我有这个数据集
DF:
Type Value Average SD Q.
S AA+ 3 1 30
S AA 2 1 30
S A 1 1 30
S B -2 1 30
S BB - -3 1 30
F AA+ 2 0.75 30
F AA 1 0.75 30
F A 0 0.75 30
F B -1 0.75 30
F BB - -2 0.75 30
我想像这样通过类型和值在循环中进行概率分布
rnorm(n, mean = 0, sd = 1)
rnorm(DF$Q., DF$Average, DF$SD)
我有唯一值列表
type_list <- unique(DF$Type)
Value_list <- unique(DF$Value)
And now I am trying to loop it
probability_distributions <- list()
for (i in 1:length(type_list)) {
for (j in 1:length(Value_list) {
pd <- rnorm(DF$Q.[i,j], DF$Average[i,j], DF$SD[i,j])
probability_distributions <- c(pd, list(probability_distributions ))
}
}
我想得到这样的东西
List:
S AA+
1 -3.837712
2 -3.690301
3 -3.837331
4 -2.302341 ....
还有另外10个列表
答案 0 :(得分：1)
您可以这样做：
setNames(lapply(seq(nrow(DF)), function(i) {
rnorm(DF$Q.[i], DF$Average[i], DF$SD[i])
}), paste(DF$Type, DF$Value))
#> $`S AA+`
#> [1] 1.9993416 2.2418212 2.3885642 1.6832777 0.9925850 3.7680021 3.4057613 2.1682841
#> [9] 3.9580758 4.8107193 0.4499855 4.1502551 4.2278937 3.1521032 2.4769162 1.8984066
#> [17] 2.1202697 2.3632997 0.8940686 2.3537416 3.9867934 2.2450999 4.5652049 2.9499507
#> [25] 3.1110287 3.4754710 2.0609961 2.2259544 2.5764415 4.7728795
#>
#> $`S AA`
#> [1] 2.42655066 1.90552983 0.88340457 1.91256485 2.39252609 1.63937783
#> [7] 1.11201564 1.05358345 1.78008844 2.34222012 2.96992413 0.66454686
#> [13] 1.19912052 -0.04679634 2.11005622 2.71610037 2.25412060 3.22876219
#> [19] 2.58340401 1.41287523 3.44666536 2.44339404 2.57794689 1.07816504
#> [25] 1.75067329 0.77810135 1.92746035 3.36490125 1.54246898 2.06520022
#>
#> $`S A`
#> [1] 1.783980336 -0.587708073 -0.369364313 0.072480197 1.300841560 1.080468946
#> [7] 2.246746831 -0.234449100 0.560930706 -1.101593953 2.618756171 2.084328491
#> [13] 0.359199093 0.747180174 0.865170727 2.795355992 1.038396717 1.412998289
#> [19] 1.699572123 1.689790945 -0.671059465 1.740048308 0.075875101 0.968311427
#> [25] -0.927792982 1.214303030 -0.005038866 -1.178953492 -0.672549131 -0.420722136
#>
#> $`S B`
#> [1] -1.44946836 -1.68956682 -1.31492609 -2.53191049 -1.81821454 -1.58840382
#> [7] -2.08505905 -3.10670620 -0.87086640 -0.33198438 0.01910293 -1.10745196
#> [13] -2.18720468 -2.12769742 -2.30533014 -2.26286684 -2.05146864 -3.97266336
#> [19] -1.98877175 -1.76465514 -2.95036985 -3.75714798 -2.35996065 -5.12158956
#> [25] -0.32745289 -1.30945018 -2.97667032 -1.98486582 -2.16545418 -3.66021337
#>
#> $`S BB-`
#> [1] -3.878434 -2.034184 -4.155821 -3.751396 -3.745084 -1.772948 -3.190858 -2.445689
#> [9] -2.228567 -3.380067 -4.128551 -2.829898 -3.358542 -1.557062 -3.519947 -3.310642
#> [17] -2.317263 -3.663578 -3.017951 -2.503409 -3.404275 -4.211649 -2.687256 -3.279862
#> [25] -5.019855 -2.730421 -2.868201 -4.678771 -3.525880 -3.175125
#>
#> $`F AA+`
#> [1] 1.5825269 2.4138863 0.8179614 1.2842804 1.6626024 3.0829298 1.8835594 1.2337108
#> [9] 3.1538523 1.8266180 2.4139429 2.8413455 2.3071590 2.9751961 1.4068090 3.1989646
#> [17] 0.6328248 1.2684777 1.5601545 2.1748322 1.6449135 2.4373332 2.3150221 2.5091457
#> [25] 3.1118458 0.9310370 2.7274812 1.8009007 1.3976708 0.6672244
#>
#> $`F AA`
#> [1] 0.363222331 0.775391336 -0.455183359 0.729975409 1.382579640 0.026522186
#> [7] 0.996364448 -0.008639176 0.961236861 1.671137345 -0.634911705 2.729812324
#> [13] 0.124187233 1.705322289 2.559326197 0.292131983 0.493409391 1.766237746
