I'm a newbie in R, trying to figure out how to find the maximum value between 2 values in a single column.
Example data:
t min most max
---------------
1 10 20 40
2 5 10 30
3 14 28 60
4 40 75 150
Result I'm looking for:
t min most max
---------------
1 10 20 40
2 14 28 60
3 40 75 150
I have tried using rowWise(), but it's not working. I am getting the maximum value row wise using:
df$new <-pmax(df$min, df$most, df$max)
df
which gives me the maximum value for the entire row.
t min most max new
-------------------
1 10 20 40 40
2 5 10 30 30
3 14 28 60 60
4 40 75 150 150
Thanks in advance.
You can do this with pmax applied to the vector against its shifted self. Putting it in a nice little helper function:
adj_max = function(x) {
pmax(x[-1], x[-length(x)])
}
as.data.frame(lapply(your_data, adj_max))
# or with dplyr
your_data %>%
summarize(across(everything(), adj_max))
Reproducible demo:
x = c(10, 5, 14, 40)
adj_max(x)
# [1] 10 14 40
Related
How can I create a matrix , with random number on row and not replace.
like this
5 29 24 20 31 33
2 18 35 4 11 21
30 40 22 14 2 28
33 14 4 18 5 10
10 33 15 2 28 18
7 22 9 25 31 20
12 29 31 22 37 26
7 31 34 28 19 23
7 34 11 6 31 28
my code :
matrix(sample(1:42, 60, replace = FALSE), ncol = 6)
But I receive this error message:
Error in sample.int(length(x), size, replace, prob) : cannot take a
sample larger than the population when 'replace = FALSE'
but it's wrong because only 1~42, it can't create a 60 matrix.
You can not generate all 60 of the numbers with one sample function as you want to allow replacement of numbers in a different row. Therefore you have to do one sample per row. #Jav provided very neat code to accomplish this in the comment to the question:
t(sapply(1:10, function(x) sample(1:42, 6, replace = FALSE)))
if you want to have a different sample in each row, then replicate can help you -- but replicate (as pretty much everything else in R) works naturally columnwise, so you have to transpose the result:
t(replicate(10, sample(1:42, 6)))
replace = FALSE is the default, so I didn't include it
after transposing, 10 becomes the number of rows and 6 becomes the number of columns
I have a data frame like this.
> abc
ID 1.x 2.x 1.y 2.y
1 4 10 20 30 40
2 16 5 10 5 10
3 42 16 17 18 19
4 91 20 20 20 20
5 103 103 42 56 84
How do I create two additional columns '1' and '2' by multiplying 1.x * 1.y and 2.x * 2.y in a generalized way?
I am trying to get a generalized solution where number of columns can be too many. So I want to multiply all x with all y. While x and y are fixed, n has to be figured out from data frame.
For simplicity lets assume n is also fixed however it is a large number.
One thing i can try is :-
abc[,c(6,7)]=abc[,c(2,3)]*abc[,c(4,5)]
It will work only if col positions are contiguous. This is good enough for me. If anyone can have more generalized solution, it will benefit us all.
If there are only couple of variables to multiply, we can do this with transform by multiplying the columns of interest
transform(abc, new1 = `1.x`*`1.y`, new2 = `2.x`*`2.y`, check.names = FALSE)
# ID 1.x 2.x 1.y 2.y new1 new2
#1 4 10 20 30 40 300 800
#2 16 5 10 5 10 25 100
#3 42 16 17 18 19 288 323
#4 91 20 20 20 20 400 400
#5 103 103 42 56 84 5768 3528
If we have lots of columns, then one approach is to split the dataset into a list of data.frames by taking the substring of names and then loop through the list and multiply the rows with do.call
abc[paste0("new", 1:2)] <- lapply(split.default(abc[-1],
sub("\\.[a-z]+$", "", names(abc)[-1])), function(x) do.call(`*`, x))
Or another option is (based on the pairwise column multiplication)
apply(aperm(array(unlist(abc[-1]), c(5, 2, 2)),
c(3, 1, 2)), 3, matrixStats::colProds)
Mutate will preserve the original variables. Mutate_all will allow you to multiply all columns in your dataframe.
abc %>%
mutate(new_vary1 = `1.x`* `2.x`,
new_vary2 = `1.y`* `2.y`) %>%
mutate_all(funs(.*`1.x`))
I have a data.table of integers with values between 1 and 60.
My question is about flooring or ceiling any number to the following values: 12 18 24 30 36 ... 60.
