I have some metabolomics data I am trying to process (validate the compounds that are actually present).
`'data.frame': 544 obs. of 48 variables:
$ X : int 1 2 3 4 5 6 7 8 9 10 ...
$ No. : int 2 32 34 95 114 141 169 234 236 278 ...
$ RT..min. : num 0.89 3.921 0.878 2.396 0.845 ...
$ Molecular.Weight : num 70 72 72 78 80 ...
$ m.z : num 103 145 114 120 113 ...
$ HMDB.ID : chr "HMDB0006804" "HMDB0031647" "HMDB0006112" "HMDB0001505" ...
$ Name : chr "Propiolic acid" "Acrylic acid" "Malondialdehyde" "Benzene" ...
$ Formula : chr "C3H2O2" "C3H4O2" "C3H4O2" "C6H6" ...
$ Monoisotopic_Mass: num 70 72 72 78 80 ...
$ Delta.ppm. : num 1.295 0.833 1.953 1.023 0.102 ...
$ X1 : num 288.3 16.7 1130.9 3791.5 33.5 ...
$ X2 : num 276.8 13.4 1069.1 3228.4 44.1 ...
$ X3 : num 398.6 19.3 794.8 2153.2 15.8 ...
$ X4 : num 247.6 100.5 1187.5 1791.4 33.4 ...
$ X5 : num 98.4 162.1 1546.4 1646.8 45.3 ...`
I tried to write a loop so that if the Delta.ppm value is larger than (m/z - molecular weight)/molecular weight, the entire row is deleted in the subsequent dataframe.
for (i in 1:nrow(rawdata)) {
ppm <- (rawdata$m.z[i] - rawdata$Molecular.Weight[i]) /
rawdata$Molecular.Weight[i]
if (ppm > rawdata$Delta.ppm[i]) {
filtered_data <- rbind(filtered_data, rawdata[i,])
}
}
Instead of giving me a new df with the validated compounds, under the 'Values' section, it generates a single number for 'ppm'.
Still very new to R, any help is super appreciated!
No need to do this row-by-row, we can remove all undesired rows in one operation:
## base R
good <- with(rawdat, (m.z - Molecular.Weight)/Molecular.Weight < Delta.ppm.)
newdat <- rawdat[good, ]
## dplyr
newdat <- filter(rawdat, (m.z - Molecular.Weight)/Molecular.Weight < Delta.ppm.)
Iteratively adding rows to a frame using rbind(old, newrow) works in practice but scales horribly, see "Growing Objects" in The R Inferno. For each row added, it makes a complete copy of all rows in old, which works but starts to slow down a lot. It is far better to produce a list of these new rows and then rbind them at one time; e.g.,
out <- list()
for (...) {
# ... newrow ...
out <- c(out, list(newrow))
}
alldat <- do.call(rbind, out)
ppm[i] <- NULL
for (i in 1:nrow(rawdata)) {
ppm[i] <- (rawdata$m.z[i] - rawdata$Molecular.Weight[i]) /
rawdata$Molecular.Weight[i]
if (ppm[i] > rawdata$Delta.ppm[i]) {
filtered_data <- rbind(filtered_data, rawdata[i,])
}
}
Related
EDIT: The problem was not within the geoMean function, but with a wrong use of aggregate(), as explained in the comments
I am trying to calculate the geometric mean of multiple measurements for several different species, which includes NAs. An example of my data looks like this:
species <- c("Ae", "Ae", "Ae", "Be", "Be")
phen <- c(2, NA, 3, 1, 2)
hveg <- c(NA, 15, 12, 60, 59)
df <- data.frame(species, phen, hveg)
When I try to calculate the geometric mean for the species Ae with the built-in function geoMean from the package EnvStats like this
library("EnvStats")
aggregate(df[, 3:3], list(df1$Sp), geoMean, na.rm=TRUE)
it works wonderful and skips the NAs to give me the geometric means per species.
Group.1 phen hveg
1 Ae 4.238536 50.555696
2 Be 1.414214 1.414214
When I do this with my large dataset, however, the function stumbles over NAs and returns NA as result even though there are e.g 10 numerical values and only one NA. This happens for example with the column SLA_mm2/mg.
My large data set looks like this:
> str(cut2trait1)
Classes ‘tbl_df’, ‘tbl’ and 'data.frame': 22 obs. of 19 variables:
$ Cut : chr "15_08" "15_08" "15_08" "15_08" ...
$ Block : num 1 1 1 1 1 1 1 1 1 1 ...
$ ID : num 451 512 431 531 591 432 551 393 511 452 ...
$ Plot : chr "1_1" "1_1" "1_1" "1_1" ...
$ Grazing : chr "n" "n" "n" "n" ...
$ Acro : chr "Leuc.vulg" "Dact.glom" "Cirs.arve" "Trif.prat" ...
$ Sp : chr "Lv" "Dg" "Ca" "Tp" ...
$ Label_neu : chr "Lv021" "Dg022" "Ca021" "Tp021" ...
$ PlantFunctionalType: chr "forb" "grass" "forb" "forb" ...
$ PlotClimate : chr "AC" "AC" "AC" "AC" ...
$ Season : chr "Aug" "Aug" "Aug" "Aug" ...
$ Year : num 2015 2015 2015 2015 2015 ...
$ Tiller : num 6 3 3 5 6 8 5 2 1 7 ...
$ Hveg : num 25 38 70 36 68 65 23 58 71 27 ...
$ Hrep : num 39 54 77 38 76 70 65 88 98 38 ...
$ Phen : num 8 8 7 8 8 7 6.5 8 8 8 ...
$ SPAD : num 40.7 42.4 48.7 43 31.3 ...
$ TDW_in_g : num 4.62 4.85 11.86 5.82 8.99 ...
$ SLA_mm2/mg : num 19.6 19.8 20.3 21.2 21.7 ...
and the result of my code
gm_cut2trait1 <- aggregate(cut2trait1[, 13:19], list(cut2trait1$Sp), geoMean, na.rm=TRUE)
is (only the first two rows):
Group.1 Tiller Hveg Hrep Phen SPAD TDW_in_g SLA_mm2/mg
1 Ae 13.521721 73.43485 106.67933 NA 28.17698 1.2602475 NA
2 Be 8.944272 43.95452 72.31182 5.477226 20.08880 0.7266361 9.309672
Here, the geometric mean of SLA for Ae is NA, even though there are 9 numeric measurements and only one NA in the column used to calculate the geometric mean.
I tried to use the geometric mean function suggested here:
Geometric Mean: is there a built-in?
But instead of NAs, this returned the value 1.000 when used with my big dataset, which doesn't solve my problem.
So my question is: What is the difference between my example df and the big dataset that throws the geoMean function off the rails?
I'm trying to apply the function RE.Johnson from the Johnson package to a whole data frame df that contains 157 observations of 16 variables and i'd like to loop trough all the dataframe instead of doing it manually.
I've tried the following code but it doesn't work.
lapply(df[1:16], function(x) RE.Johnson(x))
I know it might seem easy for you guys but I'm juste starting with R.
Thanks
EDIT
R provides me the answer Error in RE.ADT(xsl[, i]) : object 'p' not found and the data are not transformed.
And here is a summary of the data:
data.frame': 157 obs. of 16 variables:
$ X : num 786988 781045 777589 775266 786843 ...
$ Y : num 486608 488691 490089 489293 488068 ...
$ Z : num 182 128 191 80 131 ...
$ pH : num 7.93 7.69 7.49 7.66 7.92 7.08 7.24 7.19 7.44 7.37 ...
$ CE : num 0.775 3.284 3.745 4.072 0.95 ...
$ Nitrate : int 21 14 18 83 30 42 47 101 85 15 ...
$ NP : num 19.6 43.6 31.7 18.6 31.7 ...
