在不使用 R 中的 for 循环的情况下在每次迭代时模拟不同数量的观察
Simulating different number of observations at each iteration without using for loop in R
我想在 R 中针对不同的 df1 和 df2 模拟 F 分布。这是带输出的 MWE。
library(plyr)
set.seed(12345)
mdply(n = 4, expand.grid(df1=1:2, df2=1:3), rf, ncp=0)
df1 df2 V1 V2 V3 V4
1 1 1 0.1347567 8.500762e-06 0.72264208 4518.4539675
2 2 1 0.2530428 1.486142e+00 72.21221004 141.6067825
3 1 2 1.5186844 9.277733e-02 0.64339217 0.7640852
4 2 2 25.3820751 1.157754e-01 0.73888229 0.3737995
5 1 3 6.1314292 7.929352e-01 0.04016619 0.3938349
6 2 3 0.5724190 1.056585e+01 0.88576080 1.1905973
我还需要不同数量的模拟。我想知道如何在不使用 for 循环的情况下实现这一目标。
mdply(expand.grid(n=4:5, df1=1:2, df2=1:3), rf, ncp=0)
Error in list_to_dataframe(res, attr(.data, "split_labels"), .id, id_as_factor) :
Results do not have equal lengths
这是一种方法：
pars <- expand.grid(n = 4, df1 = 1:2, df2 = 1:3, ncp = 0)
set.seed(12345)
do.call("mapply", c(FUN = rf, as.list(pars), SIMPLIFY = TRUE))
这给出了
> do.call("mapply", c(FUN = rf, as.list(pars), SIMPLIFY = TRUE))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1.347567e-01 0.2530428 1.51868436 25.3820751 6.13142922 0.5724190
[2,] 8.500762e-06 1.4861425 0.09277733 0.1157754 0.79293521 10.5658537
[3,] 7.226421e-01 72.2122100 0.64339217 0.7388823 0.04016619 0.8857608
[4,] 4.518454e+03 141.6067825 0.76408522 0.3737995 0.39383494 1.1905973
对于变化的n，只需扩展原始参数：
pars <- expand.grid(n = 4:5, df1 = 1:2, df2 = 1:3, ncp = 0)
set.seed(12345)
do.call("mapply", c(FUN = rf, as.list(pars)))
给出：
> do.call("mapply", c(FUN = rf, as.list(pars)))
[[1]]
[1] 1.347567e-01 8.500762e-06 7.226421e-01 4.518454e+03
[[2]]
[1] 1.86524680 20.26270285 0.26845681 0.06860617 0.32628251
[[3]]
[1] 91.0530148 0.1471053 3.3674264 5.6419166
[[4]]
[1] 11.3468626 0.7450701 32.9943342 1.8116354 0.7173261
[[5]]
[1] 0.3686938 1.2020445 6.8348659 0.2035982
[[6]]
[1] 14.88249844 0.26743564 0.36008560 0.06546096 1.87464074
....
您可以在单个 n 情况下附加参数：
pars <- expand.grid(n = 4, df1 = 1:2, df2 = 1:3, ncp = 0)
set.seed(12345)
rands <- do.call("mapply", c(FUN = rf, as.list(pars), SIMPLIFY = TRUE))
cbind(pars, t(rands)) ## transpose `rands`
给出：
> cbind(pars, t(rands)) ## transpose `rands`
n df1 df2 ncp 1 2 3 4
1 4 1 1 0 0.1347567 8.500762e-06 0.72264208 4518.4539675
2 4 2 1 0 0.2530428 1.486142e+00 72.21221004 141.6067825
3 4 1 2 0 1.5186844 9.277733e-02 0.64339217 0.7640852
4 4 2 2 0 25.3820751 1.157754e-01 0.73888229 0.3737995
5 4 1 3 0 6.1314292 7.929352e-01 0.04016619 0.3938349
6 4 2 3 0 0.5724190 1.056585e+01 0.88576080 1.1905973
您将无法使用 n = 4:5 版本轻松完成此操作，因为您需要填写。鉴于您已加载 plyr，请使用它的 rbind.fill() 函数：
pars <- expand.grid(n = 4:5, df1 = 1:2, df2 = 1:3, ncp = 0)
set.seed(12345)
do.call("mapply", c(FUN = rf, as.list(pars)))
rands <- do.call("mapply", c(FUN = rf, as.list(pars)))
randsdf <- lapply(rands, function(x) as.data.frame(t(x)))
rbind.fill(randsdf)
这给出了
> rbind.fill(randsdf)
V1 V2 V3 V4 V5
1 0.3684304 186.73376725 3.039183e-01 1.406014e-01 NA
2 0.7545174 0.91995520 1.079430e+03 1.817998e+03 7.47710289
3 7.0747278 8.20808462 3.322224e-01 3.007743e+00 NA
4 1.9073559 4.13709509 8.722492e-01 2.089466e+00 0.81639181
5 0.3710252 0.85090566 4.398314e-01 2.546412e+00 NA
6 3.0373048 1.19982549 1.971773e-02 4.597047e-01 0.02247717
7 0.1659259 2.48738654 2.214283e+01 1.574352e+00 NA
8 3.6552135 4.08015469 1.339006e+01 1.199039e+00 0.61507504
9 0.1473347 0.01749103 5.285231e-02 6.472382e-01 NA
10 0.1102851 0.26849717 4.911630e-01 6.862164e-02 0.00373183
11 30.9371115 1.66420909 1.049709e+00 9.076381e-01 NA
12 0.2798656 0.39701996 5.146740e-02 6.545098e+00 25.91742143
结合cbind这一步我们得到了
cbind(pars, rbind.fill(randsdf))
> cbind(pars, rbind.fill(randsdf))
n df1 df2 ncp V1 V2 V3 V4 V5
1 4 1 1 0 0.3684304 186.73376725 3.039183e-01 1.406014e-01 NA
2 5 1 1 0 0.7545174 0.91995520 1.079430e+03 1.817998e+03 7.47710289
3 4 2 1 0 7.0747278 8.20808462 3.322224e-01 3.007743e+00 NA
4 5 2 1 0 1.9073559 4.13709509 8.722492e-01 2.089466e+00 0.81639181
5 4 1 2 0 0.3710252 0.85090566 4.398314e-01 2.546412e+00 NA
6 5 1 2 0 3.0373048 1.19982549 1.971773e-02 4.597047e-01 0.02247717
7 4 2 2 0 0.1659259 2.48738654 2.214283e+01 1.574352e+00 NA
8 5 2 2 0 3.6552135 4.08015469 1.339006e+01 1.199039e+00 0.61507504
9 4 1 3 0 0.1473347 0.01749103 5.285231e-02 6.472382e-01 NA
10 5 1 3 0 0.1102851 0.26849717 4.911630e-01 6.862164e-02 0.00373183
11 4 2 3 0 30.9371115 1.66420909 1.049709e+00 9.076381e-01 NA
12 5 2 3 0 0.2798656 0.39701996 5.146740e-02 6.545098e+00 25.91742143
mlply(expand.grid(n=4:5, df1=1:2, df2=1:3), rf, ncp=0)
给你一个列表。您可以将其与 rbind.fill 结合使用，如 Gavin 的回答。
我想在 R 中针对不同的 df1 和 df2 模拟 F 分布。这是带输出的 MWE。
library(plyr)
set.seed(12345)
mdply(n = 4, expand.grid(df1=1:2, df2=1:3), rf, ncp=0)
df1 df2 V1 V2 V3 V4
1 1 1 0.1347567 8.500762e-06 0.72264208 4518.4539675
2 2 1 0.2530428 1.486142e+00 72.21221004 141.6067825
3 1 2 1.5186844 9.277733e-02 0.64339217 0.7640852
4 2 2 25.3820751 1.157754e-01 0.73888229 0.3737995
5 1 3 6.1314292 7.929352e-01 0.04016619 0.3938349
6 2 3 0.5724190 1.056585e+01 0.88576080 1.1905973
我还需要不同数量的模拟。我想知道如何在不使用 for 循环的情况下实现这一目标。
mdply(expand.grid(n=4:5, df1=1:2, df2=1:3), rf, ncp=0)
Error in list_to_dataframe(res, attr(.data, "split_labels"), .id, id_as_factor) :
Results do not have equal lengths
这是一种方法：
pars <- expand.grid(n = 4, df1 = 1:2, df2 = 1:3, ncp = 0)
set.seed(12345)
do.call("mapply", c(FUN = rf, as.list(pars), SIMPLIFY = TRUE))
这给出了
> do.call("mapply", c(FUN = rf, as.list(pars), SIMPLIFY = TRUE))
[,1] [,2] [,3] [,4] [,5] [,6]
[1,] 1.347567e-01 0.2530428 1.51868436 25.3820751 6.13142922 0.5724190
[2,] 8.500762e-06 1.4861425 0.09277733 0.1157754 0.79293521 10.5658537
[3,] 7.226421e-01 72.2122100 0.64339217 0.7388823 0.04016619 0.8857608
[4,] 4.518454e+03 141.6067825 0.76408522 0.3737995 0.39383494 1.1905973
对于变化的n，只需扩展原始参数：
pars <- expand.grid(n = 4:5, df1 = 1:2, df2 = 1:3, ncp = 0)
set.seed(12345)
do.call("mapply", c(FUN = rf, as.list(pars)))
给出：
> do.call("mapply", c(FUN = rf, as.list(pars)))
[[1]]
[1] 1.347567e-01 8.500762e-06 7.226421e-01 4.518454e+03
[[2]]
[1] 1.86524680 20.26270285 0.26845681 0.06860617 0.32628251
[[3]]
[1] 91.0530148 0.1471053 3.3674264 5.6419166
[[4]]
[1] 11.3468626 0.7450701 32.9943342 1.8116354 0.7173261
[[5]]
[1] 0.3686938 1.2020445 6.8348659 0.2035982
[[6]]
[1] 14.88249844 0.26743564 0.36008560 0.06546096 1.87464074
....
