作为data.table包在R的相对缺乏经验的用户,我已经尝试处理一个文本列成大量的指示符列(虚拟变量)的,在每列中的1指示特定子串是在字符串列内找到。 例如,我要处理这个:
ID String
1 a$b
2 b$c
3 c
这个:
ID String a b c
1 a$b 1 1 0
2 b$c 0 1 1
3 c 0 0 1
我已经找到了如何做加工,但它需要运行较长时间比我想的,我怀疑我的代码是低效的。 我用虚拟数据编码A reproduceable版本如下。 请注意,在真实数据,有超过2000子搜索,每个子大约是30个字符长,并有可能达到几百万行。 如果需要,我可以并行,在这个问题抛出大量的资源,但我想对代码进行优化,尽可能。 我曾尝试运行Rprof,这表明没有明显的(对我)的改进。
set.seed(10)
elements_list <- c(outer(letters, letters, FUN = paste, sep = ""))
random_string <- function(min_length, max_length, separator) {
selection <- paste(sample(elements_list, ceiling(runif(1, min_length, max_length))), collapse = separator)
return(selection)
}
dt <- data.table(id = c(1:1000), messy_string = "")
dt[ , messy_string := random_string(2, 5, "$"), by = id]
create_indicators <- function(search_list, searched_string) {
y <- rep(0, length(search_list))
for(j in 1:length(search_list)) {
x <- regexpr(search_list[j], searched_string)
x <- x[1]
y[j] <- ifelse(x > 0, 1, 0)
}
return(y)
}
timer <- proc.time()
indicators <- matrix(0, nrow = nrow(dt), ncol = length(elements_list))
for(n in 1:nrow(dt)) {
indicators[n, ] <- dt[n, create_indicators(elements_list, messy_string)]
}
indicators <- data.table(indicators)
setnames(indicators, elements_list)
dt <- cbind(dt, indicators)
proc.time() - timer
user system elapsed
13.17 0.08 13.29
编辑
感谢伟大的反应 - 所有远优于我的方法。 下面一些速度测试每个功能的结果,稍作修改使用0L和1L在我自己的代码,通过方法来存储在单独的表的结果,并以规范的顺序。 这些经过时间从单一速度测试(而不是从许多测试中位数),但较大的运行的每个需要很长的时间。
Number of rows in dt 2K 10K 50K 250K 1M
OP 28.6 149.2 717.0
eddi 5.1 24.6 144.8 1950.3
RS 1.8 6.7 29.7 171.9 702.5
Original GT 1.4 7.4 57.5 809.4
Modified GT 0.7 3.9 18.1 115.2 473.9
GT4 0.1 0.4 2.26 16.9 86.9
漂亮清晰,GeekTrader的方法的修改版本是最好的。 我还是什么的每一步都做得有点模糊,但我可以走了过来,在我的闲暇。 虽然有些超出原质询的范围的,如果有人想解释什么GeekTrader和里卡多萨波塔的方法更有效地做,这将是双方由我和可能是由人谁在未来访问该页面的赞赏。 我特别想了解为什么一些方法规模比别人做得更好。
***** EDIT#2 *****
我试着用此评论编辑GeekTrader的回答,但似乎没有工作。 我做了两个非常小的修改到GT3功能,a)命令的列,这增加的少量时间,和b)用0L和1L,从而加快了一点东西代替0和1。 调用产生的功能GT4。 表上述编辑以在不同的表大小添加时间GT4。 显然,通过一英里的赢家,所以它具有直观的优点。
Answer 1:
更新:版本3
找到更快的方法。 此功能也具有很高的内存效率。 主要原因以前的功能,是因为拷贝/分配内部发生的缓慢lapply
环路以及rbinding
结果。
在接下来的版本中,我们预分配矩阵适当的大小,然后在适当的坐标值的变化,这使得它非常快相比其他循环版本。
funcGT3 <- function() {
#Get list of column names in result
resCol <- unique(dt[, unlist(strsplit(messy_string, split="\\$"))])
#Get dimension of result
nresCol <- length(resCol)
nresRow <- nrow(dt)
#Create empty matrix with dimensions same as desired result
mat <- matrix(rep(0, nresRow * nresCol), nrow = nresRow, dimnames = list(as.character(1:nresRow), resCol))
#split each messy_string by $
ll <- strsplit(dt[,messy_string], split="\\$")
#Get coordinates of mat which we need to set to 1
coords <- do.call(rbind, lapply(1:length(ll), function(i) cbind(rep(i, length(ll[[i]])), ll[[i]] )))
#Set mat to 1 at appropriate coordinates
mat[coords] <- 1
#Bind the mat to original data.table
return(cbind(dt, mat))
}
result <- funcGT3() #result for 1000 rows in dt
result
ID messy_string zn tc sv db yx st ze qs wq oe cv ut is kh kk im le qg rq po wd kc un ft ye if zl zt wy et rg iu
