unit.variable = "statenum",
                 time.variable = c("year"),
                 treatment.identifier = 3,
                 controls.identifier = c(1:2, 4:39),
                 time.predictors.prior = c(1980:1988),
                 time.optimize.ssr = c(1975:1988),
                 time.plot = c(1975:2000),
                 unit.names.variable = c("state")
          )


synth.out_[paste(item)]<-synth(output[paste(item)])

path.plot(dataprep.res = output[paste(item)], synth.res =     synth.out_[paste(item)])
# not sure what the last line is doing
rm(synth.out, test)

你没有给出足够的信息来试试：dataprep()是什么，test是什么？ PS曲线函数snyth.out_i上有拼写错误。

R循环列表创建

1 个答案: