I have a dataset and would like to take a lot of subsets based on various columns, values, and conditional operators. I think the most desirable output is a list containing all of these subsetted data frames as separate elements in the list. I attempted to do this by building a data frame that contains the subset conditions I would like to use, building a function, then using apply to feed that data frame to the function, but that didn't work. I'm sure there's probably a better method that uses an anonymous function or something like that, but I'm not sure how I would implement that. Below is an example code that should produce 8 subsets of data.
Original dataset, where x1 and x2 are scored on items that won't be used for subsetting and RT and LS are the variables that will be a subset on:
df <- data.frame(x1 = rnorm(100),
x2 = rnorm(100),
RT = abs(rnorm(100)),
LS = sample(1:10, 100, replace = T))
Dataframe containing the conditions for subsetting. E.g., the first subset of data should be any observations with values greater than or equal to 0.5 in the RT column, the second subset should be any observations greater than or equal to 1 in the subset column, etc. There should be 8 subsets, 4 done on the RT variable and 4 done on the LS variable.
subsetConditions <- data.frame(column = rep(c("RT", "LS"), each = 4),
operator = rep(c(">=", "<="), each = 4),
value = c(0.5, 1, 1.5, 2,
9, 8, 7, 6))
And this is the ugly function I wrote to attempt to do this:
subsetFun <- function(x){
subset(df, eval(parse(text = paste(x))))
}
subsets <- apply(subsetConditions, 1, subsetFun)
Thanks for any help!
Consider Map (wrapper to mapply) without any eval + parse. Since ==, <=, >=, and other operators can be used as functions with two arguments where 4 <= 5 can be written as `<=`(4,5) or "<="(4, 5), simply pass arguments elementwise and use get to reference the function by string:
sub_data <- function(col, op, val) {
df[get(op)(df[[col]], val),]
}
sub_dfs <- with(subsetConditions, Map(sub_data, column, operator, value))
Output
str(sub_dfs)
List of 8
$ RT:'data.frame': 62 obs. of 4 variables:
..$ x1: num [1:62] -1.12 -0.745 -1.377 0.848 1.63 ...
..$ x2: num [1:62] -0.257 -2.385 0.805 -0.313 0.662 ...
..$ RT: num [1:62] 0.693 1.662 0.731 2.145 0.543 ...
..$ LS: int [1:62] 5 5 1 2 9 1 5 9 3 10 ...
$ RT:'data.frame': 36 obs. of 4 variables:
..$ x1: num [1:36] -0.745 0.848 0.908 -0.761 0.74 ...
..$ x2: num [1:36] -2.3849 -0.3131 -2.4645 -0.0784 0.8512 ...
..$ RT: num [1:36] 1.66 2.15 1.74 1.65 1.13 ...
..$ LS: int [1:36] 5 2 1 5 9 10 2 7 1 3 ...
$ RT:'data.frame': 14 obs. of 4 variables:
..$ x1: num [1:14] -0.745 0.848 0.908 -0.761 -1.063 ...
..$ x2: num [1:14] -2.3849 -0.3131 -2.4645 -0.0784 -2.9886 ...
..$ RT: num [1:14] 1.66 2.15 1.74 1.65 2.63 ...
..$ LS: int [1:14] 5 2 1 5 5 6 9 4 8 4 ...
$ RT:'data.frame': 3 obs. of 4 variables:
..$ x1: num [1:3] 0.848 -1.063 0.197
..$ x2: num [1:3] -0.313 -2.989 0.709
..$ RT: num [1:3] 2.15 2.63 2.05
..$ LS: int [1:3] 2 5 6
$ LS:'data.frame': 92 obs. of 4 variables:
..$ x1: num [1:92] -1.12 -0.745 -1.377 0.848 0.612 ...
..$ x2: num [1:92] -0.257 -2.385 0.805 -0.313 0.958 ...
..$ RT: num [1:92] 0.693 1.662 0.731 2.145 0.489 ...
..$ LS: int [1:92] 5 5 1 2 1 9 1 5 9 3 ...
$ LS:'data.frame': 78 obs. of 4 variables:
..$ x1: num [1:78] -1.12 -0.745 -1.377 0.848 0.612 ...
..$ x2: num [1:78] -0.257 -2.385 0.805 -0.313 0.958 ...
..$ RT: num [1:78] 0.693 1.662 0.731 2.145 0.489 ...
..$ LS: int [1:78] 5 5 1 2 1 1 5 3 5 2 ...
