[英]Use a list of variables and levels to convert character data to factors using dplyr
我正在處理調查數據,我需要將響應值編碼為因素(例如,非常不同意、不同意、同意、非常同意)。 不同的問題有不同的回答選項,需要適當地編碼。 我有一個 excel 文件,其中列出了每個問題和有序的響應選項。 我已經編寫了一個for
循環來轉換所有變量,但想了解如何使用purrr
或dplyr
語法來做到這一點。
這是一個簡單的例子:
library(tidyverse)
dat <- iris %>%
mutate(
Species = as.character(Species),
second_var = as.character(round(Sepal.Length)))
factor_map <- data.frame(
var = c("Species", "second_var"),
response_opts = c("setosa,versicolor,virginica",
"4,5,6,7,8"))
# convert character string of options into lists
factor_map2 <- factor_map %>%
mutate(levels = str_split(response_opts, ","))
# simple for loop
dat2 <- dat
for (i in 1:nrow(factor_map2)) {
v <- factor_map2$var[i]
l <- factor_map2$levels[[i]]
dat2[[v]] = factor(dat2[[v]], levels = l)
rm(v, l)
}
# How to use factor_map to convert the columns in dat to factors?
# map2 doesn't seem to work, unclear why it says .x has length of 6
dat %>%
map2(factor_map2$var, factor_map2$levels,
function(x, y) factor(x, levels = y))
# Can I pass a vector of variable specific levels into across?
dat %>%
mutate(across(factor_map2$var, factor, # somehow pass in the levels
有可能
map2_dfc(factor_map2$var, factor_map2$levels,
~ factor(dat[[.x]], levels = .y))%>%
setNames(factor_map2$var)
或者不使用任何新的 package 的另一個選項,即只有 dplyr 是
dat %>%
mutate(across(all_of(factor_map2$var), ~ factor(., levels =
factor_map2$levels[match(cur_column(), factor_map2$var)])))
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