$ Cl : num 1.9 21.3 2.56 21.5 3.2 ...
$ HCO3 : num 6.65 4.85 4.4 7.72 4.1 ...
$ CO3 : num 0 0 0 0 0.0736 ...
$ Ca : num 4.12 7.52 3.48 7.58 4.8 10 4.4 4.6 4.2 7.4 ...
$ Mg : num 3.94 8.92 2.34 7.1 2.5 ...
$ K : num 0.1442 0.0759 0.0709 0.3691 0.07 ...
$ Na : num 2.41 34.55 2.51 44.01 2.1 ...
$ SO4 : num 1.45 23.6 1.2 26.66 2 ...
$ Residu_sec: num 0.496 2.102 2.397 2.606 0.608 ...
Not a complete solution, just some information for others.
I tried the Johnson::RE.Johnson manually on the columns in the iris data frame. It seems to be work fine for Sepal.Length and Petal.Length only:
lapply(iris[c(1,3)], Johnson::RE.Johnson)
... and it returns the error you mentioned for Sepal.Width and Petal.Width.
lapply(iris[c(2,4)], Johnson::RE.Johnson)
Error in RE.ADT(xsl[, i]) : object 'p' not found
This seems odd because all of those columns have a data type of num. The iris data frame doesn't appear to have any missing values or extra character values hidden anywhere, so I'm not sure why the calculation is working for those columns but not others.
Without understanding too much about what the Johnson::RE.Johnson is doing to the data, it looks like it is unable to calculate a value for p and is unable to complete the iteration for those columns.
From exploring the source code, the function appears to break down at this point:
if (xsb.valida[1, i] == 0)
xsb.adtest[1, i] <- (Johnson::RE.ADT(xsb[, i])$p) # succeeds
if (xsl.valida[1, i] == 0)
xsl.adtest[1, i] <- (Johnson::RE.ADT(xsl[, i])$p) # fails
if (xsu.valida[1, i] == 0)
xsu.adtest[1, i] <- (Johnson::RE.ADT(xsu[, i])$p) # fails
The function attempts to run Johnson::RE.ADT on xsl, which at this point is a vector of just 0's. The RE.ADT returns the same error with the p value not being found.
The problem is when the function try to perform the Anderson-Darling test to a vector of equals values. If you do this, you will get the error:
require(Johnson)
x = rep(1,n=100)
RE.ADT(x)
So, to solve this problem you could check it in the IF session inside the function RE.Johnson:
if (xsb.valida[1, i] == 0 & any(xsb[, i]!=xsb[1, i])){
xsb.adtest[1, i] <- (RE.ADT(xsb[, i])$p)
}else{
xsb.adtest[1, i] <- 0
}
if (xsl.valida[1, i] == 0 & any(xsl[, i]!=xsl[1, i])) {
xsl.adtest[1, i] <- (RE.ADT(xsl[, i])$p)
}else{
xsl.adtest[1, i] <- 0
}
if (xsu.valida[1, i] == 0 & any(xsu[, i]!=xsu[1, i])) {
xsu.adtest[1, i] <- (RE.ADT(xsu[, i])$p)
}else{
xsu.adtest[1, i] <- 0
}
Hi I'm working on SparkR. I'm try to calculate a RelativeFrequency of my Data.
SmsInt<-fread("smsCallInt.txt")
setnames(SmsInt,c("V1","V2","V3","V4","V5","V6","V7","V8"),
c("SquareID","TimeInterval","CountryCode","SmsIn","SmsOut","CallIn","CallOut","Internet"))
#Also create a dataFrame from it.
SmsInt$TimeInterval<-as.numeric(SmsInt$TimeInterval)
SmsInt.df<-createDataFrame(sqlContext,SmsInt[1:500,])
str(SmsInt)
Classes ‘data.table’ and 'data.frame': 2459324 obs. of 8 variables:
$ SquareID : int 10000 10000 10000 10000 10000 10000 10000 10000 10000 10000 ...
