我最终获得了一个很大的data.table，我必须每行进行操作。 （是的......我知道这显然不是data.table的用途）
R) set.seed(1)
R) DT=data.table(matrix(rnorm(100),nrow=10))
R) DT[,c('a','b'):=list(1:10,2:11)]
R) DT
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 a b
1: -0.6264538107 1.51178116845 0.91897737161 1.35867955153 -0.1645235963 0.3981058804 2.40161776050 0.475509528900 -0.5686687328 -0.5425200310 1 2
2: 0.1836433242 0.38984323641 0.78213630073 -0.10278772734 -0.2533616801 -0.6120263933 -0.03924000273 -0.709946430922 -0.1351786151 1.2078678060 2 3
3: -0.8356286124 -0.62124058054 0.07456498337 0.38767161156 0.6969633754 0.3411196914 0.68973936245 0.610726353489 1.1780869966 1.1604026157 3 4
4: 1.5952808021 -2.21469988718 -1.98935169586 -0.05380504058 0.5566631987 -1.1293630961 0.02800215878 -0.934097631644 -1.5235668004 0.7002136495 4 5
5: 0.3295077718 1.12493091814 0.61982574789 -1.37705955683 -0.6887556945 1.4330237017 -0.74327320888 -1.253633400239 0.5939461876 1.5868334545 5 6
6: -0.8204683841 -0.04493360902 -0.05612873953 -0.41499456330 -0.7074951570 1.9803998985 0.18879229951 0.291446235517 0.3329503712 0.5584864256 6 7
7: 0.4874290524 -0.01619026310 -0.15579550671 -0.39428995371 0.3645819621 -0.3672214765 -1.80495862889 -0.443291873218 1.0630998373 -1.2765922085 7 8
8: 0.7383247051 0.94383621069 -1.47075238390 -0.05931339671 0.7685329245 -1.0441346263 1.46555486156 0.001105351632 -0.3041839236 -0.5732654142 8 9
9: 0.5757813517 0.82122119510 -0.47815005511 1.10002537198 -0.1123462122 0.5697196274 0.15325333821 0.074341324152 0.3700188099 -1.2246126149 9 10
10: -0.3053883872 0.59390132122 0.41794156020 0.76317574846 0.8811077265 -0.1350546039 2.17261167036 -0.589520946188 0.2670987908 -0.4734006364 10 11
我希望逐行显示所有min列中的Vi，我在使用apply时曾使用过data.frame。
apply(DT[,paste0('V',1:10),with=FALSE],FUN=min,MAR=1)
[1] -0.6264538107 -0.7099464309 -0.8356286124 -2.2146998872 -1.3770595568 -0.8204683841 -1.8049586289 -1.4707523839 -1.2246126149 -0.5895209462
所以我可以轻松更新。
好的，现在说我想立刻更新min和max（当然这是一个例子，所以我只拍了两件事，但在现实生活中这将是10 ... ）
f = function(x){return(c(max=max(x),min=min(x)))}
new=apply(DT[,paste0('V',1:10),with=FALSE],FUN=f,MAR=1)
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
max 2.4016177605 1.2078678060 1.1780869966 1.595280802 1.586833455 1.9803998985 1.063099837 1.465554862 1.100025372 2.1726116704
min -0.6264538107 -0.7099464309 -0.8356286124 -2.214699887 -1.377059557 -0.8204683841 -1.804958629 -1.470752384 -1.224612615 -0.5895209462
我想写
DT[,rownames(new):=new]
但这不起作用，所以这是我的问题
new，以便我可以立即更新DT？ 编辑 ：我找到了1的解决方案，但那很难，实际上很奇怪:=无法处理matrix，我是很确定它曾经是这种情况
DT[,c('a1','a2'):=data.table(matrix(apply(DT[,paste0('V',1:10),with=FALSE],FUN=f,MAR=1),byrow=T,nrow=10))]
R) DT
V1 V2 V3 V4 V5 V6 V7 V8 V9 V10 a b
1: -0.6264538107 1.51178116845 0.91897737161 1.35867955153 -0.1645235963 0.3981058804 2.40161776050 0.475509528900 -0.5686687328 -0.5425200310 1 2
2: 0.1836433242 0.38984323641 0.78213630073 -0.10278772734 -0.2533616801 -0.6120263933 -0.03924000273 -0.709946430922 -0.1351786151 1.2078678060 2 3
3: -0.8356286124 -0.62124058054 0.07456498337 0.38767161156 0.6969633754 0.3411196914 0.68973936245 0.610726353489 1.1780869966 1.1604026157 3 4
4: 1.5952808021 -2.21469988718 -1.98935169586 -0.05380504058 0.5566631987 -1.1293630961 0.02800215878 -0.934097631644 -1.5235668004 0.7002136495 4 5
