【发布时间】:2019-06-21 13:42:38
【问题描述】:
有人能告诉我如何使用 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 ) ) )
关键是让函数(平均值或任何函数)与names(iris)[-5] 或具有变量名称的向量一起使用。
我看过这里没有成功: dplyr mutate_each_ standard evaluation ; dplyr: Standard evaluation and enquo()
我的会话信息:
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!
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