有没有一种方法可以在不使用 for 循环的情况下重复求解从数据框创建的矩阵？
Is there a way to repeatedly solve matrices created from a data frame without using for loops?
更新
我一直在尝试 apply() 函数，但输出让我很困惑。我需要 data.frame 或 data.table。 apply(smallExample, 1, SectionDev) 给出了正确的输出，但在列表中。我读过 sapply() 可以输出 data.frame 或 matrix 但我无法将整行输入函数。
新函数
SectionDev <- function(x){
parameters <- data.table("Name" = c("Head", "Foot", "Pillow", "Mattress", "Sleeping", "Restless"),
"Space" = c(8, 78, 17, 17, 3, 8),
"Portion" = c(2, 4, 2, 2, 3, 3))
RG <- as.numeric(x[which(names(x) == "Gas")])
RS <- as.numeric(x[which(names(x) == "Solid")])
A <- data.frame(MG=c(1, 0, 1, 0),
MS=c(1, 0, 0, 1),
VG=c(0, 1, -1*RG, 0),
VS=c(0, 1, 0, -1*RS))
for (j in 1:nrow(parameters)) {
SectName <- as.vector(as.matrix(parameters[j, `Name`]))
Ac <- parameters[j, `Portion`]
Vt <- parameters[j, `Space`]
Pushing <- as.numeric(x[which(names(x) == SectName)])
Driving <- as.numeric(Pushing * 249.1)
Pressing <- Driving*Ac / 9.8
b <- c(Pressing, Vt, 0, 0)
MatrixConstants <- data.table("Timestamp" = as.vector(as.matrix(x[which(names(x) == "Timestamp")])),
"Section" = SectName,
RG,
RS)
MatrixSolution <- solve(A,b)
MatrixPtResult <- cbind(MatrixConstants,t(
as.data.frame(MatrixSolution)))
if (exists("MatrixTest")){
rlist <- list(MatrixTest, MatrixPtResult)
} else {
rlist <- list(MatrixPtResult)
}
MatrixTest <- rbindlist(rlist)
}
return(MatrixTest)
}
期望输出
> MatrixOutput[1:12,]
Timestamp Portion Gas Solid MG MS VG
1: 2019-04-24 06:00:00 Head 18 234 149.2218 -67.88299 8.290098
2: 2019-04-24 06:00:00 Foot 18 234 1427.7993 -309.39116 79.322184
3: 2019-04-24 06:00:00 Pillow 18 234 318.7908 -166.28061 17.710601
4: 2019-04-24 06:00:00 Mattress 18 234 183.2262 1596.05952 10.179233
5: 2019-04-24 06:00:00 Sleeping 18 234 -316.4209 4815.47194 -17.578940
6: 2019-04-24 06:00:00 Restless 18 234 -549.3597 9013.67602 -30.519983
7: 2019-04-24 07:00:00 Head 17 239 141.7429 -80.73885 8.337819
8: 2019-04-24 07:00:00 Foot 17 239 1365.2541 -551.86634 80.309064
9: 2019-04-24 07:00:00 Pillow 17 239 303.3448 -201.67136 17.843813
10: 2019-04-24 07:00:00 Mattress 17 239 178.7719 1549.67705 10.515996
11: 2019-04-24 07:00:00 Sleeping 17 239 -277.9378 4624.47863 -16.349283
12: 2019-04-24 07:00:00 Restless 17 239 -501.7540 8966.07028 -29.514938
VS
1: -0.2900983
2: -1.3221844
3: -0.7106009
4: 6.8207672
5: 20.5789399
6: 38.5199830
7: -0.3378195
8: -2.3090642
9: -0.8438132
10: 6.4840044
11: 19.3492830
12: 37.5149384
原创
我必须为数据中的每个时间点和位置求解相同的矩阵。 for 循环有效，但在 12,000 多个时间点和 6 个位置时速度不快。
有没有办法不使用 for 来做到这一点？它只是感觉不像我现在做的那样。
我确实尝试过将 foreach 与 doParallel 一起使用，但实际上它比用 TERR 编写的 for 慢。
有两个嵌套循环。第一个提取时间点的测量值，第二个为每个位置执行 solve。
for (i in 1:nrow(smallExample)) {
RG <- as.matrix(smallExample[i, "Gas"])
RS <- as.matrix(smallExample[i, "Solid"])
A <- data.frame(MG=c(1, 0, 1, 0),
MS=c(1, 0, 0, 1),
VG=c(0, 1, -1*RG,0),
VS=c(0, 1, 0, -1*RS))
for (j in 1:nrow(parameters)) {
SectName <- as.vector(as.matrix(parameters[j, "Name"]))
Ac <- parameters[j, "Portion"]
Vt <- parameters[j, "Space"]
Pushing <- smallExample[i, get(SectName)]
Driving <- as.matrix(Pushing*249.1)