$ TimeInterval: num 1.38e+12 1.38e+12 1.38e+12 1.38e+12 1.38e+12 ...
$ CountryCode : int 0 39 49 0 39 0 39 0 39 49 ...
$ SmsIn : num 0.109 1.001 NA 0.193 0.648 ...
$ SmsOut : num NA 1.26 NA NA 1.06 ...
$ CallIn : num NA 0.0876 NA NA 0.1751 ...
$ CallOut : num 0.0219 0.2196 NA NA 0.1532 ...
$ Internet : num NA 10.1685 0.0219 NA 11.8671 ...
- attr(*, ".internal.selfref")=<externalptr>
What I want to do is create a RelativeFrequency from SmsInt$CountryCode.
When I type Country<-table(SmsInt$CountryCode)
I got this Error:
Errore: class(objId) == "jobj" is not TRUE
What can I do?There is a way to calculate it manually or with some package?
I created an algorithm but i have some trouble .
Country5<-SmsInt$CountryCode[1:90]
UniqueCountry<-unique(Country5)
VectorLen<-c()
Parsed<-c()
Freq<-c()
for(i in 1:length(UniqueCountry)){
CountryCode.i<-UniqueCountry[i]
if(CountryCode.i %in% Parsed){
Vector<-0
VectorLen[i]<-0
Freq[i]<-0
}
else{
Vector<-grep(CountryCode.i,Country5)
Parsed[i]<-CountryCode.i
VectorLen[i]<-length(Vector)
Freq[i]<-VectorLen[i]/90
Vector<-0
}
}
Vector
VectorLen #92 it needs to be 90
Freq
sum(Freq) #1.022222 needs to be 1
With 80 all works.
Ok I did it. The error was the grep function so when I was looking for number 1 it was found on number 10 once, for example.
I post the solution here.
RelativeFrequency<-function(DataSet){
UniqueCountry<-unique(DataSet)
VectorLen<-c()
Parsed<-c()
Freq<-c()
for(i in 1:length(UniqueCountry)){
CountryCode.i<-UniqueCountry[i]
if(CountryCode.i %in% Parsed){
Vector<-0
VectorLen[i]<-0
Freq[i]<-0
}
else{
Vector<-which(DataSet %in% CountryCode.i)
Parsed[i]<-CountryCode.i
VectorLen[i]<-length(Vector)
Freq[i]<-VectorLen[i]/length(DataSet)
}
}
print("Vector of RelativeFrequency")
print(Freq)
print("Frequency Sum (Needs to be 1)")
print(sum(Freq))
print("Parsed element ")
print(Parsed)
barplot(Freq,names=Parsed,space = 0.7,axisnames = TRUE,las=2)
}
I am trying to read the table from the following URL:
url <- 'http://faculty.chicagobooth.edu/ruey.tsay/teaching/introTS/m-ge3dx-4011.txt'
da <- read.table(url, header = TRUE, fill=FALSE, strip.white=TRUE)
I can look at the data using head:
> head(da)
date ge vw ew sp
1 19400131 -0.061920 -0.024020 -0.019978 -0.035228
2 19400229 -0.009901 0.013664 0.029733 0.006639
3 19400330 0.049333 0.018939 0.026168 0.009893
4 19400430 -0.041667 0.001196 0.013115 -0.004898
5 19400531 -0.197324 -0.220314 -0.269754 -0.239541
6 19400629 0.061667 0.066664 0.066550 0.076591
This works fine for the first 4 columns, for example, I can look at the column ew
> head(da$ew)
[1] -0.019978 0.029733 0.026168 0.013115 -0.269754 0.066550
but when I try to access the last one, I get some extra output which is not in the txt file.
> head(da$sp)
[1] -0.035228 0.006639 0.009893 -0.004898 -0.239541 0.076591
859 Levels: -0.000060 -0.000143 -0.000180 -0.000320 -0.000659 -0.000815 ... 0.163047
How do I get rid of the extra output? Thanks!
This is representation of a factor.
