Question

在此示例中，我有两个列表

seg.layer.borderWidth = 1
seg.layer.cornerRadius = seg.bounds.height / 2
seg.layer.masksToBounds = true
seg.apportionsSegmentWidthsByContent = true

，我想通过添加相应元素在tiers <- list("tier 1", "tier 2", "tier 3") main <- list(data.frame(a = c("this", "that")), data.frame(a = c("the other", "that too")), data.frame(a = c("once more", "kilgore trout")))中的值来mutate()中的每个列表元素（即data.frame()）。我以为main会做到：

tiers

但是我得到了意外的结果

mapply()

而我期望的是

library(dplyr)

mapply(function(x, y) y %>% mutate(tier = x), tiers, main)

我正确使用> mapply(function(x, y) y %>% mutate(tier = x), tiers, main) [,1] [,2] [,3] a factor,2 factor,2 factor,2 tier Character,2 Character,2 Character吗？如果不是，是否应该使用某些东西来获得期望的结果？我应该指出，实际数据可能最多包含[[1]] a tier 1 this tier 1 2 that tier 1 [[2]] a tier 1 the other tier 2 2 that too tier 2 [[3]] a tier 1 once more tier 3 2 kilgore trout tier 3个列表元素；我无法用mapply()来硬编码任何值。

Answer 1

您需要的是在您的SIMPLIFY = FALSE通话中添加mapply

library(dplyr)
mapply(function(x, y) y %>% mutate(tier = x), tiers, main, SIMPLIFY = FALSE)


#     a   tier
#1 this tier 1
#2 that tier 1

#[[2]]
#          a   tier
#1 the other tier 2
#2  that too tier 2

#[[3]]
#              a   tier
#1     once more tier 3
#2 kilgore trout tier 3

?mapply说

SIMPLIFY-尝试将结果简化为向量，矩阵或更高维的数组；

SIMPLIFY参数默认为TRUE中的mapply和FALSE中的Map

Map(function(x, y) y %>% mutate(tier = x), tiers, main)

如果您想将所有内容保留在基数R中，则可以使用cbind

Map(cbind, main, tiers)

Answer 2

另外，请考虑将within或transform与Map（列出mapply的版本包装器）并避免为一个功能mutate加载程序包：< / p>

Map(function(x, y) within(y, tier <- x), tiers, main)

Map(function(x, y) transform(y, tier = x), tiers, main)

Answer 3

在基础R中，如果您命名Map参数，则可以将data.frame与函数tier一起使用：

Map(data.frame, main, tier=tiers)
# [[1]]
#      a   tier
# 1 this tier 1
# 2 that tier 1
# 
# [[2]]
#           a   tier
# 1 the other tier 2
# 2  that too tier 2
# 
# [[3]]
#               a   tier
# 1     once more tier 3
# 2 kilgore trout tier 3

Answer 4

我们可以在tidyverse中使用map2

library(tidyverse)
map2(main, tiers, ~ .x %>%
           mutate(tiers = .y))
#[[1]]
#     a  tiers
#1 this tier 1
#2 that tier 1

#[[2]]
#          a  tiers
#1 the other tier 2
#2  that too tier 2

#[[3]]
#              a  tiers
#1     once more tier 3
#2 kilgore trout tier 3

mutate（）基于另一个列表的列表

4 个答案: