我有一个67MM的行data.table，人名和姓氏用空格分隔。我只需要为每个单词创建一个新列。
以下是数据的一小部分：
n <- structure(list(Subscription_Id = c("13.855.231.846.091.000",
"11.156.048.529.090.800", "24.940.584.090.830", "242.753.039.111.124",
"27.843.782.090.830", "13.773.513.145.090.800", "25.691.374.090.830",
"12.236.174.155.090.900", "252.027.904.121.210", "11.136.991.054.110.100"
), Account_Desc = c("AGUAYO CARLA", "LEIVA LILIANA", "FULLANA MARIA LAURA",
"PETREL SERGIO", "IPTICKET SRL", "LEDESMA ORLANDO", "CATTANEO LUIS RAUL",
"CABRAL CARMEN ESTELA", "ITURGOYEN HECTOR", "CASA CASILDO"),
V1 = c("AGUAYO", "LEIVA", "FULLANA", "PETREL", "IPTICKET",
"LEDESMA", "CATTANEO", "CABRAL", "ITURGOYEN", "CASA"), V2 = c("CARLA",
"LILIANA", "MARIA", "SERGIO", "SRL", "ORLANDO", "LUIS", "CARMEN",
"HECTOR", "CASILDO"), V3 = c(NA, NA, "LAURA", NA, NA, NA,
"RAUL", "ESTELA", NA, NA), `NA` = c(NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_,
NA_character_, NA_character_, NA_character_, NA_character_
)), .Names = c("Subscription_Id", "Account_Desc", "V1", "V2",
"V3", NA), class = c("data.table", "data.frame"), row.names = c(NA,
-10L), .internal.selfref = <pointer: 0x0000000000200788>)
require("data.table")
n <- data.table(n)
预期输出
# Subscription_Id Account_Desc V1 V2 V3 NA
# 1: 13.855.231.846.091.000 AGUAYO CARLA AGUAYO CARLA NA NA
# 2: 11.156.048.529.090.800 LEIVA LILIANA LEIVA LILIANA NA NA
# 3: 24.940.584.090.830 FULLANA MARIA LAURA FULLANA MARIA LAURA NA
library(stringr)
# This separates the strings, but i loose the Subscription_Id variable.
n[, str_split_fixed(Account_Desc, "[ +]", 4)]
# This doesn't work.
n[, paste0("V",1:4) := str_split_fixed(Account_Desc, "[ +]", 4)]
这有效，但我似乎正在进行3次计算。不确定是否 最有效的方式
cols = paste0("V",1:3)
for(j in 1:3){
set(n,i=NULL,j=cols[j],value = sapply(strsplit(as.character(n$Account_Desc),"[ +]"), "[", j))
}
让我们使用 big_n 进行基准测试
big_n <- data.table(Subscription_Id = rep(n[,Subscription_Id],1e7),
Account_Desc = rep(n[,Account_Desc],1e7)
)
答案 0 :(得分：7)
我不使用接近此比例的数据集，所以我不知道这是否有用。我想到的一件事是使用matrix和矩阵索引。
由于我不耐烦，我只在慢速系统上的1e5行尝试过： - ）
big_n <- data.table(Subscription_Id = rep(n[,Subscription_Id],1e5),
Account_Desc = rep(n[,Account_Desc],1e5))
StringMat <- function(input) {
Temp <- strsplit(input, " ", fixed = TRUE)
Lens <- vapply(Temp, length, 1L)
A <- unlist(Temp, use.names = FALSE)
Rows <- rep(sequence(length(Temp)), Lens)
Cols <- sequence(Lens)
m <- matrix(NA, nrow = length(Temp), ncol = max(Lens),
dimnames = list(NULL, paste0("V", sequence(max(Lens)))))
m[cbind(Rows, Cols)] <- A
m
}
system.time(outB1 <- cbind(big_n, StringMat(big_n$Account_Desc)))
# user system elapsed
# 4.524 0.000 4.533
outB1
# Subscription_Id Account_Desc V1 V2 V3
# 1: 13.855.231.846.091.000 AGUAYO CARLA AGUAYO CARLA NA
# 2: 11.156.048.529.090.800 LEIVA LILIANA LEIVA LILIANA NA
# 3: 24.940.584.090.830 FULLANA MARIA LAURA FULLANA MARIA LAURA
# 4: 242.753.039.111.124 PETREL SERGIO PETREL SERGIO NA
# 5: 27.843.782.090.830 IPTICKET SRL IPTICKET SRL NA
# ---
# 999996: 13.773.513.145.090.800 LEDESMA ORLANDO LEDESMA ORLANDO NA
# 999997: 25.691.374.090.830 CATTANEO LUIS RAUL CATTANEO LUIS RAUL
# 999998: 12.236.174.155.090.900 CABRAL CARMEN ESTELA CABRAL CARMEN ESTELA
# 999999: 252.027.904.121.210 ITURGOYEN HECTOR ITURGOYEN HECTOR NA
# 1000000: 11.136.991.054.110.100 CASA CASILDO CASA CASILDO NA
set_method功能并比较时间set_method <- function(DT){
cols = paste0("V",1:3)
for(j in 1:3){
set(DT,i=NULL,j=cols[j],
value = sapply(strsplit(as.character(DT[, Account_Desc, with = TRUE]),
"[ +]"), "[", j))
}
}
system.time(set_method(big_n))
# user system elapsed
# 25.319 0.022 25.586
str_split_fixed（哎哟！）big_n[, c("V1", "V2", "V3") := NULL]
library(stringr)
system.time(outBrodie <- cbind(big_n, as.data.table(str_split_fixed(
big_n$Account_Desc, "[ +]", 4))))
# user system elapsed
# 204.966 0.514 206.910
答案 1 :(得分：3)
编辑3：偷走阿伦的血汗：
cbind(n, as.data.table(str_split_fixed(n$Account_Desc, "[ +]", 4)))
这可以避免可能代价高昂的by并产生相同的结果（加上原始名称列）。
EDIT2：根据Arun的评论，也许：
n.2[, c(paste0("V", 1:4)):=as.list(str_split_fixed(Account_Desc, "[ +]", 4)), by=Subscription_Id]
但你还有by。旧方式：
n[, as.list(str_split_fixed(Account_Desc, "[ +]", 4)), by=Subscription_Id]
产生
# Subscription_Id V1 V2 V3 V4
# 1: 13.855.231.846.091.000 AGUAYO CARLA
# 2: 11.156.048.529.090.800 LEIVA LILIANA
# 3: 24.940.584.090.830 FULLANA MARIA LAURA
# 4: 242.753.039.111.124 PETREL SERGIO
# 5: 27.843.782.090.830 IPTICKET SRL
# 6: 13.773.513.145.090.800 LEDESMA ORLANDO
# 7: 25.691.374.090.830 CATTANEO LUIS RAUL
# 8: 12.236.174.155.090.900 CABRAL CARMEN ESTELA
# 9: 252.027.904.121.210 ITURGOYEN HECTOR
# 10: 11.136.991.054.110.100 CASA CASILDO
编辑：警告，一些stringr函数可能很慢（不确定是否这个）。如果这对于您的过程来说仍然很慢，您可能希望使用strsplit编写自己的函数，并将其填充到适当的长度。