5: 0.3295077718 1.12493091814 0.61982574789 -1.37705955683 -0.6887556945 1.4330237017 -0.74327320888 -1.253633400239 0.5939461876 1.5868334545 5 6
6: -0.8204683841 -0.04493360902 -0.05612873953 -0.41499456330 -0.7074951570 1.9803998985 0.18879229951 0.291446235517 0.3329503712 0.5584864256 6 7
7: 0.4874290524 -0.01619026310 -0.15579550671 -0.39428995371 0.3645819621 -0.3672214765 -1.80495862889 -0.443291873218 1.0630998373 -1.2765922085 7 8
8: 0.7383247051 0.94383621069 -1.47075238390 -0.05931339671 0.7685329245 -1.0441346263 1.46555486156 0.001105351632 -0.3041839236 -0.5732654142 8 9
9: 0.5757813517 0.82122119510 -0.47815005511 1.10002537198 -0.1123462122 0.5697196274 0.15325333821 0.074341324152 0.3700188099 -1.2246126149 9 10
10: -0.3053883872 0.59390132122 0.41794156020 0.76317574846 0.8811077265 -0.1350546039 2.17261167036 -0.589520946188 0.2670987908 -0.4734006364 10 11
a1 a2
1: 2.401617761 -0.6264538107
2: 1.207867806 -0.7099464309
3: 1.178086997 -0.8356286124
4: 1.595280802 -2.2146998872
5: 1.586833455 -1.3770595568
6: 1.980399899 -0.8204683841
7: 1.063099837 -1.8049586289
8: 1.465554862 -1.4707523839
9: 1.100025372 -1.2246126149
10: 2.172611670 -0.5895209462
EDIT2 ：在我的数据中查看使用DT[, (newColnames):=f(.DT), by=IDX, .SDcols=someIdx]比应用方式慢得多，是预期的吗？
答案 0 :(得分：10)
在每一行上创建.SD可能是一项非常昂贵的操作，尤其是当您的data.table由rows >> columns组成时。我建议在带有循环的列中使用pmin和pmax。我将用更大的数据（沿着行）来说明这一点。
set.seed(1)
require(data.table)
DT1 <- data.table(matrix(rnorm(1e6),ncol=10))
DT1[, a := 1:1e5]
DT2 <- copy(DT1)
DT3 <- copy(DT1)
arun <- function(DT) {
# assign first column (dummy)
DT[, `:=`(min = DT[, V1], max = DT[, V1])]
# get all other column names and use pmin and pmax
# and replace min and max columns
cols <- names(DT)[2:10]
for (i in cols) {
DT[, `:=`(min = pmin(min, DT[[i]]), max = pmax(max, DT[[i]]))]
}
DT
}
eddi <- function(DT) {
DT[, `:=`(min = min(.SD), max = max(.SD)), by = a, .SDcols = paste0("V", 1:10)]
}
frank <- function(DT) {
cols <- names(DT)[grepl('^V[[:digit:]]+$',names(DT))]
newcols <- c("min","max")
myfun <- range
DT[,(newcols):=as.list(myfun(.SD)),.SDcols=cols,by=1:nrow(DT)]
}
require(microbenchmark)
microbenchmark(o1 <- arun(DT1), o2 <- eddi(DT2), o3 <- frank(DT3), times=2)
Unit: milliseconds
expr min lq median uq max neval
o1 <- arun(DT1) 204.4417 204.4417 250.5205 296.5992 296.5992 2
o2 <- eddi(DT2) 92343.5321 92343.5321 96706.1622 101068.7923 101068.7923 2
o3 <- frank(DT3) 49083.7000 49083.7000 49521.9296 49960.1592 49960.1592 2
identical(o1, o2) # TRUE
identical(o1, o3) # TRUE
-
正如@Frank在评论中指出的那样，您可以将for循环替换为do.call：
DT[, c("min", "max") := { z <- dt[, 1:10, with=FALSE];
list(do.call(pmin, z), do.call(pmax, z))}]
答案 1 :(得分：5)
由于您已在data.table *中将行号作为一列，您可以这样做：
DT[, `:=`(a1 = max(.SD), a2 = min(.SD)), by = a, .SDcols = paste0("V", 1:10)]
或
setkey(DT, a)
DT[J(a), `:=`(a1 = max(.SD), a2 = min(.SD)), .SDcols = paste0("V", 1:10)]
第二个选项使用静默 by-without-by 。
*当然您也可以使用row.names或1:nrow(DT)
答案 2 :(得分：2)
如果您想使用其他功能，请说明步骤：
cols <- names(DT)[grepl('^V[[:digit:]]+$',names(DT))]
newcols <- c("min","max")
myfun <- range
DT[,(newcols):=as.list(myfun(.SD)),.SDcols=cols,by=1:nrow(DT)]
答案 3 :(得分：1)
我是否遗漏了某些东西，这不是给了行的最小值
set.seed(1)
DT=data.table(matrix(rnorm(100),nrow=10))
DT[,c('a','b'):=list(1:10,2:11)]
DT
cols<-c("V1", "V2", "V3", "V4", "V5", "V6", "V7", "V8", "V9", "V10")
方法1
DT[,Min_Vi:=do.call(pmin, c(.SD, na.rm=TRUE)), .SDcols=cols]
方法2
transform(DT,Min_Vi=pmin(get(cols)))