$ LS:'data.frame': 75 obs. of 4 variables:
..$ x1: num [1:75] -1.12 -0.745 -1.377 0.848 0.612 ...
..$ x2: num [1:75] -0.257 -2.385 0.805 -0.313 0.958 ...
..$ RT: num [1:75] 0.693 1.662 0.731 2.145 0.489 ...
..$ LS: int [1:75] 5 5 1 2 1 1 5 3 5 2 ...
$ LS:'data.frame': 62 obs. of 4 variables:
..$ x1: num [1:62] -1.12 -0.745 -1.377 0.848 0.612 ...
..$ x2: num [1:62] -0.257 -2.385 0.805 -0.313 0.958 ...
..$ RT: num [1:62] 0.693 1.662 0.731 2.145 0.489 ...
..$ LS: int [1:62] 5 5 1 2 1 1 5 3 5 2 ...
You were actually pretty close with your function, but just needed to make an adjustment. So, with paste for each row, you need to collapse all 3 columns so that it is only 1 string rather than 3, then it can properly evaluate the expression.
subsetFun <- function(x){
subset(df, eval(parse(text = paste(x, collapse = ""))))
}
subsets <- apply(subsetConditions, 1, subsetFun)
Output
Then, it will return the 8 subsets.
str(subsets)
List of 8
$ :'data.frame': 67 obs. of 4 variables:
..$ x1: num [1:67] -1.208 0.606 -0.17 0.728 -0.424 ...
..$ x2: num [1:67] 0.4058 -0.3041 -0.3357 0.7904 -0.0264 ...
..$ RT: num [1:67] 1.972 0.883 0.598 0.633 1.517 ...
..$ LS: int [1:67] 8 9 2 10 8 5 3 4 7 2 ...
$ :'data.frame': 35 obs. of 4 variables:
..$ x1: num [1:35] -1.2083 -0.4241 -0.0906 0.9851 -0.8236 ...
..$ x2: num [1:35] 0.4058 -0.0264 1.0054 0.0653 1.4647 ...
..$ RT: num [1:35] 1.97 1.52 1.05 1.63 1.47 ...
..$ LS: int [1:35] 8 8 5 4 7 3 1 6 8 6 ...
$ :'data.frame': 16 obs. of 4 variables:
..$ x1: num [1:16] -1.208 -0.424 0.985 0.99 0.939 ...
..$ x2: num [1:16] 0.4058 -0.0264 0.0653 0.3486 -0.7562 ...
..$ RT: num [1:16] 1.97 1.52 1.63 1.85 1.8 ...
..$ LS: int [1:16] 8 8 4 6 10 2 6 6 3 9 ...
$ :'data.frame': 7 obs. of 4 variables:
..$ x1: num [1:7] 0.963 0.423 -0.444 0.279 0.417 ...
..$ x2: num [1:7] 0.6612 0.0354 0.0555 0.1253 -0.3056 ...
..$ RT: num [1:7] 2.71 2.15 2.05 2.01 2.07 ...
..$ LS: int [1:7] 2 6 9 9 7 7 4
$ :'data.frame': 91 obs. of 4 variables:
..$ x1: num [1:91] -0.952 -1.208 0.606 -0.17 -0.048 ...
..$ x2: num [1:91] -0.645 0.406 -0.304 -0.336 -0.897 ...
..$ RT: num [1:91] 0.471 1.972 0.883 0.598 0.224 ...
..$ LS: int [1:91] 6 8 9 2 1 8 4 5 3 4 ...
$ :'data.frame': 75 obs. of 4 variables:
..$ x1: num [1:75] -0.952 -1.208 -0.17 -0.048 -0.424 ...
..$ x2: num [1:75] -0.6448 0.4058 -0.3357 -0.8968 -0.0264 ...
..$ RT: num [1:75] 0.471 1.972 0.598 0.224 1.517 ...
..$ LS: int [1:75] 6 8 2 1 8 4 5 3 4 1 ...
$ :'data.frame': 65 obs. of 4 variables:
..$ x1: num [1:65] -0.9517 -0.1698 -0.048 0.2834 -0.0906 ...
..$ x2: num [1:65] -0.645 -0.336 -0.897 -2.072 1.005 ...
..$ RT: num [1:65] 0.471 0.598 0.224 0.486 1.053 ...
..$ LS: int [1:65] 6 2 1 4 5 3 4 1 7 4 ...