您可以在单个 n 情况下附加参数：
pars <- expand.grid(n = 4, df1 = 1:2, df2 = 1:3, ncp = 0)
set.seed(12345)
rands <- do.call("mapply", c(FUN = rf, as.list(pars), SIMPLIFY = TRUE))
cbind(pars, t(rands)) ## transpose `rands`
给出：
> cbind(pars, t(rands)) ## transpose `rands`
n df1 df2 ncp 1 2 3 4
1 4 1 1 0 0.1347567 8.500762e-06 0.72264208 4518.4539675
2 4 2 1 0 0.2530428 1.486142e+00 72.21221004 141.6067825
3 4 1 2 0 1.5186844 9.277733e-02 0.64339217 0.7640852
4 4 2 2 0 25.3820751 1.157754e-01 0.73888229 0.3737995
5 4 1 3 0 6.1314292 7.929352e-01 0.04016619 0.3938349
6 4 2 3 0 0.5724190 1.056585e+01 0.88576080 1.1905973
您将无法使用 n = 4:5 版本轻松完成此操作，因为您需要填写。鉴于您已加载 plyr，请使用它的 rbind.fill() 函数：
pars <- expand.grid(n = 4:5, df1 = 1:2, df2 = 1:3, ncp = 0)
set.seed(12345)
do.call("mapply", c(FUN = rf, as.list(pars)))
rands <- do.call("mapply", c(FUN = rf, as.list(pars)))
randsdf <- lapply(rands, function(x) as.data.frame(t(x)))
rbind.fill(randsdf)
这给出了
> rbind.fill(randsdf)
V1 V2 V3 V4 V5
1 0.3684304 186.73376725 3.039183e-01 1.406014e-01 NA
2 0.7545174 0.91995520 1.079430e+03 1.817998e+03 7.47710289
3 7.0747278 8.20808462 3.322224e-01 3.007743e+00 NA
4 1.9073559 4.13709509 8.722492e-01 2.089466e+00 0.81639181
5 0.3710252 0.85090566 4.398314e-01 2.546412e+00 NA
6 3.0373048 1.19982549 1.971773e-02 4.597047e-01 0.02247717
7 0.1659259 2.48738654 2.214283e+01 1.574352e+00 NA
8 3.6552135 4.08015469 1.339006e+01 1.199039e+00 0.61507504
9 0.1473347 0.01749103 5.285231e-02 6.472382e-01 NA
10 0.1102851 0.26849717 4.911630e-01 6.862164e-02 0.00373183
11 30.9371115 1.66420909 1.049709e+00 9.076381e-01 NA
12 0.2798656 0.39701996 5.146740e-02 6.545098e+00 25.91742143
结合cbind这一步我们得到了
cbind(pars, rbind.fill(randsdf))
> cbind(pars, rbind.fill(randsdf))
n df1 df2 ncp V1 V2 V3 V4 V5
1 4 1 1 0 0.3684304 186.73376725 3.039183e-01 1.406014e-01 NA
2 5 1 1 0 0.7545174 0.91995520 1.079430e+03 1.817998e+03 7.47710289
3 4 2 1 0 7.0747278 8.20808462 3.322224e-01 3.007743e+00 NA
4 5 2 1 0 1.9073559 4.13709509 8.722492e-01 2.089466e+00 0.81639181
5 4 1 2 0 0.3710252 0.85090566 4.398314e-01 2.546412e+00 NA
6 5 1 2 0 3.0373048 1.19982549 1.971773e-02 4.597047e-01 0.02247717
7 4 2 2 0 0.1659259 2.48738654 2.214283e+01 1.574352e+00 NA
8 5 2 2 0 3.6552135 4.08015469 1.339006e+01 1.199039e+00 0.61507504
9 4 1 3 0 0.1473347 0.01749103 5.285231e-02 6.472382e-01 NA
10 5 1 3 0 0.1102851 0.26849717 4.911630e-01 6.862164e-02 0.00373183
11 4 2 3 0 30.9371115 1.66420909 1.049709e+00 9.076381e-01 NA
12 5 2 3 0 0.2798656 0.39701996 5.146740e-02 6.545098e+00 25.91742143
mlply(expand.grid(n=4:5, df1=1:2, df2=1:3), rf, ncp=0)
给你一个列表。您可以将其与 rbind.fill 结合使用，如 Gavin 的回答。