I have a data frame m with:
>m
id w y z
1 2 5 8
2 18 5 98
3 1 25 5
4 52 25 8
5 5 5 4
6 3 3 5
Below is a general function for normally transforming a variable that I need to apply to columns w,y,z.
y<-qnorm((rank(x,na.last="keep")-0.5)/sum(!is.na(x))
For example, if I wanted to run this function on "column w" to get the output column appended to dataframe "m" then:
m$w_n<-qnorm((rank(m$w,na.last="keep")-0.5)/sum(!is.na(m$w))
Can someone help me automate this to run on multiple columns in data frame m?
Ideally, I would want an output data frame with the following columns:
id w y z w_n y_n z_n
Note this is a sample data frame, the one I have is much larger and I have more letter columns to run this function on other than w, y,z.
Thanks!
Probably a way to do it in a single step, but what about:
df <- data.frame(id = 1:6, w = sample(50, 6), z = sample(50, 6) )
df
id w z
1 1 39 40
2 2 20 26
3 3 43 11
4 4 4 37
5 5 36 24
6 6 27 14
transCols <- function(x) qnorm((rank(x,na.last="keep")-0.5)/sum(!is.na(x)))
tmpdf <- lapply(df[, -1], transCols)
names(tmpdf) <- paste0(names(tmpdf), "_n")
df_final <- cbind(df, tmpdf)
df_final
df_final
id w z w_n z_n
1 1 39 40 -0.2104284 -1.3829941
2 2 20 26 1.3829941 1.3829941
3 3 43 11 0.2104284 0.6744898
4 4 4 37 -1.3829941 0.2104284
5 5 36 24 0.6744898 -0.6744898
6 6 27 14 -0.6744898 -0.2104284
Related
I have a list of lists that looks like this:
list(list("A[1]" = data.frame(W = 1:5),
"A[2]" = data.frame(X = 6:10),
B = data.frame(Y = 11:15),
C = data.frame(Z = 16:20)),
list("A[1]" = data.frame(W = 21:25),
"A[2]" = data.frame(X = 26:30),
B = data.frame(Y = 31:35),
C = data.frame(Z = 36:40)),
list("A[1]" = data.frame(W = 41:45),
"A[2]" = data.frame(X = 46:50),
B = data.frame(Y = 51:55),
C = data.frame(Z = 56:60))) -> dflist
I need my output to also be a list of list with length 3 so that each sublist retains elements whose names start with A[ while dropping other elements.
Based on some previous questions, I am trying to use this:
dflist %>%
map(keep, names(.) %in% "A[")
but that gives the following error:
Error in probe(.x, .p, ...) : length(.p) == length(.x) is not TRUE
Trying to select a single element, for example just A[1] like this:
dflist %>%
map(keep, names(.) %in% "A[1]")
also doesn't work. How can I achieve the desired output?
I think you want:
purrr::map(dflist, ~.[stringr::str_starts(names(.), "A\\[")])
What this does is:
For each sublist (purrr::map)
Select all elements of that sublist (.[], where . is the sublist)
Whose names start with A[ (stringr::str_starts(names(.), "A\\["))
You got the top level map correct, since you want to modify the sublists. However, map(keep, names(.) %in% "A[") has some issues:
names(.) %in% "A[" should be a function or a formula (starting with ~
purrr::keep applies the filtering function to each element of the sublist, namely to the data frames directly. It never "sees" the names of each data frame. Actually I don't think you can use keep for this problem at all
Anyway this produces:
[[1]]
[[1]]$`A[1]`
W
1 1
2 2
3 3
4 4
5 5
[[1]]$`A[2]`
X
1 6
2 7
3 8
4 9
5 10
[[2]]
[[2]]$`A[1]`
W
1 21
2 22
3 23
4 24
5 25
[[2]]$`A[2]`
X
1 26
2 27
3 28
4 29
5 30
[[3]]
[[3]]$`A[1]`
W
1 41
2 42
3 43
4 44
5 45
[[3]]$`A[2]`
X
1 46
2 47
3 48
4 49
5 50
If we want to use keep, use
library(dplyr)
library(purrr)
library(stringr)
map(dflist, ~ keep(.x, str_detect(names(.x), fixed("A["))))
Here a base R solution:
lapply(dflist, function(x) x[grep("A\\[",names(x))] )
[[1]]
[[1]]$`A[1]`
W
1 1
2 2
3 3
4 4
5 5
[[1]]$`A[2]`
X
1 6
2 7
3 8
4 9
5 10
[[2]]
[[2]]$`A[1]`
W
1 21
2 22
3 23
4 24
5 25
[[2]]$`A[2]`
X
1 26
2 27
3 28
4 29
5 30
[[3]]
[[3]]$`A[1]`
W
1 41
2 42
3 43
4 44
5 45
[[3]]$`A[2]`
X
1 46
2 47
3 48
4 49
5 50
This should be relatively simple but I am new to R and cannot quite figure this out.
I will illustrate what I am trying to do.