Pressing <- Driving * as.matrix(Ac) / 9.8
b <- c(Pressing,Vt,0,0)
MatrixConstants <- data.table("Timestamp" = as.vector(as.matrix(smallExample[i, "Timestamp"])),
"Portion" = SectName,
RG,
RS)
MatrixSolution <- solve(A,b)
MatrixPtResult <- cbind(MatrixConstants,t(
as.data.frame(MatrixSolution)))
if (exists("MatrixOutput")){
rlist <- list(MatrixOutput, MatrixPtResult)
} else {
rlist <- list(MatrixPtResult)
}
MatrixOutput <- rbindlist(rlist)
cat(paste("\r",i, "Observations of ", nrow(smallExample),"\t",nrow(MatrixOutput),"Matrix results",sep = " "))
}
}
结果是时间、位置、测量值和解决方案的四个部分的非透视 table。
数据：
library(data.table)
smallExample <- structure(list(Gas = c(18, 17, 15, 14, 12, 11),
Solid = c(234, 239, 237, 238, 233, 239),
Timestamp = structure(c(1556085600,
1556089200,
1556092800,
1556096400,
1556100000,
1556103600),
class = c("POSIXct", "POSIXt"), tzone = "UTC"),
Head = c(1.6, 1.2, 1.1, 1.0, 0.8, 0.7),
Foot = c(11, 8, 7, 6, 5, 4),
Pillow = c(3, 2, 2, 2, 1, 1),
Mattress = c(35, 34, 31, 30, 28, 26),
Sleeping = c(59, 57, 59, 58, 59, 59),
Restless = c(111, 111, 110, 111, 112, 113)),
class = c("data.table", "data.frame"),
row.names = c(NA_integer_, -6L))
parameters <- data.table("Name" = c("Head", "Foot", "Pillow", "Mattress", "Sleeping", "Restless"),
"Space" = c(8, 78, 17, 17, 3, 8),
"Portion" = c(2, 4, 2, 2, 3, 3))
解决方案是将 apply 和 do.call 与 rbind 一起使用（thanks to fotNelton). Benchmarking with microbenchmark using apply came in 60% faster. Just left wishing I could get do.call(rbindlist,)) 。
MatrixAnswer <- as.data.table(do.call(rbind, apply(MatrixSubTest, 1, SectionMasses)))
更新
我一直在尝试 apply() 函数，但输出让我很困惑。我需要 data.frame 或 data.table。 apply(smallExample, 1, SectionDev) 给出了正确的输出，但在列表中。我读过 sapply() 可以输出 data.frame 或 matrix 但我无法将整行输入函数。
新函数
SectionDev <- function(x){
parameters <- data.table("Name" = c("Head", "Foot", "Pillow", "Mattress", "Sleeping", "Restless"),
"Space" = c(8, 78, 17, 17, 3, 8),
"Portion" = c(2, 4, 2, 2, 3, 3))
RG <- as.numeric(x[which(names(x) == "Gas")])
RS <- as.numeric(x[which(names(x) == "Solid")])
A <- data.frame(MG=c(1, 0, 1, 0),
MS=c(1, 0, 0, 1),
VG=c(0, 1, -1*RG, 0),
VS=c(0, 1, 0, -1*RS))
for (j in 1:nrow(parameters)) {
SectName <- as.vector(as.matrix(parameters[j, `Name`]))
Ac <- parameters[j, `Portion`]
Vt <- parameters[j, `Space`]
Pushing <- as.numeric(x[which(names(x) == SectName)])
Driving <- as.numeric(Pushing * 249.1)
Pressing <- Driving*Ac / 9.8
b <- c(Pressing, Vt, 0, 0)
MatrixConstants <- data.table("Timestamp" = as.vector(as.matrix(x[which(names(x) == "Timestamp")])),
"Section" = SectName,
RG,
RS)
MatrixSolution <- solve(A,b)
MatrixPtResult <- cbind(MatrixConstants,t(
as.data.frame(MatrixSolution)))
if (exists("MatrixTest")){
rlist <- list(MatrixTest, MatrixPtResult)
} else {
rlist <- list(MatrixPtResult)
}
MatrixTest <- rbindlist(rlist)
}
return(MatrixTest)
}
期望输出
> MatrixOutput[1:12,]
Timestamp Portion Gas Solid MG MS VG
1: 2019-04-24 06:00:00 Head 18 234 149.2218 -67.88299 8.290098
2: 2019-04-24 06:00:00 Foot 18 234 1427.7993 -309.39116 79.322184