> str(da)
'data.frame': 861 obs. of 5 variables:
$ date: int 19400131 19400229 19400330 19400430 19400531 19400629 19400731 19400831 19400930 19401031 ...
$ ge : num -0.0619 -0.0099 0.0493 -0.0417 -0.1973 ...
$ vw : num -0.024 0.0137 0.0189 0.0012 -0.2203 ...
$ ew : num -0.02 0.0297 0.0262 0.0131 -0.2698 ...
$ sp : Factor w/ 859 levels "-0.000060","-0.000143",..: 226 411 445 42 353 828 613 585 441 684 ...
Row 58 has a dot instead of a number. This is sufficient information for R to handle the variable as a factor. Once you change the dot to NA or fix the error, you will be able to read in the data fine.
Another option would be to change the point to something meaningful after the data has been read in, and coercing to numeric afterwards. The following statement will coerce . to NA.
da$sp <- as.numeric(as.character(da$sp))
> str(da)
'data.frame': 861 obs. of 5 variables:
$ date: int 19400131 19400229 19400330 19400430 19400531 19400629 19400731 19400831 19400930 19401031 ...
$ ge : num -0.0619 -0.0099 0.0493 -0.0417 -0.1973 ...
$ vw : num -0.024 0.0137 0.0189 0.0012 -0.2203 ...
$ ew : num -0.02 0.0297 0.0262 0.0131 -0.2698 ...
$ sp : num -0.03523 0.00664 0.00989 -0.0049 -0.23954 ...
I have the following data frame and I want to break it up into 10 different data frames. I want to break the initial 100 row data frame into 10 data frames of 10 rows. I could do the following and get the desired results.
df = data.frame(one=c(rnorm(100)), two=c(rnorm(100)), three=c(rnorm(100)))
df1 = df[1:10,]
df2 = df[11:20,]
df3 = df[21:30,]
df4 = df[31:40,]
df5 = df[41:50,]
...
Of course, this isn't an elegant way to perform this task when the initial data frames are larger or if there aren't an easy number of segments that it can be broken down into.
So given the above, let's say we have the following data frame.
df = data.frame(one=c(rnorm(1123)), two=c(rnorm(1123)), three=c(rnorm(1123)))
Now I want to split it into new data frames comprised of 200 rows, and the final data frame with the remaining rows. What would be a more elegant (aka 'quick') way to perform this task.
> str(split(df, (as.numeric(rownames(df))-1) %/% 200))
List of 6
$ 0:'data.frame': 200 obs. of 3 variables:
..$ one : num [1:200] -1.592 1.664 -1.231 0.269 0.912 ...
..$ two : num [1:200] 0.639 -0.525 0.642 1.347 1.142 ...
..$ three: num [1:200] -0.45 -0.877 0.588 1.188 -1.977 ...
$ 1:'data.frame': 200 obs. of 3 variables:
..$ one : num [1:200] -0.0017 1.9534 0.0155 -0.7732 -1.1752 ...
..$ two : num [1:200] -0.422 0.869 0.45 -0.111 0.073 ...
..$ three: num [1:200] -0.2809 1.31908 0.26695 0.00594 -0.25583 ...
$ 2:'data.frame': 200 obs. of 3 variables:
..$ one : num [1:200] -1.578 0.433 0.277 1.297 0.838 ...
..$ two : num [1:200] 0.913 0.378 0.35 -0.241 0.783 ...
..$ three: num [1:200] -0.8402 -0.2708 -0.0124 -0.4537 0.4651 ...
$ 3:'data.frame': 200 obs. of 3 variables:
..$ one : num [1:200] 1.432 1.657 -0.72 -1.691 0.596 ...
..$ two : num [1:200] 0.243 -0.159 -2.163 -1.183 0.632 ...
..$ three: num [1:200] 0.359 0.476 1.485 0.39 -1.412 ...
$ 4:'data.frame': 200 obs. of 3 variables:
..$ one : num [1:200] -1.43 -0.345 -1.206 -0.925 -0.551 ...