I have the following:
names <- c("A","B","C")
values <- c(3,6,9)
values2 <- c(5,10,15)
y <- c("2019")
r <- c("1")
t <- c("Team A", "Team B", "Team C")
mgn <- c(33, 56, 63)
df1 <- data.frame(names,y,r,t,values,values2,mgn)
I also have a matrix:
numbers <- matrix(1:6, nrow = 3, ncol = 2)
I am trying to loop through each of the values and values2 in my df1 and multiply these by the values in my numbers matrix like so:
3 x 1 = 3
5 x 4 = 20
6 x 2 = 12
10 x 5 = 50
9 x 3 = 27
15 x 6 = 90
I would then like to print each of these values like:
values values2
[1] 3 20
[2] 12 50
[3] 18 90
I tried the following (just for the first values col):
for(col in 1:ncol(numbers)){
df1$values %*% numbers[col]
print(df1$values)
}
But this is the ouput I get:
[1] 3 6 9
[1] 6 12 18
[1] 6 12 18
[1] 12 24 36
[1] 12 24 36
[1] 24 48 72
I then would like to repeat the process, so that the next row of values and values2 is multiplied by the first row again in numbers (2 and 5) so that:
3 x 2 = 6
5 x 5 = 25
and so on, until all the combinations are calculated.
This would give me the output like so:
3 x 1 = 3
5 x 4 = 20
6 x 1 = 6
10 x 4 = 40
9 x 1 = 9
15 x 4 = 60
Then it should go to the next line of values and values2 and repeat:
3 x 2 = 6
5 x 5 = 25
6 x 2 = 12
10 x 5 = 50
9 x 2 = 18
15 x 5 = 75
And finally the last line:
3 x 3 = 9
5 x 6 = 30
6 x 3 = 18
10 x 6 = 60
9 x 3 = 27
15 x 6 = 90
Finally, I would like to loop through each of these, add them together like:
sumvalues = values + values2
create a total column like:
df1%>%group_by(y, r, t)%>%dplyr::mutate(total=sum(sumvalues)
then obtain the pearson correlation for each by:
cor(mgn, sumvalues, method = "pearson")
So I can have the output like so:
sumvalues total mgn pearson
[1]
[2]
[3]
Here's how I did it:
#### make the two objects to have the same dimensions:
df2<-df1[ ,c(2:3)]
#### multiply and create new object:
new<-df2*numbers
#### if you want to return the first column to df1:
df3<-cbind(df1[1],x)
print(df3)
Your first output can be reached by :
df1[-1] * numbers
# values values2
#1 3 20
#2 12 50
#3 27 90
To get all possible combinations you can use apply with sweep :
apply(numbers, 1, function(x) sweep(df1[-1], 2, x, `*`))
#[[1]]
# values values2
#1 3 20
#2 6 40
#3 9 60
#[[2]]
# values values2
#1 6 25
#2 12 50
#3 18 75
#[[3]]
# values values2
#1 9 30
#2 18 60
#3 27 90
I have a simple 9 element dataframe.
A B C
1 8 21 1
2 40 25 32
3 10 15 49
I want to extract the diagonal elements and store it in a variable. Is there an easier way to do this other than taking one number out at a time to store to a variable?
In this case as they are all numeric you can use:
df <- data.frame(a=c(4,8,10), b = c(25,24,15), c = c(1,32,49))
df
df
a b c
1 4 25 1
2 8 24 32
3 10 15 49
Where this takes the diagonal.
diag(as.matrix(df))
[1] 4 24 49
You can use the diag function which extracts the diagonal of a matrix:
Data <- data.frame(a = c(1,2,3), b= c(11,12,13), c = c(111,112,113))
Data2 <- as.matrix(Data)
Result <- diag(Data2)
Result #Returns 1 12 113
Specifically, say I had three data frames d1, d2, d3:
d1:
X Y Z value
1 0 20 135 43
2 0 4 105 50
3 5 18 20 10
...
d2:
X Y Z value
1 0 20 135 15
2 0 4 105 14
3 2 9 12 16
...
d3:
X Y Z value
1 0 20 135 29
2 2 9 14 16
...
I want to be able to combine these data frames such that each row of the combined data frame consists of three values, based on all unique X, Y, Z combinations. If such an X, Y, Z combination does not exist in one of the original data frames then I just want it to have a value of null (or some arbitrarily low number if that isn't possible). So I'd want an output of:
dfinal:
X Y Z value1 value2 value3
1 0 20 135 43 15 29
2 0 4 105 50 14 null
3 5 18 20 10 null null
4 2 9 12 null 16 null
5 2 9 14 null null 16
...
Is there any efficient way of doing this? I've tried doing this instead using data.table which seemed more suited for this but have yet to figure out how.
?merge
Should do the trick?
By default the data frames are merged on the columns with names they both have, but separate specifications of the columns can be given by by.x and by.y.
So:
merge(d1,d2, by=c("X","Y","Z"))
And you can include all=TRUE, to have complete rows.