For example, let's say my data.table contains the number 13. I want R to "transform" this number into 12 and 18 as 13 lies in between those numbers. Moreover, if I have 18 I want R to keep it at 18.
If my data.table contains the value 50, I want R to convert that number into 48 and 54 and so on.
My goal is to get two different data.tables. One where the floored values are saved and one where the ceiled values are saved.
Any idea how one could do this in R?
EDIT: Numbers smaller than 12 should always be transformed to 12.
Example output:
If have the following data.table data.table(c(1,28,29,41,53,53,17,41,41,53))
I want the following two output data.tables: floored values data.table(c(12,24,24,36,48,48,12,36,36,48))
I want the following two output data.tables: ceiled values data.table(c(12,30,30,42,54,54,18,42,42,54))
Here is a fairly direct way (edited to round up to 12 if any values are below):
df <- data.frame(nums = 10:20)
df$floors <- with(df,pmax(12,6*floor(nums/6)))
df$ceils <- with(df,pmax(12,6*ceiling(nums/6)))
Leading to:
> df
nums floors ceils
1 10 12 12
2 11 12 12
3 12 12 12
4 13 12 18
5 14 12 18
6 15 12 18
7 16 12 18
8 17 12 18
9 18 18 18
10 19 18 24
11 20 18 24
Here's a way we could do this, using sapply and the which.min functions. From your question, it's not immediately clear how values < 12 should be handled.
x <- 1:60
num_list <- seq(12, 60, 6)
floorr <- sapply(x, function(x){
diff_vec <- x - num_list
diff_vec <- ifelse(diff_vec < 0, Inf, diff_vec)
num_list[which.min(diff_vec)]
})
ceill <- sapply(x, function(x){
diff_vec <- num_list - x
diff_vec <- ifelse(diff_vec < 0, Inf, diff_vec)
num_list[which.min(diff_vec)]
})
tail(cbind(x, floorr, ceill))
x floorr ceill
[55,] 55 54 60
[56,] 56 54 60
[57,] 57 54 60
[58,] 58 54 60
[59,] 59 54 60
[60,] 60 60 60
I am new to R. I have a data frame like following
>df=data.frame(Id=c("Entry_1","Entry_1","Entry_1","Entry_2","Entry_2","Entry_2","Entry_3","Entry_4","Entry_4","Entry_4","Entry_4"),Start=c(20,20,20,37,37,37,68,10,10,10,10),End=c(50,50,50,78,78,78,200,94,94,94,94),Pos=c(14,34,21,50,18,70,101,35,2,56,67),Hits=c(12,34,17,89,45,87,1,5,6,3,26))
Id Start End Pos Hits
Entry_1 20 50 14 12
Entry_1 20 50 34 34
Entry_1 20 50 21 17
Entry_2 37 78 50 89
Entry_2 37 78 18 45
Entry_2 37 78 70 87
Entry_3 68 200 101 1
Entry_4 10 94 35 5
Entry_4 10 94 2 6
Entry_4 10 94 56 3
Entry_4 10 94 67 26
For each entry I would like to iterate the data.frame in 3 different modes. For an example, for Entry_1 mode_1 =seq(20,50,3)and mode_2=seq(21,50,3) and mode_3=seq(22,50,3). I would like sum all the Values in Column "Hits" whose corresponding values in Column "Pos" that falls in mode_1 or_mode_2 or mode_3 and generate a data.frame like follow:
Id Mode_1 Mode_2 Mode_3
Entry_1 0 17 34
Entry_2 87 89 0
Entry_3 1 0 0
Entry_4 26 8 0
I tried the following code:
mode_1=0
mode_2=0
mode_3=0
mode_1_sum=0
mode_2_sum=0
mode_3_sum=0
for(i in dim(df)[1])
{
if(df$Pos[i] %in% seq(df$Start[i],df$End[i],3))
{
mode_1_sum=mode_1_sum+df$Hits[i]
print(mode_1_sum)
}
mode_1=mode_1_sum+counts
print(mode_1)
ifelse(df$Pos[i] %in% seq(df$Start[i]+1,df$End[i],3))
{
mode_2_sum=mode_2_sum+df$Hits[i]
print(mode_2_sum)
}
mode_2_sum=mode_2_sum+counts
print(mode_2)
ifelse(df$Pos[i] %in% seq(df$Start[i]+2,df$End[i],3))
{
mode_3_sum=mode_3_sum+df$Hits[i]
print(mode_3_sum)
}
mode_3_sum=mode_3_sum+counts
print(mode_3_sum)
}
But the above code only prints 26. Can any one guide me how to generate my desired output, please. I can provide much more details if needed. Thanks in advance.
