Row Wise Sorting in R

Row wise Sorting in R

If we need only to sort by rows, use apply with MARGIN=1 and assign the output back to the original columns after transposing the output.

df1[-1] <- t(apply(df1[-1], 1, 
        FUN=function(x) sort(x, decreasing=TRUE)))

df1
#   Name English Math French
# 1 John      86   78     56
# 2  Sam      97   86     79
# 3 Viru      93   44     34

NOTE: But we may need to change the column names as sorting by row gives the new sorted values.

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Another option will be use apply separately to get the column names and the values, with Map we get the corresponding columns, cbind with the first column to have the output.

nMat <- `dim<-`(names(df1)[-1][t(apply(df1[-1], 1,  
        order, decreasing=TRUE))], dim(df1[-1]))
vMat <- t(apply(df1[-1], 1,  sort, decreasing=TRUE))
cbind(df1[1], data.frame(Map(cbind, as.data.frame(nMat,
   stringsAsFactors=FALSE), as.data.frame(vMat))))
#  Name    V1.1 V1.2   V2.1 V2.2    V3.1 V3.2
#1 John  French   86   Math   78 English   56
#2  Sam    Math   97 French   86 English   79
#3 Viru English   93   Math   44  French   34

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Or another option is data.table. We melt the 'wide' format to 'long' format, grouped by 'Name', we order the 'value' in decreasing order in 'i', get the Subset of Data.table (.SD), create a new column ('N'), grouped by 'Name' and use dcast to convert from 'long' to 'wide'.

library(data.table)
dcast(melt(setDT(df1), id.var='Name')[order(-value), 
  .SD, Name][, N:=paste0("Col", 1:.N) , .(Name)],
    Name~N, value.var=c("variable", "value"))
#   Name variable_Col1 variable_Col2 variable_Col3 value_Col1 value_Col2 value_Col3
#1: John        French          Math       English         86         78         56
#2:  Sam          Math        French       English         97         86         79
#3: Viru       English          Math        French         93         44         34

EDIT:
The above data.table solution will not work in case you have 10 or more columns with values, because then col10 will preceed col2 in the ordering, even though higher values will be stored in col2. To resolve this issue, you can use just number for the names of your new columns as in:

dcast(melt(setDT(df1), id.var='Name')[order(-value), 
      .SD, Name][, N:=1:.N , .(Name)],
        Name~N, value.var=c("variable", "value"))

R row-wise sort on specific columns

Instead of assigning to df, only assign to the columns you want to sort.

df[1:2] <- t(apply(df[1:2], 1, 
              FUN=function(x) sort(x, decreasing=FALSE)))

Or written more simply:

to_sort <- 1:2
df[to_sort] <- t(apply(df[to_sort], 1, sort, decreasing = FALSE))

Sorting dates row-wise

The apply approach also works with dates. They just get coerced to a character matrix, but we can coerce as.data.frame and lapply as.Date over it.

my_data[-1] <- as.data.frame(t(apply(my_data[-1], 1, sort))) |> lapply(as.Date)

Gives

my_data
#   id         d1         d2         d3         d4         d5
# 1  1 1999-03-14 2009-08-31 2010-01-19 2013-01-01 2015-11-25
# 2  2 2000-09-10 2001-02-22 2007-01-29 2010-04-10 2019-09-11
# 3  3 2001-04-05 2007-12-26 2008-09-12 2012-07-15 2015-10-14
# 4  4 1999-03-15 2007-01-18 2009-12-19 2014-02-08 2016-07-19
# 5  5 2003-07-03 2004-07-22 2006-11-05 2009-05-31 2011-05-25

Where

str(my_data)
# 'data.frame': 5 obs. of  6 variables:
# $ id: int  1 2 3 4 5
# $ d1: Date, format: "1999-03-14" "2000-09-10" "2001-04-05" "1999-03-15" ...
# $ d2: Date, format: "2009-08-31" "2001-02-22" "2007-12-26" "2007-01-18" ...
# $ d3: Date, format: "2010-01-19" "2007-01-29" "2008-09-12" "2009-12-19" ...
# $ d4: Date, format: "2013-01-01" "2010-04-10" "2012-07-15" "2014-02-08" ...
# $ d5: Date, format: "2015-11-25" "2019-09-11" "2015-10-14" "2016-07-19" ...

