你如何在mutate_at()中使用approx()？

Question

我在让mut()在mutate_at()中工作时遇到问题.我确实设法使用一个很长的mutate()函数来获得我想要的东西,但是为了将来的参考,我想知道是否有更优雅且更少复制粘贴的mutate_at()方法来做到这一点.

最重要的问题是将数据集与1年间隔的数据合并为3年间隔的数据,并插入年份,数据集中没有数据,间隔为3年.这些年中缺少值,而一年需要某种形式的推断.

library("tidyverse")

demodf <- data.frame(groupvar = letters[rep(1:15, each = 6)],
                     timevar = c(2000, 2003, 2006, 2009, 2012, 2015),
                     x1 = runif(n = 90, min = 0, max = 3),
                     x2 = runif(n = 90, min = -1, max = 4),
                     x3 = runif(n = 90, min = 1, max = 12),
                     x4 = runif(n = 90, min = 0, max = 30),
                     x5 = runif(n = 90, min = -2, max = 5),
                     x6 = runif(n = 90, min = 20, max = 50),
                     x7 = runif(n = 90, min = 1, max = 37),
                     x8 = runif(n = 90, min = 0.3, max = 0.5))

demotbl <- tbl_df(demodf)

masterdf <- data.frame(groupvar = letters[rep(1:15, each = 17)],
                      timevar = 2000:2016,
                      z1 = runif(n = 255, min = 0, max = 1E6))

mastertbl <- tbl_df(masterdf)

joineddemotbls <- mastertbl %>% left_join(demotbl, by = c("groupvar", "timevar"))

View(joineddemotbls)

joineddemotblswithinterpolation <- joineddemotbls %>% group_by(groupvar) %>%
  mutate(x1i = approx(timevar, x1, timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]],
         x2i = approx(timevar, x2, timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]],
         x3i = approx(timevar, x3, timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]],
         x4i = approx(timevar, x4, timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]],
         x5i = approx(timevar, x5, timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]],
         x6i = approx(timevar, x6, timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]],
         x7i = approx(timevar, x7, timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]],
         x8i = approx(timevar, x8, timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]])

View(joineddemotblswithinterpolation)

# this is what I want

这非常有效.但我已经尝试了所有这些mutate_at()变体,并没有让它们工作.我确定某处的语法有错误...

joineddemotblswithinterpolation2 <- joineddemotblswithinterpolation %>% group_by(groupvar) %>%
  mutate_at(vars(x1, x2, x3, x4, x5, x6, x7, x8), approx(timevar, ., timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]])

# error

joineddemotblswithinterpolation2 <- joineddemotblswithinterpolation %>% group_by(groupvar) %>%
  mutate_at(vars(x1, x2, x3, x4, x5, x6, x7, x8), approxfun(timevar, ., timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]])

# error

joineddemotblswithinterpolation2 <- joineddemotblswithinterpolation %>% group_by(groupvar) %>%
  mutate_at(vars(x1, x2, x3, x4, x5, x6, x7, x8), funs(approxfun(timevar, ., timevar, rule = 2, f = 0, ties = mean, method = "linear")[["y"]]))

# error

joineddemotblswithinterpolation2 <- joineddemotblswithinterpolation %>% group_by(groupvar) %>%
  mutate_at(vars(x1, x2, x3, x4, x5, x6, x7, x8), funs(approxfun(timevar, ., rule = 2, f = 0, ties = mean, method = "linear")[["y"]]))

我甚至尝试过na.approx(),但也无济于事......

library("zoo")
joineddemotblswithinterpolation2 <- joineddemotblswithinterpolation %>% group_by(groupvar) %>%
  mutate_at(vars(x1, x2, x3, x4, x5, x6, x7, x8), na.approx(., timevar, na.rm = FALSE))

我从以下相关问题构建了这些不同的试验:

在dplyr中使用约

使用dplyr的线性插值

在dplyr中使用approx()和组

使用dplyr进行线性插值,但跳过具有所有缺失值的组

R:按组插入NA

谢谢你的帮助!

Answer 1

你很近.这对我有用:

joineddemotblswithinterpolation <- joineddemotbls %>%
  group_by(groupvar) %>%
  mutate_at(vars(starts_with("x")), # easier than listing each column separately
            funs("i" = approx(timevar, ., timevar, rule = 2, f = 0, ties = mean,
                              method = "linear")[["y"]]))

这将创建列x1_i,x2_i等与内插值.