1: 1 zn$tc$sv$db$yx 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
2: 2 st$ze$qs$wq 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
3: 3 oe$cv$ut$is 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4: 4 kh$kk$im$le$qg 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5: 5 rq$po$wd$kc 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 1 1 1 0 0 0 0 0 0 0 0 0 0
---
996: 996 rp$cr$tb$sa 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
997: 997 cz$wy$rj$he 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0
998: 998 cl$rr$bm 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
999: 999 sx$hq$zy$zd 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
1000: 1000 bw$cw$pw$rq 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0
基准againt里卡多建议版本2(这是在数据250K行):
Unit: seconds
expr min lq median uq max neval
GT2 104.68672 104.68672 104.68672 104.68672 104.68672 1
GT3 15.15321 15.15321 15.15321 15.15321 15.15321 1
VERSION 1以下是建议答案1版
set.seed(10)
elements_list <- c(outer(letters, letters, FUN = paste, sep = ""))
random_string <- function(min_length, max_length, separator) {
selection <- paste(sample(elements_list, ceiling(runif(1, min_length, max_length))), collapse = separator)
return(selection)
}
dt <- data.table(ID = c(1:1000), messy_string = "")
dt[ , messy_string := random_string(2, 5, "$"), by = ID]
myFunc <- function() {
ll <- strsplit(dt[,messy_string], split="\\$")
COLS <- do.call(rbind,
lapply(1:length(ll),
function(i) {
data.frame(
ID= rep(i, length(ll[[i]])),
COL = ll[[i]],
VAL= rep(1, length(ll[[i]]))
)
}
)
)
res <- as.data.table(tapply(COLS$VAL, list(COLS$ID, COLS$COL), FUN = length ))
dt <- cbind(dt, res)
for (j in names(dt))
set(dt,which(is.na(dt[[j]])),j,0)
return(dt)
}
create_indicators <- function(search_list, searched_string) {
y <- rep(0, length(search_list))
for(j in 1:length(search_list)) {
x <- regexpr(search_list[j], searched_string)
x <- x[1]
y[j] <- ifelse(x > 0, 1, 0)
}
return(y)
}
OPFunc <- function() {
indicators <- matrix(0, nrow = nrow(dt), ncol = length(elements_list))
for(n in 1:nrow(dt)) {
indicators[n, ] <- dt[n, create_indicators(elements_list, messy_string)]
}
indicators <- data.table(indicators)
setnames(indicators, elements_list)
dt <- cbind(dt, indicators)
return(dt)
}
library(plyr)
plyrFunc <- function() {
indicators = do.call(rbind.fill, sapply(1:dim(dt)[1], function(i)
dt[i,
data.frame(t(as.matrix(table(strsplit(messy_string,
split = "\\$")))))
]))
dt = cbind(dt, indicators)
#dt[is.na(dt)] = 0 #THIS DOESN'T WORK. USING FOLLOWING INSTEAD
for (j in names(dt))
set(dt,which(is.na(dt[[j]])),j,0)
return(dt)
}
BENCHMARK
system.time(res <- myFunc())
## user system elapsed
## 1.01 0.00 1.01
system.time(res2 <- OPFunc())
## user system elapsed
## 21.58 0.00 21.61
system.time(res3 <- plyrFunc())
## user system elapsed
## 1.81 0.00 1.81
版本2:里卡多建议
我在这里,而不是在我的答案张贴这作为框架是真的@ GeekTrader的 -Rick_
myFunc.modified <- function() {
ll <- strsplit(dt[,messy_string], split="\\$")