..$ two : num [1:200] -1.343 1.322 0.208 0.444 -0.861 ...
..$ three: num [1:200] 0.00807 -0.20209 -0.56865 1.06983 -0.29673 ...
$ 5:'data.frame': 123 obs. of 3 variables:
..$ one : num [1:123] -1.269 1.555 -0.19 1.434 -0.889 ...
..$ two : num [1:123] 0.558 0.0445 -0.0639 -1.934 -0.8152 ...
..$ three: num [1:123] -0.0821 0.6745 0.6095 1.387 -0.382 ...
If some code might have changed the rownames it would be safer to use:
split(df, (seq(nrow(df))-1) %/% 200)
require(ff)
df <- data.frame(one=c(rnorm(1123)), two=c(rnorm(1123)), three=c(rnorm(1123)))
for(i in chunk(from = 1, to = nrow(df), by = 200)){
print(df[min(i):max(i), ])
}
If you can generate a vector that defines the groups, you can split anything:
f <- rep(seq_len(ceiling(1123 / 200)),each = 200,length.out = 1123)
> df1 <- split(df,f = f)
> lapply(df1,dim)
$`1`
[1] 200 3
$`2`
[1] 200 3
$`3`
[1] 200 3
$`4`
[1] 200 3
$`5`
[1] 200 3
$`6`
[1] 123 3
Chops df into 1 million row groups and pushes and appends a million at a time to df in SQL
batchsize = 1000000 # vary to your liking
# cycles through data by batchsize
for (i in 1:ceiling(nrow(df)/batchsize))
{
print(i) # just to show the progress
# below shows how to cycle through data
batch <- df[(((i-1)*batchsize)+1(batchsize*i),,drop=FALSE] # drop = FALSE keeps it from being converted to a vector
# if below not done then the last batch has Nulls above the number of rows of actual data
batch <- batch[!is.na(batch$ID),] # ID is a variable I presume is in every row
#in this case the table already existed, if new table overwrite = TRUE
(dbWriteTable(con, "df", batch, append = TRUE,row.names = FALSE))
}
Something like this...?
b <- seq(10, 100, 10)
lapply(seq_along(b), function(i) df[(b-9)[i]:b[i], ])
[[1]]
one two three
1 -2.4157992 -0.6232517 1.0531358
2 0.6769020 0.3908089 -1.9543895
3 0.9804026 -2.5167334 0.7120919
4 -1.2200089 0.5108479 0.5599177
5 0.4448290 -1.2885275 -0.7665413
6 0.8431848 -0.9359947 0.1068137
7 -1.8168134 -0.2418887 1.1176077
8 1.4475904 -0.8010347 2.3716663
9 0.7264027 -0.3573623 -1.1956806
10 0.2736119 -1.5553148 0.2691115
[[2]]
one two three
11 -0.3273536 -1.92475496 -0.08031696
12 1.5558892 -1.20158371 0.09104958
13 1.9202047 -0.13418754 0.32571632
14 -0.0515136 -2.15669216 0.23099397
15 0.1909732 -0.30802742 -1.28651457
16 0.8545580 -0.18238266 1.57093844
17 0.4903039 0.02895376 -0.47678196
18 0.5125400 0.97052082 -0.70541908
19 -1.9324370 0.22093545 -0.34436105
20 -0.5763433 0.10442551 -2.05597985
[[3]]
one two three
21 0.7168771 -1.22902943 -0.18728871
22 1.2785641 0.14686576 -1.74738091
23 -1.1856173 0.43829361 0.41269975
24 0.0220843 1.57428924 -0.80163986
25 -1.0012255 0.05520813 0.50871603
26 -0.1842323 -1.61195239 0.04843504
27 0.2328831 -0.38432225 0.95650710
28 0.8821687 -1.32456215 -1.33367967
