R 变换列表中的字符向量，以向量的内容为条件答案

【问题标题】：R transform character vectors in a list, conditional on the content of the vectorR 变换列表中的字符向量，以向量的内容为条件
【发布时间】：2022-08-19 01:48:11
【问题描述】：

问题是我有一个字符向量列表。

例子：

mylist <- list( c(\"once\",\"upon\",\"a\",\"time\"),
                c(\"once\", \"in\", \"olden\", \"times\"),
                c(\"Let\",\"all\",\"good\",\"men\"),
                c(\"Let\",\"This\"),
                c(\"once\", \"is\",\"never\",\"enough\"),
                c(\"in\",\"the\"),
                c(\"Come\",\"dance\",\"all\",\"around\"))

我想将 c(\"one\", \"two\") 添加到那些以 \"once\" 开头的向量以结束列表

mylist <- list( c(\"one\", \"two\", \"once\",\"upon\",\"a\",\"time\"),
                c(\"one\", \"two\", \"once\", \"in\", \"olden\", \"times\"),
                c(\"Let\",\"all\",\"good\",\"men\"),
                c(\"Let\",\"This\"),
                c(\"one\", \"two\", \"once\", \"is\",\"never\",\"enough\"),
                c(\"in\",\"the\"),
                c(\"Come\",\"dance\",\"all\",\"around\"))

至今

我可以选择相关的向量

mylist[grep(\"once\",mylist)]

我可以在前面加上 \"one\" 和 \"two\" 来创建一个结果列表

resultlist <- lapply(mylist[grep(\"once\",mylist)],FUN = function(listrow) prepend(listrow,c(\"One\",\"Two\")))

但是将结果放在 mylist 中的正确位置？

不，这逃脱了我！

提示，提示和解决方案最受欢迎:-)

标签： r list conditional-statements character transform

【解决方案1】：

我们可以使用

lapply(mylist , \(x) if(grepl("once" , x[1])) 
      append(x,  c("one", "two") , 0) else x)

输出

[[1]]
[1] "one"  "two"  "once" "upon" "a"    "time"

[[2]]
[1] "one"   "two"   "once"  "in"    "olden" "times"

[[3]]
[1] "Let"  "all"  "good" "men" 

[[4]]
[1] "Let"  "This"

[[5]]
[1] "one"    "two"    "once"   "is"     "never"  "enough"

[[6]]
[1] "in"  "the"

[[7]]
[1] "Come"   "dance"  "all"    "around"

【讨论】：

【解决方案2】：

我认为你根本不需要grep。遍历列表，检查"once" 的第一个值并通过c() 附加额外的值：

lapply(mylist, \(x) if(x[1] == "once") c("one", "two", x) else x)
##[[1]]
##[1] "one"  "two"  "once" "upon" "a"    "time"
##
##[[2]]
##[1] "one"   "two"   "once"  "in"    "olden" "times"
##
##[[3]]
##[1] "Let"  "all"  "good" "men" 
##
##[[4]]
##[1] "Let"  "This"
##
##[[5]]
##[1] "one"    "two"    "once"   "is"     "never"  "enough"
##
##[[6]]
##[1] "in"  "the"
##
##[[7]]
##[1] "Come"   "dance"  "all"    "around"

【讨论】：

完美运行。谢谢你。只是为了我的启迪，\（x）是什么意思？
\(x) 是function(x) 的简写，其中x 是mylist 的每个元素的占位符，因为它被lapply 循环。

【解决方案3】：

map_if 的另一个选择

library(purrr)
map_if(mylist, .p = ~ first(.x) == "once", .f = ~ c("one", "two", .x))

-输出

[[1]]
[1] "one"  "two"  "once" "upon" "a"    "time"

[[2]]
[1] "one"   "two"   "once"  "in"    "olden" "times"

[[3]]
[1] "Let"  "all"  "good" "men" 

[[4]]
[1] "Let"  "This"

[[5]]
[1] "one"    "two"    "once"   "is"     "never"  "enough"

[[6]]
[1] "in"  "the"

[[7]]
[1] "Come"   "dance"  "all"    "around"

【讨论】：