我遇到问题,让mut()在mutate_at()内部工作。我确实设法使用一个非常长的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mutate_at()的方法来做到这一点。
总体问题是将数据集与1年间隔的数据合并为3年间隔的数据,并插入数据集中没有数据的年份,间隔为3年。这些年间缺少值,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
效果很好。但是我已经尝试了所有这些mutate_at()变体并且还没有让它们起作用。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))
我从以下相关问题中构建了这些不同的试验:
Linear Interpolation using dplyr
Using approx() with groups in dplyr
linear interpolation with dplyr but skipping groups with all missing values
R: Interpolation of NAs by group
感谢您的帮助!
答案 0 :(得分:5)
你非常接近。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"]]))
这将使用插值创建列x1_i
,x2_i
等。