我需要像你这样的专家帮助解决问题，这对我的R技能来说太大了。
我有一个矢量和一个data.frame：
vec = c("v1;v2","v3","v4","v5;v6")
vecNames = c("v1","v2","v3","v4","v5","v6")
vecNames
## [1] "v1" "v2" "v3" "v4" "v5" "v6"
vecDescription = c("descr1","descr2","descr3","descr4","descr5","descr6")
vecDescription
## [1] "descr1" "descr2" "descr3" "descr4" "descr5" "descr6"
df = data.frame(vecNames, vecDescription)
df
vecNames vecDescription
1 v1 descr1
2 v2 descr2
3 v3 descr3
4 v4 descr4
5 v5 descr5
6 v6 descr6
data.frame用于注释。
mapping = df$vecDescription[match(vec, df$vecNames)]
输出符合预期：
as.vector(mapping)
## [1] NA "descr3" "descr4" NA
但我想：
## [1] "descr1;descr2" "descr3" "descr4" "descr5;descr6"
我使用for循环成功了，但是当应用于500k行时，这种方法非常慢。
答案 0 :(得分：6)
另一个基础R解决方案：
L <- strsplit(vec,split = ';')
R <- as.character(df$vecDescription)[match(unlist(L),df$vecNames)]
sapply(relist(R, L), paste, collapse=';')
和基准：
f.m <- function(vec,df) {
L <- strsplit(vec,split = ';')
R <- with(df,vecDescription[match(unlist(L),vecNames)])
sapply(relist(R, L), paste, collapse=';')
}
f.m2 <- function(vec,df) {
L <- strsplit(vec,split = ';')
R <- as.character(df$vecDescription)[match(unlist(L),df$vecNames)]
sapply(relist(R, L), paste, collapse=';')
}
f.j <- function(vec,df) {
elts = strsplit(vec, ";")
mapping = df$vecDescription[match(do.call(c, elts), df$vecNames)]
tapply(mapping, rep(1:length(elts), sapply(elts, length)),
paste, collapse = ';')
}
f.da <- function(vec,df) {
vec <- strsplit(vec, ";")
sapply(vec, function(x) with(df, paste(vecDescription[vecNames %in% x], collapse = ";")))
}
f.da2 <- function(vec,df) {
vapply(vec, function(x) with(df, paste(vecDescription[vecNames %in% x], collapse = ";")), character(1))
}
library(data.table)
library(reshape2)
f.eddi <- function(vec,df) {
dt = as.data.table(df) # or use setDT to convert in place
setkey(dt, vecNames)
dt[melt(strsplit(vec, split = ";"))][,
paste(vecDescription, collapse = ";"), by = L1][, V1]
}
f.eddi2 <- function(vec,df) {
setkey(dt, vecNames)
melt2 = function(l) data.table(value = unlist(l, use.names = F),
L1 = unlist(lapply(seq_along(l),
function(i) rep(i, length(l[[i]]))),
use.names = F))
dt[melt2(strsplit(vec, split = ";"))][,
paste(vecDescription, collapse = ";"), by = L1][, V1]
}
f.Metrics <- function(vec,df) {
x1<-strsplit(vec,";")
x2<-data.frame(do.call(rbind,x1))
x3<-df$vecDescription[df$vecNames %in% x2[,1]]
x4<-df$vecDescription[df$vecNames %in% x2[,2]]
sapply(1:length(x1),function(i){ifelse(x3[i]!=x4[i],paste(x3[i],x4[i],sep=";"),paste(x3[i]))})
}
df2 = data.frame(vecNames, vecDescription, stringsAsFactors = FALSE)
library('microbenchmark')
microbenchmark(f.m(vec,df), f.j(vec,df2), f.da(vec,df), f.da2(vec,df), f.eddi(vec,df))
结果：
Unit: microseconds
expr min lq mean median uq max neval cld
f.m(vec, df) 186.414 218.6155 263.8829 231.8240 248.3900 2506.887 100 b
f.m2(vec, df) 94.751 113.4995 124.3000 122.1635 134.3795 195.045 100 a
f.j(vec, df2) 211.411 231.2145 254.2509 242.9275 261.9220 481.501 100 b
f.da(vec, df) 145.689 176.9130 199.1804 185.8020 195.6595 1383.394 100 ab
f.da2(vec, df) 117.027 140.6245 153.2124 150.5025 157.9735 298.111 100 ab
f.eddi(vec, df) 3396.690 3586.1695 3799.5835 3648.2905 3762.6335 6468.448 100 d
f.Metrics(vec, df) 748.323 789.5460 881.9349 809.0135 833.5465 3335.045 100 c
<强> [更新]
正如@eddi正确指出的那样，应该使用一个更大的数据集来实现更真实的基准测试，所以我们在这里：
n <- 1000
set.seed(1)
sample1 <- sample(n)
sample2 <- sample(n)
vec <- sapply(sample1, function(i) if (runif(1)>0.5) paste0('v',c(i,sample(n,size=1)),collapse=';') else paste0('v',i))
vecNames <- paste0('v', sample2)
vecDescription <- paste0('descr', sample2)
df = data.frame(vecNames, vecDescription)
df2 = data.frame(vecNames, vecDescription, stringsAsFactors = FALSE)
library('microbenchmark')