29 -0.8902177 0.86414661 -1.39629358
30 -0.6586293 -2.27325919 0.27367902
[[4]]
one two three
31 1.3810437 -1.0178835 0.07779591
32 0.6102753 0.3538498 1.92316801
33 -1.5034439 0.7926925 2.21706284
34 0.8251638 0.3992922 0.56781321
35 -1.0832114 0.9878058 -0.16820827
36 -0.4132375 -0.9214491 1.06681472
37 -0.6787631 1.3497766 2.18327887
38 -3.0082585 -1.3047024 -0.04913214
39 -0.3433300 1.1008951 -2.02065141
40 0.6009334 1.2334421 0.15623298
[[5]]
one two three
41 -1.8608051 -0.08589437 0.02370983
42 -0.1829953 0.91139017 -0.01356590
43 1.1146731 0.42384993 -0.68717391
44 1.9039900 -1.70218225 0.06100297
45 -0.4851939 1.38712015 -1.30613414
46 -0.4661664 0.23504099 -0.29335162
47 0.5807227 -0.87821946 -0.14816121
48 -2.0168910 -0.47657382 0.90503226
49 2.5056404 0.27574224 0.10326333
50 0.2238735 0.34441325 -0.17186115
[[6]]
one two three
51 1.51613140 -2.5630782 -0.6720399
52 0.03859537 -2.6688365 0.3395574
53 -0.08695292 -0.5114117 -0.1378789
54 -0.51878363 -0.5401962 0.3946324
55 -2.20482710 0.1716744 0.1786546
56 -0.28133749 -0.4497112 0.5936497
57 -2.38269088 -0.4625695 1.0048914
58 0.37865952 0.5055141 0.3337986
59 0.09329172 0.1560469 0.2835735
60 -1.10818863 -0.2618910 0.3650042
[[7]]
one two three
61 -1.2507208 -1.5050083 -0.63871084
62 0.1379394 0.7996674 -1.80196762
63 0.1582008 -0.3208973 0.40863693
64 -0.6224605 0.1416938 -0.47174711
65 1.1556149 -1.4083576 -1.12619693
66 -0.6956604 0.7994991 1.16073748
67 0.6576676 1.4391007 0.04134445
68 1.4610598 -1.0066840 -1.82981058
69 1.1951788 -0.4005535 1.57256648
70 -0.1994519 0.2711574 -1.04364396
[[8]]
one two three
71 1.23897065 0.4473611 -0.35452535
72 0.89015916 2.3747385 0.87840852
73 -1.17339703 0.7433220 0.40232381
74 -0.24568490 -0.4776862 1.24082294
75 -0.47187443 -0.3271824 0.38542703
76 -2.20899136 -1.1131712 -0.33663075
77 -0.05968035 -0.6023045 -0.23747388
78 1.19687199 -1.3390960 -1.37884241
79 -1.29310506 0.3554548 -0.05936756
80 -0.17470891 1.6198307 0.69170207
[[9]]
one two three
81 -1.06792315 0.04801998 0.08166394
82 0.84152560 -0.45793907 0.27867619
83 0.07619456 -1.21633682 -2.51290495
84 0.55895466 -1.01844178 -0.41887672
85 0.33825508 -1.15061381 0.66206732
86 -0.36041720 0.32808609 -1.83390913
87 -0.31595401 -0.87081019 0.45369366
88 0.92331087 1.22055348 -1.91048757
89 1.30491142 1.22582353 -1.32244004
90 -0.32906839 1.76467263 1.84479228
[[10]]
one two three
91 2.80656707 -0.9708417 0.25467304
92 0.35770119 -0.6132523 -1.11467041
93 0.09598908 -0.5710063 -0.96412216
94 -1.08728715 0.3019572 -0.04422049
95 0.14317455 0.1452287 -0.46133199
96 -1.00218917 -0.1360570 0.88864256
97 -0.25316855 0.6341925 -1.37571664
98 0.36375921 1.2244921 0.12718650
99 0.13345555 0.5330221 -0.29444683
100 2.28548261 -2.0413222 -0.53209956