dplyr ::如何使用函数中的变量引用进行变异？

Question

有人可以告诉我如何使用dplyr将带有参数名称的向量传递给函数吗？

library("dplyr", quietly = TRUE, warn.conflicts = FALSE) # version 0.8.0.1

# Does not work
iris %>% rowwise() %>%  mutate(v1 = mean( as.name(names(iris)[-5]) ) )
iris %>% rowwise() %>%  mutate(v1 = mean( !!(names(iris)[-5]) ) )
iris %>% rowwise() %>%  mutate(v1 = mean( enquo(names(iris)[-5]) ) )
iris %>% rowwise() %>%  
mutate(v1 = mean( c("Sepal.Length", "Sepal.Width", "Petal.Length", "Petal.Width")  ) )

# This works and is the intended result
iris %>% rowwise() %>%  
mutate(v1 = mean( c(Sepal.Length, Sepal.Width, Petal.Length, Petal.Width )  ) )

关键是要让函数（均值或任何函数）与names(iris)[-5]变量名称一起使用或使用向量。

我看这里没有成功： dplyr mutate_each_标准评估； dplyr：标准评估和enquo（）

我的会议信息：

R version 3.5.3 (2019-03-11)
Platform: x86_64-w64-mingw32/x64 (64-bit)
Running under: Windows 10 x64 (build 17763)

Matrix products: default

locale:
[1] LC_COLLATE=French_France.1252  LC_CTYPE=French_France.1252   
[3] LC_MONETARY=French_France.1252 LC_NUMERIC=C                  
[5] LC_TIME=French_France.1252    

attached base packages:
[1] stats     graphics  grDevices utils     datasets  methods   base     

other attached packages:
[1] ggplot2_3.1.0   visdat_0.5.3    lubridate_1.7.4 naniar_0.4.2   
[5] dplyr_0.8.0.1  

loaded via a namespace (and not attached):
 [1] Rcpp_1.0.1       rstudioapi_0.10  magrittr_1.5     tidyselect_0.2.5
 [5] munsell_0.5.0    colorspace_1.4-0 R6_2.4.0         rlang_0.3.4     
 [9] fansi_0.4.0      stringr_1.4.0    plyr_1.8.4       tools_3.5.3     
[13] grid_3.5.3       packrat_0.5.0    gtable_0.2.0     utf8_1.1.4      
[17] cli_1.1.0        withr_2.1.2      digest_0.6.18    lazyeval_0.2.2  
[21] assertthat_0.2.0 tibble_2.1.1     crayon_1.3.4     tidyr_0.8.3     
[25] purrr_0.3.2      glue_1.3.1       labeling_0.3     stringi_1.4.3   
[29] compiler_3.5.3   pillar_1.3.1     scales_1.0.0     pkgconfig_2.0.2 

提前致谢 ！

Answer 1

使用 map2_dbl

library(tidyverse)
iris %>% mutate(v1 = map2_dbl(Sepal.Length, Sepal.Width, ~mean(c(.x, .y)))) %>% head

#  Sepal.Length Sepal.Width Petal.Length Petal.Width Species   v1
#1          5.1         3.5          1.4         0.2  setosa 4.30
#2          4.9         3.0          1.4         0.2  setosa 3.95
#3          4.7         3.2          1.3         0.2  setosa 3.95
#4          4.6         3.1          1.5         0.2  setosa 3.85
#5          5.0         3.6          1.4         0.2  setosa 4.30
#6          5.4         3.9          1.7         0.4  setosa 4.65

或者，如果您想获取mean某些列。

cols <- c("Sepal.Length", "Sepal.Width")

iris %>% mutate(v1 = rowMeans(.[cols])) %>% head