#> [19] 0.386872427 0.282159449 2.185839460 2.324832101 0.829723631 2.710832646
#> [25] 2.427810412 0.948533848 0.389605646 0.495058514 2.051522848 1.405012456
#>
#> $`F A`
#> [1] 0.481705627 -0.009539824 0.137159665 -0.366385935 0.851427552 -0.244538267
#> [7] 1.493896900 0.440079671 0.741249918 1.106717951 -0.035215035 -1.325648324
#> [13] -0.457225479 0.444942684 0.902540415 0.156192620 0.629354519 0.707281075
#> [19] 0.771069839 0.560672883 -0.143570299 0.768517623 -0.378166481 0.261411645
#> [25] -0.382030406 0.358368343 0.375739047 -0.079185388 0.020481554 0.325286853
#>
#> $`F B`
#> [1] -1.84829836 -0.23448482 -0.79804428 -0.58858852 -1.12706587 -2.40883019
#> [7] -0.43876960 -1.19507511 -0.53630451 0.53595272 -1.86671863 0.01470606
#> [13] -1.27564149 -1.04373285 -1.39916357 -1.37387536 -1.86260468 -0.90531931
#> [19] -0.64535208 -1.13989391 -2.21446484 -1.30206928 -0.69039082 -1.54053955
#> [25] -1.44254892 -1.87721996 -0.55640752 -1.50147921 1.37595324 -0.24022044
#>
#> $`F BB-`
#> [1] -2.1244905 -1.9981876 -2.5379765 -1.7889965 -1.9945386 -3.2972752 -3.3826052
#> [8] -1.5490977 -3.7384532 -2.2852341 -0.7586263 -3.2471413 -1.7789528 -0.8225739
#> [15] -1.6160398 -1.0355732 -2.7256361 -1.5257080 -1.8626910 -0.9874129 -0.1345042
#> [22] -1.9140037 -1.1781947 -2.3020324 -2.2693023 -2.4145912 -3.0160062 -1.9449959
#> [29] -2.0292414 -1.4585976
答案 1 :(得分：1)
请始终提供我们可以复制粘贴为reproducible example的内容。在这里，我为您创建了一些东西。
library(dplyr)
library(data.table)
Type <- 1:3
Value <- 11:13
SD <- seq(0.1,0.3, by=0.1)
mean <- c(1,11,21)
df <- data.frame(Type, Value, SD, mean)
> df
Type Value SD mean
1 1 11 0.1 1
2 2 12 0.2 11
3 3 13 0.3 21
您想为 all 所有可能的组合创建一个rnorm系列，以我的最小示例为9个值。 expand.grid会很繁重。
> expand.grid(unique(types), unique(values)) %>% nrow
[1] 9
首先，您需要使用正确的方法和SD来构建唯一的data.frame。然后，您只需致电lapply即可获得包含您预期结果的list。
# get unique combinations
unique.df <- expand.grid(types, values)
colnames(unique.df) <- c("Type", "Value")
# I suppose the mean and SD are determined by the type
unique.df$mean <- df$mean[match(unique.df$Type, df$Type)]
unique.df$SD <- df$SD[match(unique.df$Type, df$Type)]
#convert to list by keeping names
unique.list <- setNames(split(unique.df, seq(nrow(unique.df))), rownames(unique.df))
probability_distributions <- lapply(unique.list, function(x) rnorm(x[["Value"]], x[["mean"]], x[["SD"]]))
请注意，首先在这里创建一个数据框，其中包含您的功能所需的所有正确信息。
> head(unique.df)
Type Value probability_distributions mean SD
1 1 11 11.046879 1 0.1
2 2 11 9.803352 11 0.2
3 3 11 11.799637 21 0.3
然后我将unique.df data.frame转换为包含所有行的列表，通过保留名称，对此进行解释here
最后，没有任何显式循环，我可以在此列表上调用lapply。最困难的是要知道如何在构建结果之前安排数据，以便您避免循环并使用相对优雅的代码，例如lapply。
完整代码：
library(dplyr)
Type <- 1:3
Value <- 11:13
SD <- seq(0.1,0.3, by=0.1)
mean <- c(1,11,21)
df <- data.frame(Type, Value, SD, mean)
# get unique combinations
unique.df <- expand.grid(types, values)
colnames(unique.df) <- c("Type", "Value")
# I suppose the mean and SD are determined by the type
unique.df$mean <- df$mean[match(unique.df$Type, df$Type)]
unique.df$SD <- df$SD[match(unique.df$Type, df$Type)]
#convert to list by keeping names
unique.list <- setNames(split(unique.df, seq(nrow(unique.df))), rownames(unique.df))
probability_distributions <- lapply(unique.list, function(x) rnorm(x[["Value"]], x[["mean"]], x[["SD"]]))