3: 2019-04-24 06:00:00 Pillow 18 234 318.7908 -166.28061 17.710601
4: 2019-04-24 06:00:00 Mattress 18 234 183.2262 1596.05952 10.179233
5: 2019-04-24 06:00:00 Sleeping 18 234 -316.4209 4815.47194 -17.578940
6: 2019-04-24 06:00:00 Restless 18 234 -549.3597 9013.67602 -30.519983
7: 2019-04-24 07:00:00 Head 17 239 141.7429 -80.73885 8.337819
8: 2019-04-24 07:00:00 Foot 17 239 1365.2541 -551.86634 80.309064
9: 2019-04-24 07:00:00 Pillow 17 239 303.3448 -201.67136 17.843813
10: 2019-04-24 07:00:00 Mattress 17 239 178.7719 1549.67705 10.515996
11: 2019-04-24 07:00:00 Sleeping 17 239 -277.9378 4624.47863 -16.349283
12: 2019-04-24 07:00:00 Restless 17 239 -501.7540 8966.07028 -29.514938
VS
1: -0.2900983
2: -1.3221844
3: -0.7106009
4: 6.8207672
5: 20.5789399
6: 38.5199830
7: -0.3378195
8: -2.3090642
9: -0.8438132
10: 6.4840044
11: 19.3492830
12: 37.5149384
原创
我必须为数据中的每个时间点和位置求解相同的矩阵。 for 循环有效，但在 12,000 多个时间点和 6 个位置时速度不快。
有没有办法不使用 for 来做到这一点？它只是感觉不像我现在做的那样。
我确实尝试过将 foreach 与 doParallel 一起使用，但实际上它比用 TERR 编写的 for 慢。
有两个嵌套循环。第一个提取时间点的测量值，第二个为每个位置执行 solve。
for (i in 1:nrow(smallExample)) {
RG <- as.matrix(smallExample[i, "Gas"])
RS <- as.matrix(smallExample[i, "Solid"])
A <- data.frame(MG=c(1, 0, 1, 0),
MS=c(1, 0, 0, 1),
VG=c(0, 1, -1*RG,0),
VS=c(0, 1, 0, -1*RS))
for (j in 1:nrow(parameters)) {
SectName <- as.vector(as.matrix(parameters[j, "Name"]))
Ac <- parameters[j, "Portion"]
Vt <- parameters[j, "Space"]
Pushing <- smallExample[i, get(SectName)]
Driving <- as.matrix(Pushing*249.1)
Pressing <- Driving * as.matrix(Ac) / 9.8
b <- c(Pressing,Vt,0,0)
MatrixConstants <- data.table("Timestamp" = as.vector(as.matrix(smallExample[i, "Timestamp"])),
"Portion" = SectName,
RG,
RS)
MatrixSolution <- solve(A,b)
MatrixPtResult <- cbind(MatrixConstants,t(
as.data.frame(MatrixSolution)))
if (exists("MatrixOutput")){
rlist <- list(MatrixOutput, MatrixPtResult)
} else {
rlist <- list(MatrixPtResult)
}
MatrixOutput <- rbindlist(rlist)
cat(paste("\r",i, "Observations of ", nrow(smallExample),"\t",nrow(MatrixOutput),"Matrix results",sep = " "))
}
}
结果是时间、位置、测量值和解决方案的四个部分的非透视 table。
数据：
library(data.table)
smallExample <- structure(list(Gas = c(18, 17, 15, 14, 12, 11),
Solid = c(234, 239, 237, 238, 233, 239),
Timestamp = structure(c(1556085600,
1556089200,
1556092800,
1556096400,
1556100000,
1556103600),
class = c("POSIXct", "POSIXt"), tzone = "UTC"),
Head = c(1.6, 1.2, 1.1, 1.0, 0.8, 0.7),
Foot = c(11, 8, 7, 6, 5, 4),
Pillow = c(3, 2, 2, 2, 1, 1),
Mattress = c(35, 34, 31, 30, 28, 26),
Sleeping = c(59, 57, 59, 58, 59, 59),
Restless = c(111, 111, 110, 111, 112, 113)),
class = c("data.table", "data.frame"),
row.names = c(NA_integer_, -6L))
parameters <- data.table("Name" = c("Head", "Foot", "Pillow", "Mattress", "Sleeping", "Restless"),
"Space" = c(8, 78, 17, 17, 3, 8),
"Portion" = c(2, 4, 2, 2, 3, 3))
解决方案是将 apply 和 do.call 与 rbind 一起使用（thanks to fotNelton). Benchmarking with microbenchmark using apply came in 60% faster. Just left wishing I could get do.call(rbindlist,))
MatrixAnswer <- as.data.table(do.call(rbind, apply(MatrixSubTest, 1, SectionMasses)))