R: Get the Min/Max of Each Item of a Vector Compared to Single Value

R: Get the min/max of each item of a vector compared to single value

I'm guessing you're looking for pmin/pmax:

> pmin(a$v, n)
[1] 3 1 4
> pmax(a$v, n)
[1] 4 4 5

Creating a loop in R to find a minimum value of a vector by using comparison operators (, , ==, etc.)

This should do the trick. First create a variable called "low" with a high value (as @alistaire suggested). Then, loop through your values of aVector. At each value, check if the value is less than low. If it is, update low to that value.

set.seed(42)

aVector <- sample(0:100, 100, replace=FALSE)

low <- Inf # initialize with high value
for (i in aVector) {

  if(i < low){
    low <- i
  }

}
print(low)

# Confirm we got correct answer
min(aVector)

Fastest way to find second (third...) highest/lowest value in vector or column

Rfast has a function called nth_element that does exactly what you ask.

Further the methods discussed above that are based on partial sort, don't support finding the k smallest values

Update (28/FEB/21) package kit offers a faster implementation (topn) see https://stackoverflow.com/a/66367996/4729755, https://stackoverflow.com/a/53146559/4729755

Disclaimer: An issue appears to occur when dealing with integers which can by bypassed by using as.numeric (e.g. Rfast::nth(as.numeric(1:10), 2)), and will be addressed in the next update of Rfast.

Rfast::nth(x, 5, descending = T)

Will return the 5th largest element of x, while

Rfast::nth(x, 5, descending = F)

Will return the 5th smallest element of x

Benchmarks below against most popular answers.

For 10 thousand numbers:

N = 10000
x = rnorm(N)

maxN <- function(x, N=2){
    len <- length(x)
    if(N>len){
        warning('N greater than length(x).  Setting N=length(x)')
        N <- length(x)
    }
    sort(x,partial=len-N+1)[len-N+1]
}

microbenchmark::microbenchmark(
Rfast = Rfast::nth(x,5,descending = T),
maxn = maxN(x,5),
order = x[order(x, decreasing = T)[5]])

Unit: microseconds
  expr      min       lq      mean   median        uq       max neval
 Rfast  160.364  179.607  202.8024  194.575  210.1830   351.517   100
  maxN  396.419  423.360  559.2707  446.452  487.0775  4949.452   100
 order 1288.466 1343.417 1746.7627 1433.221 1500.7865 13768.148   100

For 1 million numbers:

N = 1e6
x = rnorm(N)

microbenchmark::microbenchmark(
Rfast = Rfast::nth(x,5,descending = T),
maxN = maxN(x,5),
order = x[order(x, decreasing = T)[5]]) 

Unit: milliseconds
  expr      min        lq      mean   median        uq       max neval
 Rfast  89.7722  93.63674  114.9893 104.6325  120.5767  204.8839   100
  maxN 150.2822 207.03922  235.3037 241.7604  259.7476  336.7051   100
 order 930.8924 968.54785 1005.5487 991.7995 1031.0290 1164.9129   100

Minimum and maximum sequential values of a vector

You can do:

range(v[which.min(v):length(v)])

[1] 1 7

Test for equality among all elements of a single numeric vector

I use this method, which compares the min and the max, after dividing by the mean:

# Determine if range of vector is FP 0.
zero_range <- function(x, tol = .Machine$double.eps ^ 0.5) {
  if (length(x) == 1) return(TRUE)
  x <- range(x) / mean(x)
  isTRUE(all.equal(x[1], x[2], tolerance = tol))
}

If you were using this more seriously, you'd probably want to remove missing values before computing the range and mean.

R set column to maximum of current entry and specified value in an elegant way

max gives you maximum of the whole column but for your case you need pmax(parallel maximum) so it gives you maximum of 1 or each number in the vector.

data$foo <- pmax(data$foo, 1)
data

#   foo
#1    1
#2    2
#3    1
#4    2
#5    1
#6    2
#7    1
#8    2
#9    1
#10   2

Is there a vectorized parallel max() and min()?

Sounds like you're looking for pmax and pmin ("parallel" max/min):

Extremes                 package:base                  R Documentation

Maxima and Minima

Description:

     Returns the (parallel) maxima and minima of the input values.

Usage:

     max(..., na.rm = FALSE)
     min(..., na.rm = FALSE)

     pmax(..., na.rm = FALSE)
     pmin(..., na.rm = FALSE)

     pmax.int(..., na.rm = FALSE)
     pmin.int(..., na.rm = FALSE)

Arguments:

     ...: numeric or character arguments (see Note).

   na.rm: a logical indicating whether missing values should be
          removed.

Details:

     ‘pmax’ and ‘pmin’ take one or more vectors (or matrices) as
     arguments and return a single vector giving the ‘parallel’ maxima
     (or minima) of the vectors.  The first element of the result is
     the maximum (minimum) of the first elements of all the arguments,
     the second element of the result is the maximum (minimum) of the
     second elements of all the arguments and so on.  Shorter inputs
     are recycled if necessary.  ‘attributes’ (such as ‘names’ or
     ‘dim’) are transferred from the first argument (if applicable).

Creating a function that calculates the min and max without using min() | max()

Consider head() or tail() after sorting:

minmax <- function(x) {
    sorted_vec <- sort(x)
    c(min=head(sorted_vec, 1), max=tail(sorted_vec, 1))
}

Alternatively, by indexing after sorting:

minmax <- function(x) {
  sorted_vec <- sort(x)
  c(min=sorted_vec[1], max=sorted_vec[length(x)])
}

Finding maximum from variables in r

You can use Map. Map applies a function to the corresponding elements of given vectors.

probsA <- c(2.634872e-02, 6.709075e-03, 1.107573e-04, 1.708307e-03, 2.820171e-05)
probsB <- c(0.0013311712, 0.0012295459, 0.0009688963, 0.0011356790, 0.0008949280)

Map(max, probsA, probsB)

[[1]]
[1] 0.02634872

[[2]]
[1] 0.006709075

[[3]]
[1] 0.0009688963

[[4]]
[1] 0.001708307

[[5]]
[1] 0.000894928

Excluding both the minimum and maximum value

Update

Remembered after looking at @Rich Pauloo's answer, we can directly use which.max and which.min to get index of minimum and maximum value

as.data.frame(t(apply(df, 1, function(x) x[-c(which.max(x), which.min(x))])))

#  V1 V2 V3
#1 13 11  6
#2 15  8 18
#3  5 10 21
#4 14 12 17
#5 19  9 20

Here which.max/which.min will ensure that you get the index of first minimum and maximum respectively for each row.

---

Some other variations could be

as.data.frame(t(apply(df, 1, function(x) 
            x[-c(which.max(x == min(x)), which.max(x == max(x)))])))

If you want to use which we can do

as.data.frame(t(apply(df, 1, function(x) 
                x[-c(which(x == min(x)[1]), which(x == max(x)[1]))])))

data

set.seed(1234)
df <- as.data.frame(matrix(sample(25), 5, 5))
df

#  V1 V2 V3 V4 V5
#1  3 13 11 16  6
#2 15  1  8 25 18
#3 24  5  4 10 21
#4 14 12 17  2 22
#5 19  9 20  7 23

---

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