$ :'data.frame': 58 obs. of 4 variables:
..$ x1: num [1:58] -0.9517 -0.1698 -0.048 0.2834 -0.0906 ...
..$ x2: num [1:58] -0.645 -0.336 -0.897 -2.072 1.005 ...
..$ RT: num [1:58] 0.471 0.598 0.224 0.486 1.053 ...
..$ LS: int [1:58] 6 2 1 4 5 3 4 1 4 2 ...
Related
This question already has answers here:
Can lists be created that name themselves based on input object names?
(4 answers)
Closed 2 years ago.
I have several lists (ListA, ListB, ListC...) with the same internal structure as the example below. I would like to combine all of them, keeping their structure, and have one list with all lists (ListAll). How can I do this?
Example:
I have:
ListA
$ data :'data.frame': 1 obs. of 2 variables:
..$ mean: num -0.128
..$ sd : num 1.11
$ simulations :'data.frame': 1000 obs. of 2 variables:
..$ mean: num [1:1000] -0.0116 -0.0156 0.0336 -0.0502 -0.0427 ...
..$ sd : num [1:1000] 1.003 1.014 0.963 1.036 1.051 ...
$ values:'data.frame': 35 obs. of 2 variables:
..$ C: num [1:35] 3.45 2.91 2.62 2.06 1.87 ...
..$ D: num [1:35] 5.42 2.89 3.34 1.68 1.43 ...
and several lists with the same structure.
I would like to get:
ListAll
$ ListA
$ data :'data.frame': 1 obs. of 2 variables:
..$ mean: num -0.128
..$ sd : num 1.11
$ simulations :'data.frame': 1000 obs. of 2 variables:
..$ mean: num [1:1000] -0.0116 -0.0156 0.0336 -0.0502 -0.0427 ...
..$ sd : num [1:1000] 1.003 1.014 0.963 1.036 1.051 ...
$ values:'data.frame': 35 obs. of 2 variables:
..$ C: num [1:35] 3.45 2.91 2.62 2.06 1.87 ...
..$ D: num [1:35] 5.42 2.89 3.34 1.68 1.43 ...
$ ListB
$ data :'data.frame': 1 obs. of 2 variables:
..$ mean: num -0.132
..$ sd : num 1.01
$ simulations :'data.frame': 1000 obs. of 2 variables:
..$ mean: num [1:1000] -0.0114 -0.0123 0.0378 -0.0102 -0.0340 ...
..$ sd : num [1:1000] 1.013 1.011 0.876 1.012 1.023 ...
$ values:'data.frame': 35 obs. of 2 variables:
..$ C: num [1:35] 4.41 1.61 1.42 1.96 2.07 ...
..$ D: num [1:35] 2.41 2.19 2.54 2.08 2.53 ...
** and names(listAll) would be:**
ListaA, ListB, ListC...
You can create a list of lists in base R.
ListAll <- list(ListA, ListB, ListC)
This question already has answers here:
Combining elements of list of lists by index
(3 answers)
in r combine a list of lists into one list
(3 answers)
Closed 4 years ago.
I have 2 list of lists in R with the same list name as follow :
str(total_delta_final[[1]])
List of 4
$ sector1_T02 :'data.frame': 24 obs. of 3 variables:
..$ DeltaF_1: num [1:24] 0.737 0.737 0.693 0.738 0.738 ...
..$ DeltaF_2: num [1:24] 0.24 0.24 0.279 0.239 0.239 ...
..$ DeltaF_3: num [1:24] 0.0233 0.0233 0.0275 0.0232 0.0232 ...
$ sector2_T03 :'data.frame': 24 obs. of 3 variables:
..$ DeltaF_1: num [1:24] 0.582 0.582 0.568 0.69 0.69 ...
..$ DeltaF_2: num [1:24] 0.377 0.377 0.39 0.282 0.282 ...
..$ DeltaF_3: num [1:24] 0.0406 0.0406 0.0426 0.0278 0.0278 ...
$ sector3_T03 :'data.frame': 24 obs. of 3 variables:
..$ DeltaF_1: num [1:24] 0.607 0.607 0.495 0.409 0.375 ...
..$ DeltaF_2: num [1:24] 0.356 0.356 0.451 0.519 0.544 ...
..$ DeltaF_3: num [1:24] 0.0373 0.0373 0.0541 0.072 0.0809 ...
$ sector12_T02:'data.frame': 24 obs. of 3 variables:
..$ DeltaF_1: num [1:24] 0.743 0.743 0.758 0.689 0.705 ...