Sorting each row of a data frame

You could use the plain apply function with MARGIN = 1 to apply over rows and then transpose the result.

t(apply(df, 1, sort))

Fastest way to sort and desort rows of a matrix [r]

Row-wise sorting seems to be straightforward. To get the original order back (un-sort) we need the row-wise ranks rather than their order. Thereafter, what works for column sorting in @Josh O'Brien's answer we can adapt for rows.

Base R solution:

rr <- t(apply(m, 1, rank))  ## get initial RANKS by row
sm <- t(apply(m, 1, sort))  ## sort m

##  DOING STUFF HERE  ##

sm[] <- sm[cbind(as.vector(row(rr)), as.vector(rr))]  ## un-sort
all(m == sm)  ## check
# [1] TRUE

Seems to work.

In your linked answer, the rowSort function of the Rfast package stands out well in terms of performance, which may cover the sorting issue. Moreover there's also a rowRanks function that will cover our ranking issue. So we can avoid apply.

Let's try it out.

m[1:3, ]
#           [,1]      [,2]      [,3]        [,4]
# [1,] 0.9148060 0.5142118 0.3334272 0.719355838
# [2,] 0.9370754 0.3902035 0.3467482 0.007884739
# [3,] 0.2861395 0.9057381 0.3984854 0.375489965

library(Rfast)
rr <- rowRanks(m)  ## get initial RANKS by row
sm <- rowSort(m)   ## sort m
sm[1:3, ]  # check
#            [,1]      [,2]      [,3]      [,4]
# [1,] 0.36106962 0.4112159 0.6262453 0.6311956
# [2,] 0.01405302 0.2171577 0.5459867 0.6836634
# [3,] 0.07196981 0.2165673 0.5739766 0.6737271

##  DOING STUFF HERE  ##

sm[] <- sm[cbind(as.vector(row(rr)), as.vector(rr))]  ## un-sort
all(sm == m)  ## check
# [1] TRUE

Dito.

Benchmark

m.test <- matrix(runif(4e6), ncol = 4)
dim(m.test)
# [1] 1000000       4

# Unit: milliseconds
#     expr        min       lq       mean     median         uq        max neval cld
#    Rfast   897.6286   910.91   956.6259   924.1914   986.1246   1048.058     3 a  
#    baseR 87931.2824 88004.73 95659.8671 88078.1737 99524.1594 110970.145     3   c
#  forloop 58927.7784 59434.54 60317.3903 59941.2930 61012.1963  62083.100     3  b 

Not so bad!!

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Data/Code:

set.seed(42)

m <- matrix(runif(100), nrow = 25, ncol = 4)

## benchmark
m.test <- matrix(runif(4e6), ncol = 4)

microbenchmark::microbenchmark(
  Rfast={
    rr <- rowRanks(m.test)
    sm <- rowSort(m.test)
    sm[] <- sm[cbind(as.vector(row(rr)), as.vector(rr))]},
  baseR={
    rr <- t(apply(m.test, 1, rank))
    sm <- t(apply(m.test, 1, sort))
    sm[] <- sm[cbind(as.vector(row(rr)), as.vector(rr))]
  },
  forloop={
    om <- t(apply(m.test, 1, order, decreasing = T))
    sm <- m.test
    for (i in seq_len(nrow(m.test))) {
      sm[i, ] <- sm[i, om[i, ]]
    }
    for (i in seq_len(nrow(m.test))) {
      sm[i, ] <- sm[i, order(om[i, ])]
    }
  }, times=3L
)

How to sort each row of a data frame WITHOUT losing the column names

Store the names and apply them:

nm = names(df)
sorted_df <- as.data.frame(t(apply(df, 1, sort)))
names(sorted_df) = nm

You could compress this down to a single line if you prefer:

sorted_df = setNames(as.data.frame(t(apply(df, 1, sort))), names(df))

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