## MODIFICATIONS:
# using `rbindlist` instead of `do.call(rbind.. )`
COLS <- rbindlist( lapply(1:length(ll),
function(i) {
data.frame(
ID= rep(i, length(ll[[i]])),
COL = ll[[i]],
VAL= rep(1, length(ll[[i]])),
# MODICIATION: Not coercing to factors
stringsAsFactors = FALSE
)
}
)
)
# MODIFICATION: Preserve as matrix, the output of tapply
res2 <- tapply(COLS$VAL, list(COLS$ID, COLS$COL), FUN = length )
# FLATTEN into a data.table
resdt <- data.table(r=c(res2))
# FIND & REPLACE NA's of single column
resdt[is.na(r), r:=0L]
# cbind with dt, a matrix, with the same attributes as `res2`
cbind(dt,
matrix(resdt[[1]], ncol=ncol(res2), byrow=FALSE, dimnames=dimnames(res2)))
}
### Benchmarks:
orig = quote({dt <- copy(masterDT); myFunc()})
modified = quote({dt <- copy(masterDT); myFunc.modified()})
microbenchmark(Modified = eval(modified), Orig = eval(orig), times=20L)
# Unit: milliseconds
# expr min lq median uq max
# 1 Modified 895.025 971.0117 1011.216 1189.599 2476.972
# 2 Orig 1953.638 2009.1838 2106.412 2230.326 2356.802
Answer 2:
# split the `messy_string` and create a long table, keeping track of the id
DT2 <- setkey(DT[, list(val=unlist(strsplit(messy_string, "\\$"))), by=list(ID, messy_string)], "val")
# add the columns, initialize to 0
DT2[, c(elements_list) := 0L]
# warning expected, re:adding large ammount of columns
# iterate over each value in element_list, assigning 1's ass appropriate
for (el in elements_list)
DT2[el, c(el) := 1L]
# sum by ID
DT2[, lapply(.SD, sum), by=list(ID, messy_string), .SDcols=elements_list]
请注意,我们沿着输送messy_string
列,因为它比离开它背后更便宜,然后join
荷兰国际集团的ID把它找回来。 如果你不需要它在最终的输出,只是删除它上面。
基准:
创建示例数据:
# sample data, using OP's exmple
set.seed(10)
N <- 1e6 # number of rows
elements_list <- c(outer(letters, letters, FUN = paste, sep = ""))
messy_string_vec <- random_string_fast(N, 2, 5, "$") # Create the messy strings in a single shot.
masterDT <- data.table(ID = c(1:N), messy_string = messy_string_vec, key="ID") # create the data.table
侧面说明它是显著更快地一次性生成随机字符串和分配结果作为一列,而不是调用函数N次,并通过一个分配给每个企业之一。
# Faster way to create the `messy_string` 's
random_string_fast <- function(N, min_length, max_length, separator) {
ints <- seq(from=min_length, to=max_length)
replicate(N, paste(sample(elements_list, sample(ints)), collapse=separator))
}
比较四种方法:
- 这个答案 - “DT.RS”
- @ EDDI的答案 - “Plyr.eddi”
- @ GeekTrader的答案 - DT.GT
- GeekTrader的回答了一些修改 - DT.GT_Mod
下面是设置:
library(data.table); library(plyr); library(microbenchmark)
# data.table method - RS
usingDT.RS <- quote({DT <- copy(masterDT);
DT2 <- setkey(DT[, list(val=unlist(strsplit(messy_string, "\\$"))), by=list(ID, messy_string)], "val"); DT2[, c(elements_list) := 0L]
for (el in elements_list) DT2[el, c(el) := 1L]; DT2[, lapply(.SD, sum), by=list(ID, messy_string), .SDcols=elements_list]})
# data.table method - GeekTrader
usingDT.GT <- quote({dt <- copy(masterDT); myFunc()})
# data.table method - GeekTrader, modified by RS
usingDT.GT_Mod <- quote({dt <- copy(masterDT); myFunc.modified()})