..$ DeltaF_2: num [1:24] 0.234 0.234 0.22 0.283 0.269 ...
..$ DeltaF_3: num [1:24] 0.0226 0.0226 0.0213 0.028 0.0263 ...
> str(total_TI_final[[1]])
List of 4
$ sector1_T02 :'data.frame': 24 obs. of 3 variables:
..$ I_1: num [1:24] NA 0.0756 0.083 0.0799 0.0799 ...
..$ I_2: num [1:24] 0.122 NA 0.163 0.172 0.172 ...
..$ I_3: num [1:24] 0.212 0.211 NA 0.266 0.273 ...
$ sector2_T03 :'data.frame': 24 obs. of 3 variables:
..$ I_1: num [1:24] NA 0.0986 0.1013 0.1011 0.101 ...
..$ I_2: num [1:24] 0.15 NA 0.184 0.211 0.211 ...
..$ I_3: num [1:24] 0.249 0.249 NA 0.331 0.337 ...
$ sector3_T03 :'data.frame': 24 obs. of 3 variables:
..$ I_1: num [1:24] NA 0.119 0.115 0.113 0.105 ...
..$ I_2: num [1:24] 0.193 NA 0.2 0.193 0.177 ...
..$ I_3: num [1:24] 0.323 0.323 NA 0.277 0.256 ...
$ sector12_T02:'data.frame': 24 obs. of 3 variables:
..$ I_1: num [1:24] NA 0.0825 0.0681 0.0723 0.0706 ...
..$ I_2: num [1:24] 0.138 NA 0.146 0.145 0.144 ...
..$ I_3: num [1:24] 0.24 0.24 NA 0.22 0.226 ...
How could I merge these 2 list of lists in a way that my final output looks like total_TI_final[[1]][1] and the second list total_delta_final[[1]][1] then total_TI_final[[1]][2] and total_delta_final[[1]][2] and so on ...
We can use Map
Map(c, total_delta_final, total_TI_final)
I am new to UCI Machine Learning Repository datasets
I have tried to download the data into R, but I can not do it.
Could someone please help with this?
Note, I am using MacBook Pro.
data capture
data capture
This is the data I want to use
You need to look at the data first to understand its arrangement and whether there is any metadata like a header. Your browser should be sufficient for this. The first two lines of the ionosphere.data file are:
1,0,0.99539,-0.05889,0.85243,0.02306,0.83398,-0.37708,1,0.03760,0.85243,-0.17755,0.59755,-0.44945,0.60536,-0.38223,0.84356,-0.38542,0.58212,-0.32192,0.56971,-0.29674,0.36946,-0.47357,0.56811,-0.51171,0.41078,-0.46168,0.21266,-0.34090,0.42267,-0.54487,0.18641,-0.45300,g
1,0,1,-0.18829,0.93035,-0.36156,-0.10868,-0.93597,1,-0.04549,0.50874,-0.67743,0.34432,-0.69707,-0.51685,-0.97515,0.05499,-0.62237,0.33109,-1,-0.13151,-0.45300,-0.18056,-0.35734,-0.20332,-0.26569,-0.20468,-0.18401,-0.19040,-0.11593,-0.16626,-0.06288,-0.13738,-0.02447,b
So, no header, but it is a CSV file. Can use either read.table with sep="," or read.csv with header=FALSE. You might (incorrectly as did I) assume the column names are in the other file, but this is a machine learning task where there are no labels, so the read.* functions will assign generic names to the columns of the dataframe created.
You copy the link address with your browser to the datafile, then paste it into read.table in quotes and add the separator argument (since read.table's default separator values (whitespace) does not include commas:
ionosphere <- read.table( "https://archive.ics.uci.edu/ml/machine-learning-databases/ionosphere/ionosphere.data",
sep=",") # header=FALSE is default for read.table
> str(ionosphere)
'data.frame': 351 obs. of 35 variables:
$ V1 : int 1 1 1 1 1 1 1 0 1 1 ...
$ V2 : int 0 0 0 0 0 0 0 0 0 0 ...
$ V3 : num 0.995 1 1 1 1 ...
$ V4 : num -0.0589 -0.1883 -0.0336 -0.4516 -0.024 ...
$ V5 : num 0.852 0.93 1 1 0.941 ...
$ V6 : num 0.02306 -0.36156 0.00485 1 0.06531 ...
$ V7 : num 0.834 -0.109 1 0.712 0.921 ...
$ V8 : num -0.377 -0.936 -0.121 -1 -0.233 ...