microbenchmark（f.m2（vec，df2），fj（vec，df2），f.da2（vec，df2），f.eddi2（vec，df2），f。Metrics（vec，df2））
结果：
Unit: milliseconds
expr min lq mean median uq max neval cld
f.m(vec, df) 31.679775 35.682250 38.813526 38.53798 41.278268 50.94508 100 b
f.m2(vec, df) 8.384308 9.596091 10.833422 10.32222 10.954757 18.33386 100 a
f.j(vec, df2) 4.665586 5.216920 6.003011 5.65613 6.184318 12.32919 100 a
f.da(vec, df) 87.810338 94.419069 98.369134 96.63011 101.004672 165.76800 100 c
f.da2(vec, df) 84.199736 89.024529 94.053774 91.57543 94.448173 171.84077 100 c
f.eddi(vec, df) 276.079649 299.699244 314.580860 311.82896 329.421674 352.73114 100 d
f.Metrics(vec, df) 482.671849 496.465168 507.629372 505.23325 513.390346 594.13570 100 e
现在冠军是f.j()，比f.m2()快两倍，而其他功能则慢一个数量级。
[更新2]
在此基准测试中，n = 5000，并且所有函数都以df2作为输入（字符串是字符）：
Unit: milliseconds
expr min lq mean median uq max neval cld
f.m2(vec, df2) 44.97854 47.12005 51.13561 48.58260 55.11687 85.57911 100 b
f.j(vec, df2) 24.03023 26.03697 28.10994 27.09699 28.45757 39.77269 100 a
f.da2(vec, df2) 1150.06311 1236.57530 1276.34064 1269.03829 1296.79251 1583.44486 100 d
f.eddi2(vec, df2) 65.88291 68.06959 72.89662 70.05462 76.19301 178.73181 100 c
f.Metrics(vec, df2) 54.54662 57.37777 59.95356 58.41737 62.15440 69.84452 100 b
另一个基准，n = 50000：
Unit: milliseconds
expr min lq mean median uq max neval cld
f.m2(vec, df2) 551.7985 602.0489 659.5792 638.6707 685.9923 1135.1548 100 b
f.j(vec, df2) 340.2615 415.2678 454.9885 447.5994 494.9217 661.5898 100 a
f.eddi2(vec, df2) 833.3205 920.6528 979.3859 963.0641 1018.2014 1519.3684 100 c
f.Metrics(vec, df2) 795.4200 895.8132 970.6516 954.8318 1001.6742 1427.0432 100 c
和最后一个，n = 500000：
Unit: seconds
expr min lq mean median uq max neval cld
f.m2(vec, df2) 7.420941 7.645800 8.047706 7.978916 8.301547 9.134872 10 b
f.j(vec, df2) 5.043295 5.316371 5.925725 5.514834 6.288766 8.289737 10 a
f.eddi2(vec, df2) 11.190716 11.373425 12.144147 11.935814 12.487354 14.798366 10 c
f.Metrics(vec, df2) 13.086297 13.859301 14.143273 14.149004 14.524544 15.151098 10 d
答案 1 :(得分：5)
您需要执行以下操作：
df = data.frame(vecNames, vecDescription, stringsAsFactors = FALSE)
elts = strsplit(vec, ";")
mapping = df$vecDescription[match(do.call(c, elts), df$vecNames)]
tapply(mapping, rep(1:length(elts), sapply(elts, length)),
paste, collapse = ';')
请注意data.frame定义中的stringsAsFactors = FALSE。从根本上说，仍然有一个使用tapply的循环，但我不认为它可以被矢量化。
答案 2 :(得分：5)
library(data.table)
library(reshape2)
dt = as.data.table(df) # or use setDT to convert in place
setkey(dt, vecNames)
dt[melt(strsplit(vec, split = ";"))][,
paste(vecDescription, collapse = ";"), by = L1][, V1]
#[1] "descr1;descr2" "descr3" "descr4" "descr5;descr6"
对于大数据，melt将成为瓶颈，您可以使用以下功能：
melt2 = function(l) data.table(value = unlist(l, use.names = F),
L1 = unlist(lapply(seq_along(l),
function(i) rep(i, length(l[[i]]))),
use.names = F))
答案 3 :(得分：5)
这是另一个基础R快速解决方案
vec <- strsplit(vec, ";")
sapply(vec, function(x) with(df, paste(vecDescription[vecNames %in% x], collapse = ";")))
## [1] "descr1;descr2" "descr3" "descr4" "descr5;descr6"
或者我们可以在
中使用vapply加快速度
vapply(vec, function(x) with(df, paste(vecDescription[vecNames %in% x], collapse = ";")), character(1))
答案 4 :(得分：2)
x1<-strsplit(vec,";")
x2<-data.frame(do.call(rbind,x1))
x3<-df$vecDescription[df$vecNames %in% x2[,1]]
x4<-df$vecDescription[df$vecNames %in% x2[,2]]
x5<-lapply(1:length(x1),function(i){ifelse(x3[i]!=x4[i],paste(x3[i],x4[i],sep=";"),paste(x3[i]))})
> x5
[[1]]
[1] "descr1;descr2"
[[2]]
[1] "descr3"
[[3]]
[1] "descr4"
[[4]]
[1] "descr5;descr6"
答案 5 :(得分：2)
Simpe使用我坚持的 qdap ：
library(qdap)
mgsub(vecNames, vecDescription, vec)
## [1] "descr1;descr2" "descr3" "descr4" "descr5;descr6"
如果您对 qdap 的开发版本进行基准测试，mgsub显着降低内存负担并且更快。这个简短的脚本将下载开发版：
if (!require("pacman")) install.packages("pacman")
pacman::p_load_gh("trinker/qdap")