# ply method from below
usingPlyr.eddi <- quote({dt <- copy(masterDT); indicators = do.call(rbind.fill, sapply(1:dim(dt)[1], function(i) dt[i, data.frame(t(as.matrix(table(strsplit(messy_string, split = "\\$"))))) ]));
dt = cbind(dt, indicators); dt[is.na(dt)] = 0; dt })
以下是测试结果:
microbenchmark( usingDT.RS=eval(usingDT.RS), usingDT.GT=eval(usingDT.GT), usingDT.GT_Mod=eval(usingDT.GT_Mod), usingPlyr.eddi=eval(usingPlyr.eddi), times=5L)
On smaller data:
N = 600
Unit: milliseconds
expr min lq median uq max
1 usingDT.GT 1189.7549 1198.1481 1200.6731 1202.0972 1203.3683
2 usingDT.GT_Mod 581.7003 591.5219 625.7251 630.8144 650.6701
3 usingDT.RS 2586.0074 2602.7917 2637.5281 2819.9589 3517.4654
4 usingPlyr.eddi 2072.4093 2127.4891 2225.5588 2242.8481 2349.6086
N = 1,000
Unit: seconds
expr min lq median uq max
1 usingDT.GT 1.941012 2.053190 2.196100 2.472543 3.096096
2 usingDT.RS 3.107938 3.344764 3.903529 4.010292 4.724700
3 usingPlyr 3.297803 3.435105 3.625319 3.812862 4.118307
N = 2,500
Unit: seconds
expr min lq median uq max
1 usingDT.GT 4.711010 5.210061 5.291999 5.307689 7.118794
2 usingDT.GT_Mod 2.037558 2.092953 2.608662 2.638984 3.616596
3 usingDT.RS 5.253509 5.334890 6.474915 6.740323 7.275444
4 usingPlyr.eddi 7.842623 8.612201 9.142636 9.420615 11.102888
N = 5,000
expr min lq median uq max
1 usingDT.GT 8.900226 9.058337 9.233387 9.622531 10.839409
2 usingDT.GT_Mod 4.112934 4.293426 4.460745 4.584133 6.128176
3 usingDT.RS 8.076821 8.097081 8.404799 8.800878 9.580892
4 usingPlyr.eddi 13.260828 14.297614 14.523016 14.657193 16.698229
# dropping the slower two from the tests:
microbenchmark( usingDT.RS=eval(usingDT.RS), usingDT.GT=eval(usingDT.GT), usingDT.GT_Mod=eval(usingDT.GT_Mod), times=6L)
N = 10,000
Unit: seconds
expr min lq median uq max
1 usingDT.GT_Mod 8.426744 8.739659 8.750604 9.118382 9.848153
2 usingDT.RS 15.260702 15.564495 15.742855 16.024293 16.249556
N = 25,000
... (still running)
-----------------
用于标杆功能:
# original random string function
random_string <- function(min_length, max_length, separator) {
selection <- paste(sample(elements_list, ceiling(runif(1, min_length, max_length))), collapse = separator)
return(selection)
}
# GeekTrader's function
myFunc <- function() {
ll <- strsplit(dt[,messy_string], split="\\$")
COLS <- do.call(rbind,
lapply(1:length(ll),
function(i) {
data.frame(
ID= rep(i, length(ll[[i]])),
COL = ll[[i]],
VAL= rep(1, length(ll[[i]]))
)
}
)
)
res <- as.data.table(tapply(COLS$VAL, list(COLS$ID, COLS$COL), FUN = length ))
dt <- cbind(dt, res)
for (j in names(dt))
set(dt,which(is.na(dt[[j]])),j,0)
return(dt)
}
# Improvements to @GeekTrader's `myFunc` -RS '
myFunc.modified <- function() {
ll <- strsplit(dt[,messy_string], split="\\$")
## MODIFICATIONS:
# using `rbindlist` instead of `do.call(rbind.. )`
COLS <- rbindlist( lapply(1:length(ll),
function(i) {
data.frame(
ID= rep(i, length(ll[[i]])),
COL = ll[[i]],
VAL= rep(1, length(ll[[i]])),
# MODICIATION: Not coercing to factors
stringsAsFactors = FALSE
)
}
)
)