$ V9 : num 1 1 0.89 0 0.772 ...
$ V10: num 0.0376 -0.0455 0.012 0 -0.164 ...
$ V11: num 0.852 0.509 0.731 0 0.528 ...
$ V12: num -0.1776 -0.6774 0.0535 0 -0.2028 ...
$ V13: num 0.598 0.344 0.854 0 0.564 ...
$ V14: num -0.44945 -0.69707 0.00827 0 -0.00712 ...
$ V15: num 0.605 -0.517 0.546 -1 0.344 ...
$ V16: num -0.38223 -0.97515 0.00299 0.14516 -0.27457 ...
$ V17: num 0.844 0.055 0.838 0.541 0.529 ...
$ V18: num -0.385 -0.622 -0.136 -0.393 -0.218 ...
$ V19: num 0.582 0.331 0.755 -1 0.451 ...
$ V20: num -0.3219 -1 -0.0854 -0.5447 -0.1781 ...
$ V21: num 0.5697 -0.1315 0.7089 -0.6997 0.0598 ...
$ V22: num -0.297 -0.453 -0.275 1 -0.356 ...
$ V23: num 0.3695 -0.1806 0.4339 0 0.0231 ...
$ V24: num -0.474 -0.357 -0.121 0 -0.529 ...
$ V25: num 0.5681 -0.2033 0.5753 1 0.0329 ...
$ V26: num -0.512 -0.266 -0.402 0.907 -0.652 ...
$ V27: num 0.411 -0.205 0.59 0.516 0.133 ...
$ V28: num -0.462 -0.184 -0.221 1 -0.532 ...
$ V29: num 0.2127 -0.1904 0.431 1 0.0243 ...
$ V30: num -0.341 -0.116 -0.174 -0.201 -0.622 ...
$ V31: num 0.4227 -0.1663 0.6044 0.2568 -0.0571 ...
$ V32: num -0.5449 -0.0629 -0.2418 1 -0.5957 ...
$ V33: num 0.1864 -0.1374 0.5605 -0.3238 -0.0461 ...
$ V34: num -0.453 -0.0245 -0.3824 1 -0.657 ...
$ V35: Factor w/ 2 levels "b","g": 2 1 2 1 2 1 2 1 2 1 ...
I want to add a column to each of my data frames in my list table after I do this code :
#list of my dataframes
df <- list(df1,df2,df3,df4)
#compute stats
stats <- function(d) do.call(rbind, lapply(split(d, d[,2]), function(x) data.frame(Nb= length(x$Year), Mean=mean(x$A), SD=sd(x$A) )))
#Apply to list of dataframes
table <- lapply(df, stats)
This column which I call Source for example, include the names of my dataframes along with Nb, Mean and SD variables. So the variable Source should contain df1,df1,df1... for my table[1], and so on.
Is there anyway I can add it in my code above?
Here's a different way of doing things:
First, let's start with some reproducible data:
set.seed(1)
n = 10
dat <- list(data.frame(a=rnorm(n), b=sample(1:3,n,TRUE)),
data.frame(a=rnorm(n), b=sample(1:3,n,TRUE)),
data.frame(a=rnorm(n), b=sample(1:3,n,TRUE)),
data.frame(a=rnorm(n), b=sample(1:3,n,TRUE)))
Then, you want a function that adds columns to a data.frame. The obvious candidate is within. The particular things you want to calculate are constant values for each observation within a particular category. To do that, use ave for each of the columns you want to add. Here's your new function:
stat <- function(d){
within(d, {
Nb = ave(a, b, FUN=length)
Mean = ave(a, b, FUN=mean)
SD = ave(a, b, FUN=sd)
})
}
Then just lapply it to your list of data.frames:
lapply(dat, stat)
As you can see, columns are added as appropriate:
> str(lapply(dat, stat))
List of 4
$ :'data.frame': 10 obs. of 5 variables:
..$ a : num [1:10] -0.626 0.184 -0.836 1.595 0.33 ...
..$ b : int [1:10] 3 1 2 1 1 2 1 2 3 2
..$ SD : num [1:10] 0.85 0.643 0.738 0.643 0.643 ...
..$ Mean: num [1:10] -0.0253 0.649 -0.3058 0.649 0.649 ...
..$ Nb : num [1:10] 2 4 4 4 4 4 4 4 2 4
$ :'data.frame': 10 obs. of 5 variables:
..$ a : num [1:10] -0.0449 -0.0162 0.9438 0.8212 0.5939 ...