It's not an elegant solution, but it works.
m <- 3 # Number of modes you want
foo <- ((df$Pos - df$Start)%%m + 1) * (df$Start < df$Pos) * (df$End > df$Pos)
tab <- matrix(0,nrow(df),m)
for(i in 1:m) tab[foo==i,i] <- df$Hits[foo==i]
aggregate(tab,list(df$Id),FUN=sum)
# Group.1 V1 V2 V3
# 1 Entry_1 0 17 34
# 2 Entry_2 87 89 0
# 3 Entry_3 1 0 0
# 4 Entry_4 26 8 0
-- EXPLANATION --
First, we find the indices of df$Pos That are both bigger than df$Start and smaller than df$End. These should return 1 if TRUE and 0 if FALSE. Next, we take the difference between df$Pos and df$Start, we take mod 3 (which will give a vector of 0s, 1s and 2s), and then we add 1 to get the right mode. We multiply these two things together, so that the values that fall within the interval retain the right mode, and the values that fall outside the interval become 0.
Next, we create an empty matrix that will contain the values. Then, we use a for-loop to fill in the matrix. Finally, we aggregate the matrix.
I tried looking for a quicker solution, but the main problem I cannot work around is the varying intervals for each row.
I have set of data (of 5000 points with 4 dimensions) that I have clustered using kmeans in R.
I want to order the points in each cluster by their distance to the center of that cluster.
Very simply, the data looks like this (I am using a subset to test out various approaches):
id Ans Acc Que Kudos
1 100 100 100 100
2 85 83 80 75
3 69 65 30 29
4 41 45 30 22
5 10 12 18 16
6 10 13 10 9
7 10 16 16 19
8 65 68 100 100
9 36 30 35 29
10 36 30 26 22
Firstly, I used the following method to cluster the dataset into 2 clusters:
(result <- kmeans(data, 2))
This returns a kmeans object that has the following methods:
cluster, centers etc.
But I cannot figure out how to compare each point and produce an ordered list.
Secondly, I tried the seriation approach as suggested by another SO user here
I use these commands:
clus <- kmeans(scale(x, scale = FALSE), centers = 3, iter.max = 50, nstart = 10)
mns <- sapply(split(x, clus$cluster), function(x) mean(unlist(x)))
result <- dat[order(order(mns)[clus$cluster]), ]
Which seems to produce an ordered list but if I bind it to the labeled clusters (using the following cbind command):
result <- cbind(x[order(order(mns)[clus$cluster]), ],clus$cluster)
I get the following result, which does not appear to be ordered correctly:
id Ans Acc Que Kudos clus
1 3 69 65 30 29 1
2 4 41 45 30 22 1
3 5 10 12 18 16 2
4 6 10 13 10 9 2
5 7 10 16 16 19 2
6 9 36 30 35 29 2
7 10 36 30 26 22 2
8 1 100 100 100 100 1
9 2 85 83 80 75 2
10 8 65 68 100 100 2
I don't want to be writing commands willy-nilly but understand how the approach works. If anyone could help out or spread some light on this, it would be really great.
EDIT:::::::::::
As the clusters can be easily plotted, I'd imagine there is a more straightforward way to get and rank the distances between points and the center.
The centers for the above clusters (when using k = 2) are as follows. But I do not know how to get and compare this with each individual point.
Ans Accep Que Kudos
1 83.33333 83.66667 93.33333 91.66667
2 30.28571 30.14286 23.57143 20.85714
NB::::::::
I don't need top use kmeans but I want to specify the number of clusters and retrieve an ordered list of points from those clusters.
Here is an example that does what you ask, using the first example from ?kmeans. It is probably not terribly efficient, but is something to build upon.
#Taken straight from ?kmeans
x <- rbind(matrix(rnorm(100, sd = 0.3), ncol = 2),
matrix(rnorm(100, mean = 1, sd = 0.3), ncol = 2))
colnames(x) <- c("x", "y")
cl <- kmeans(x, 2)
x <- cbind(x,cl = cl$cluster)
#Function to apply to each cluster to
# do the ordering
orderCluster <- function(i,data,centers){
#Extract cluster and center
dt <- data[data[,3] == i,]
ct <- centers[i,]
#Calculate distances
dt <- cbind(dt,dist = apply((dt[,1:2] - ct)^2,1,sum))
#Sort
dt[order(dt[,4]),]
}
do.call(rbind,lapply(sort(unique(cl$cluster)),orderCluster,data = x,centers = cl$centers))