# MODIFICATION: Preserve as matrix, the output of tapply
res2 <- tapply(COLS$VAL, list(COLS$ID, COLS$COL), FUN = length )
# FLATTEN into a data.table
resdt <- data.table(r=c(res2))
# FIND & REPLACE NA's of single column
resdt[is.na(r), r:=0L]
# cbind with dt, a matrix, with the same attributes as `res2`
cbind(dt,
matrix(resdt[[1]], ncol=ncol(res2), byrow=FALSE, dimnames=dimnames(res2)))
}
### Benchmarks comparing the two versions of GeekTrader's function:
orig = quote({dt <- copy(masterDT); myFunc()})
modified = quote({dt <- copy(masterDT); myFunc.modified()})
microbenchmark(Modified = eval(modified), Orig = eval(orig), times=20L)
# Unit: milliseconds
# expr min lq median uq max
# 1 Modified 895.025 971.0117 1011.216 1189.599 2476.972
# 2 Orig 1953.638 2009.1838 2106.412 2230.326 2356.802
Answer 3:
这里有一个新的有些方法,使用cSplit_e()
从splitstackshape
包。
library(splitstackshape)
cSplit_e(dt, split.col = "String", sep = "$", type = "character",
mode = "binary", fixed = TRUE, fill = 0)
# ID String String_a String_b String_c
#1 1 a$b 1 1 0
#2 2 b$c 0 1 1
#3 3 c 0 0 1
Answer 4:
这里有一个〜10倍的速度更快的版本使用rbind.fill
。
library(plyr)
indicators = do.call(rbind.fill, sapply(1:dim(dt)[1], function(i)
dt[i,
data.frame(t(as.matrix(table(strsplit(messy_string,
split = "\\$")))))
]))
dt = cbind(dt, indicators)
# dt[is.na(dt)] = 0
# faster NA replace (thanks geektrader)
for (j in names(dt))
set(dt, which(is.na(dt[[j]])), j, 0L)
Answer 5:
下面是使用的方法rapply
和table
。 我相信会有比这里使用表稍快的做法,但它仍然比略快myfunc.Modified
从@ricardo; S答案
# a copy with enough column pointers available
dtr <- alloc.col(copy(dt) ,1000L)
rapplyFun <- function(){
ll <- strsplit(dtr[, messy_string], '\\$')
Vals <- rapply(ll, classes = 'character', f= table, how = 'replace')
Names <- unique(rapply(Vals, names))
dtr[, (Names) := 0L]
for(ii in seq_along(Vals)){
for(jj in names(Vals[[ii]])){
set(dtr, i = ii, j = jj, value =Vals[[ii]][jj])
}
}
}
microbenchmark(myFunc.modified(), rapplyFun(),times=5)
Unit: milliseconds
# expr min lq median uq max neval
# myFunc.modified() 395.1719 396.8706 399.3218 400.6353 401.1700 5
# rapplyFun() 308.9103 309.5763 309.9368 310.2971 310.3463 5
Answer 6:
这里是另一种解决方案,即构建一个稀疏矩阵对象,而不是你拥有什么。 该刮胡子了大量的时间和内存。
它产生有序的结果,甚至与转换data.table
它比GT3快0L
和1L
和没有重新排序(这可能是因为我用不同的方法在所需的坐标值到达-我没有通过GT3算法中去)然而,如果你不转换,并将其作为一个稀疏矩阵是约10-20x比GT3快(并具有更小的内存占用)。
library(Matrix)
strings = strsplit(dt$messy_string, split = "$", fixed = TRUE)
element.map = data.table(el = elements_list, n = seq_along(elements_list), key = "el")
tmp = data.table(n = seq_along(strings), each = unlist(lapply(strings, length)))
rows = tmp[, rep(n, each = each), by = n][, V1]
cols = element.map[J(unlist(strings))][,n]
dt.sparse = sparseMatrix(rows, cols, x = 1,
dims = c(max(rows), length(elements_list)))
# optional, should be avoided until absolutely necessary
dt = cbind(dt, as.data.table(as.matrix(dt.sparse)))
setnames(dt, c('id', 'messy_string', elements_list))
我们的想法是分割为字符串,然后使用data.table
作为地图对象的每个子映射到其正确的列位置。 从那里,它只是一个正确搞清楚行和矩阵中的填充物。
文章来源: Split a string column into several dummy variables