..$ b : int [1:10] 2 3 2 1 1 1 1 2 2 2
..$ SD : num [1:10] 1.141 NA 1.141 0.136 0.136 ...
..$ Mean: num [1:10] -0.0792 -0.0162 -0.0792 0.7791 0.7791 ...
..$ Nb : num [1:10] 5 1 5 4 4 4 4 5 5 5
$ :'data.frame': 10 obs. of 5 variables:
..$ a : num [1:10] 1.3587 -0.1028 0.3877 -0.0538 -1.3771 ...
..$ b : int [1:10] 2 3 2 1 3 1 3 1 1 1
..$ SD : num [1:10] 0.687 0.668 0.687 0.635 0.668 ...
..$ Mean: num [1:10] 0.873 -0.625 0.873 0.267 -0.625 ...
..$ Nb : num [1:10] 2 3 2 5 3 5 3 5 5 5
$ :'data.frame': 10 obs. of 5 variables:
..$ a : num [1:10] -0.707 0.365 0.769 -0.112 0.881 ...
..$ b : int [1:10] 3 3 2 2 1 1 3 1 2 2
..$ SD : num [1:10] 0.593 0.593 1.111 1.111 0.297 ...
..$ Mean: num [1:10] -0.318 -0.318 0.24 0.24 0.54 ...
..$ Nb : num [1:10] 3 3 4 4 3 3 3 3 4 4
I am analyzing degradation of certain substances over time. Goal of the script is to get a list containing the variance of the observations at a certain time after application ("Dag" variable).
Output of the script below is a list containing NULL values. I think the problem lies with my assignment of the variable aux1 to the list item but the line works when I do it via command line.
There is probably a faster way of calculating this; even after 2 months I still feel overwhelmed by R.
mydata1<-as.data.frame(matrix(rnorm(600),ncol=6))
names(mydata1)=c("a","b","c","d","e","f")
substance1<-names(mydata1)
times1<-as.data.frame(rep(seq_len(10),10),ncol=1)
names(times1)<-"Dag"
times2<-unique(times1)
mydata1<-cbind(times1,mydata1)
vartijd<-function(times,mydata,substance){
varlist<<-vector("list",length(substance))
for (j in 1:length(substance))
aux<-sapply(times,function(i)var(mydata[mydata$Dag==i,substance[j]],na.rm=TRUE))
aux1<-cbind(times,aux)
varlist[[j]]<-aux1
}
vartijd(times2,mydata1,substance1)
With basic fixes to your code, this works fine.
vartijd<-function(times,mydata,substance){
varlist<-vector("list",length(substance)) # local `<-` assignment
for (j in 1:length(substance)){ # opening bracket
aux<-sapply(times,function(i)var(mydata[mydata$Dag==i,substance[j]],na.rm=TRUE))
aux1<-cbind(times,aux)
varlist[[j]]<-aux1
} # closing bracket
return(varlist) # explicit return
}
Result:
> out <- vartijd(times2,mydata1,substance1)
> str(out)
List of 6
$ :'data.frame': 10 obs. of 2 variables:
..$ Dag: int [1:10] 1 2 3 4 5 6 7 8 9 10
..$ aux: num [1:10] 0.997 0.997 0.997 0.997 0.997 ...
$ :'data.frame': 10 obs. of 2 variables:
..$ Dag: int [1:10] 1 2 3 4 5 6 7 8 9 10
..$ aux: num [1:10] 0.891 0.891 0.891 0.891 0.891 ...
$ :'data.frame': 10 obs. of 2 variables:
..$ Dag: int [1:10] 1 2 3 4 5 6 7 8 9 10
..$ aux: num [1:10] 1.08 1.08 1.08 1.08 1.08 ...
$ :'data.frame': 10 obs. of 2 variables:
..$ Dag: int [1:10] 1 2 3 4 5 6 7 8 9 10
..$ aux: num [1:10] 0.927 0.927 0.927 0.927 0.927 ...
$ :'data.frame': 10 obs. of 2 variables:
..$ Dag: int [1:10] 1 2 3 4 5 6 7 8 9 10
..$ aux: num [1:10] 0.86 0.86 0.86 0.86 0.86 ...
$ :'data.frame': 10 obs. of 2 variables:
..$ Dag: int [1:10] 1 2 3 4 5 6 7 8 9 10
..$ aux: num [1:10] 0.874 0.874 0.874 0.874 0.874 ...