我在R的数据框中有一列，其值为“-1”，“0”，“1”。我想分别用“否”，“也许”和"is"替换这些值。我将使用sub来做到这一点。
我可以编写一个条件函数，然后编写代码:
df[col] <- lapply(df[col], conditional_function_substitution)
我也可以一次进行一次替换(三个例子中的第一个):
df[col] <- lapply(df[col], sub, pattern = '-1', replacement = "no")
我想知道是否可以在一行中完成？就像是:
df[col] <- lapply(df[col], sub, pattern = c('-1','0','1'), replacement = c('no','maybe','yes')
感谢您的见解!
最佳答案
通过将-1、0和1加2，您可以将索引放入所需结果的向量中:
c("no", "maybe", "yes")[dat + 2]
# [1] "no" "yes" "maybe" "yes" "yes" "no"
一个相关的选项可以利用
match函数来找出索引:c("no", "maybe", "yes")[match(dat, -1:1)]
# [1] "no" "yes" "maybe" "yes" "yes" "no"
或者，您可以使用命名向量进行重新编码:
unname(c("-1"="no", "0"="maybe", "1"="yes")[as.character(dat)])
# [1] "no" "yes" "maybe" "yes" "yes" "no"
您还可以使用嵌套的
ifelse:ifelse(dat == -1, "no", ifelse(dat == 0, "maybe", "yes"))
# [1] "no" "yes" "maybe" "yes" "yes" "no"
如果您不介意加载新程序包，则
Recode程序包中的car函数将执行以下操作:library(car)
Recode(dat, "-1='no'; 0='maybe'; 1='yes'")
# [1] "no" "yes" "maybe" "yes" "yes" "no"
数据:
dat <- c(-1, 1, 0, 1, 1, -1)
请注意，如果将
dat存储为字符串，则除第一个以外的所有内容都将起作用；首先，您需要使用as.numeric(dat)。如果代码清晰是您的主要目标，则应选择最容易理解的代码-我个人将选择第二个或最后一个，但这是个人喜好。
如果您对代码速度感兴趣，则可以对解决方案进行基准测试。这是我介绍的五个选项的基准，还包括当前作为其他答案发布的其他两个解决方案，以长度为100k的随机向量为基准:
set.seed(144)
dat <- sample(c(-1, 0, 1), replace=TRUE, 100000)
opt1 <- function(dat) c("no", "maybe", "yes")[dat + 2]
opt2 <- function(dat) c("no", "maybe", "yes")[match(dat, -1:1)]
opt3 <- function(dat) unname(c("-1"="no", "0"="maybe", "1"="yes")[as.character(dat)])
opt4 <- function(dat) ifelse(dat == -1, "no", ifelse(dat == 0, "maybe", "yes"))
opt5 <- function(dat) Recode(dat, "-1='no'; 0='maybe'; 1='yes'")
AnandaMahto <- function(dat) factor(dat, levels = c(-1, 0, 1), labels = c("no", "maybe", "yes"))
hrbrmstr <- function(dat) sapply(as.character(dat), switch, `-1`="no", `0`="maybe", `1`="yes", USE.NAMES=FALSE)
library(microbenchmark)
microbenchmark(opt1(dat), opt2(dat), opt3(dat), opt4(dat), opt5(dat), AnandaMahto(dat), hrbrmstr(dat))
# Unit: milliseconds
# expr min lq mean median uq max neval
# opt1(dat) 1.513500 2.553022 2.763685 2.656010 2.837673 4.384149 100
# opt2(dat) 2.153438 3.013502 3.251850 3.117058 3.269230 5.851234 100
# opt3(dat) 59.716271 61.890470 64.978685 62.509046 63.723048 144.708757 100
# opt4(dat) 62.934734 64.715815 71.181477 65.652195 71.123384 123.840577 100
# opt5(dat) 82.976441 84.849147 89.071808 85.752429 88.473162 155.347273 100
# AnandaMahto(dat) 57.267227 58.643889 60.508402 59.065642 60.368913 80.852157 100
# hrbrmstr(dat) 137.883307 148.626496 158.051220 153.441243 162.594752 228.271336 100
前两个选项似乎比其他任何一个选项都快一个数量级，尽管矢量可能必须非常大，或者您需要多次重复该操作才能使该选项生效有所不同。
如@AnandaMahto所指出的，如果我们使用字符输入而不是数字输入，那么这些结果在质量上是不同的:
set.seed(144)
dat <- sample(c("-1", "0", "1"), replace=TRUE, 100000)
opt1 <- function(dat) c("no", "maybe", "yes")[as.numeric(dat) + 2]
opt2 <- function(dat) c("no", "maybe", "yes")[match(dat, -1:1)]
opt3 <- function(dat) unname(c("-1"="no", "0"="maybe", "1"="yes")[as.character(dat)])
opt4 <- function(dat) ifelse(dat == -1, "no", ifelse(dat == 0, "maybe", "yes"))
opt5 <- function(dat) Recode(dat, "-1='no'; 0='maybe'; 1='yes'")
AnandaMahto <- function(dat) factor(dat, levels = c(-1, 0, 1), labels = c("no", "maybe", "yes"))
hrbrmstr <- function(dat) sapply(dat, switch, `-1`="no", `0`="maybe", `1`="yes", USE.NAMES=FALSE)
library(microbenchmark)
microbenchmark(opt1(dat), opt2(dat), opt3(dat), opt4(dat), opt5(dat), AnandaMahto(dat), hrbrmstr(dat))
# Unit: milliseconds
# expr min lq mean median uq max neval
# opt1(dat) 8.397194 9.519075 10.784108 9.693706 10.163203 55.78417 100
# opt2(dat) 2.281438 3.091418 4.231162 3.210794 3.436038 49.39879 100
# opt3(dat) 3.606863 5.481115 6.466393 5.720282 6.344651 48.47924 100
# opt4(dat) 66.819638 69.996704 74.596960 71.290522 73.404043 127.52415 100
# opt5(dat) 32.897019 35.701401 38.488489 36.336489 38.950272 88.20915 100
# AnandaMahto(dat) 1.329443 2.114504 2.824306 2.275736 2.493907 46.19333 100
# hrbrmstr(dat) 81.898572 91.043729 154.331766 100.006203 141.425717 1594.17447 100
现在，@ AnandaMahto提出的
factor解决方案是最快的，其次是使用match和命名向量查找进行向量索引。同样，所有运行时都足够快，以至于您需要一个大的向量或多次运行才能使这一切变得重要。关于r - 一行R中的多个替换，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/32170417/