The missing data will be NA
merge(d1,d2, by=c("X","Y","Z"), all=TRUE)
Take a look at dplyr and its join methods. I wrote a small example:
library(dplyr)
library(data.table)
d1 <- data.table(X = c(1,2,3), Y = c(2,3,4), Z = c(8,3,9), value = c(22,3,44))
d2 <- data.table(X = c(1,4,3), Y = c(2,6,4), Z = c(8,9,9), value = c(44,22,11))
d2 <- rename(d2, value2 = value)
full_join(d1,d2)
output:
X Y Z value value2
1 1 2 8 22 44
2 2 3 3 3 NA
3 3 4 9 44 11
4 4 6 9 NA 22
Seems simple but I can't figure it out.
I have a bunch of animal location data (217 individuals) as a single dataframe. I'm trying to randomly select X locations per individual for further analysis with the caveat that X is within the range of 6-156.
So I'm trying to set up a loop that first randomly selects a value within the range of 6-156 then use that value (say 56) to randomly extract 56 locations from the first individual animal and so on.
for(i in unique(ANIMALS$ID)){
sub<-sample(6:156,1)
sub2<-i([sample(nrow(i),sub),])
}
This approach didn't seem to work so I tried tweaking it...
for(i in unique(ANIMALS$ID)){
sub<-sample(6:156,1)
rand<-i[sample(1:nrow(i),sub,replace=FALSE),]
}
This did not work either.. Any suggestions or previous postings would be helpful!
Head of the datafile...ANIMALS is the name of the df, ID indicates unique individuals
> FID X Y MONTH DAY YEAR HOUR MINUTE SECOND ELKYR SOURCE ID animalid
1 0 510313 4813290 9 5 2008 22 30 0 342008 FG 1 1
2 1 510382 4813296 9 6 2008 1 30 0 342008 FG 1 1
3 2 510385 4813311 9 6 2008 2 0 0 342008 FG 1 1
4 3 510385 4813394 9 6 2008 3 30 0 342008 FG 1 1
5 4 510386 4813292 9 6 2008 2 30 0 342008 FG 1 1
6 5 510386 4813431 9 6 2008 4 1 0 342008 FG 1 1
Here's one way using mapply. This function takes two lists (or something that can be coerced into a list) and applies function FUN to corresponding elements.
# simulate some data
xy <- data.frame(animal = rep(1:10, each = 10), loc = runif(100))
# calculate number of samples for individual animal
num.samples.per.animal <- sample(3:6, length(unique(xy$animal)), replace = TRUE)
num.samples.per.animal
[1] 6 3 4 4 6 3 3 6 3 5
# subset random x number of rows from each animal
result <- do.call("rbind",
mapply(num.samples.per.animal, split(xy, f = xy$animal), FUN = function(x, y) {
y[sample(1:nrow(y), x),]
}, SIMPLIFY = FALSE)
)
result
animal loc
7 1 0.99483999
1 1 0.50951321
10 1 0.36505294
6 1 0.34058842
8 1 0.26489107
9 1 0.47418823
13 2 0.27213396
12 2 0.28087775
15 2 0.22130069
23 3 0.33646632
21 3 0.02395097
28 3 0.53079981
29 3 0.85287600
35 4 0.84534073
33 4 0.87370167
31 4 0.85646813
34 4 0.11642335
46 5 0.59624723
48 5 0.15379729
45 5 0.57046122
42 5 0.88799675
44 5 0.62171858
49 5 0.75014593
60 6 0.86915983
54 6 0.03152932
56 6 0.66128549
64 7 0.85420774
70 7 0.89262455
68 7 0.40829671
78 8 0.19073661
72 8 0.20648832
80 8 0.71778913
73 8 0.77883677
75 8 0.37647108
74 8 0.65339300
82 9 0.39957202
85 9 0.31188471
88 9 0.10900795
100 10 0.55282999
95 10 0.10145296
96 10 0.09713218
93 10 0.64900866
94 10 0.76099256
EDIT
Here is another (more straightforward) approach that also handles cases when number of rows is less than the number of samples that should be allocated.
set.seed(357)
result <- do.call("rbind",
by(xy, INDICES = xy$animal, FUN = function(x) {
avail.obs <- nrow(x)
num.rows <- sample(3:15, 1)
while (num.rows > avail.obs) {
message("Sample to be larger than available data points, repeating sampling.")
num.rows <- sample(3:15, 1)
}
x[sample(1:avail.obs, num.rows), ]
}))
result
I like Stackoverflow because I learn so much. #RomanLustrik provided a simple solution; mine is straight-froward as well:
# simulate some data
xy <- data.frame(animal = rep(1:10, each = 10), loc = runif(100))
newVec <- NULL #Create a blank dataFrame
for(i in unique(xy$animal)){
#Sample a number between 1 and 10 (or 6 and 156, if you need)
samp <- sample(1:10, 1)
#Determine which rows of dataFrame xy correspond with unique(xy$animal)[i]
rows <- which(xy$animal == unique(xy$animal)[i])
#From xy, sample samp times from the rows associated with unique(xy$animal)[i]
newVec1 <- xy[sample(rows, samp, replace = TRUE), ]
#append everything to the same new dataFrame
newVec <- rbind(newVec, newVec1)
}