How to Reverse the Order of a Dataframe in R

Reverse row order in a data frame

You could use rev to reverse the row names

df[rev(rownames(df)),]
#   A B
# 3 5 6
# 2 3 4
# 1 1 2

If you want to correct the new reversed row names, you could write a little function

flip <- function(data) {
    new <- data[rev(rownames(data)), ]
    rownames(new) <- NULL
    new
}
flip(df)
#   A B
# 1 5 6
# 2 3 4
# 3 1 2

How to reverse the order of dataframe in a pattern in R

Another base R option using order + ave

df[with(df, order(Stage, ave(1:nrow(df), Stage, FUN = seq_along))), ]

which gives

  Stage        Score
5   W-1           75
6   W-1 1 Min 04 Sec
3   W-2           45
4   W-2 1 Min 34 Sec
1   W-3           25
2   W-3 2 Min 10 Sec

Data

> dput(df)
structure(list(Stage = c("W-3", "W-3", "W-2", "W-2", "W-1", "W-1"
), Score = c("25", "2 Min 10 Sec", "45", "1 Min 34 Sec", "75",
"1 Min 04 Sec")), row.names = c(NA, -6L), class = "data.frame")

How to reverse the order of values in one column in a dataframe grouped by another column?

You can try ave if you are with base R:

df <- within(df,B <- ave(B,A,FUN = function(x) sort(x,decreasing = TRUE)))

which gives

> df
   A B
1  A 3
2  A 2
3  A 1
4  B 2
5  B 1
6  C 5
7  C 4
8  C 3
9  C 2
10 C 1

How to reverse rows of a data.frame or data.table in R

We can use apply row-wise and reverse the non-NA values in each row.

df[-1] <- t(apply(df[-1], 1, function(x) {x[!is.na(x)] <- rev(x[!is.na(x)]);x}))

How to reverse the order of observations in a data frame with individuals that are of different lengths

We can arrange by ID as well

library(dplyr)
df %>% 
    arrange(ID, desc(num))

Is there a simple way of reversing the dates in a dataframe in R?

For ordering rows, you can use this:

BTC <- BTC[order(BTC$date),]

To remove several columns at once, you can use the package dplyr:

library('dplyr')

BTC <- BTC %>% select(-high, -low, -unix, -symbol, -Volume.USDT, -tradecount)

If the column are ordered from high to tradecount, you can shorten it to:

BTC <- BTC %>% select(-(high:tradecount))

Reverse the order of the rows for every 'id' in data in R

Here's an dplyr example:

library(tidyverse)
data <- tibble(id = rep(letters[1:3], each=3),
                   date = rep(as.Date(c("2022-07-01", "2022-06-25", "2022-05-15")), 3))

data %>% arrange(id, date)

# A tibble: 9 x 2
  id    date      
  <chr> <date>    
1 a     2022-05-15
2 a     2022-06-25
3 a     2022-07-01
4 b     2022-05-15
5 b     2022-06-25
6 b     2022-07-01
7 c     2022-05-15
8 c     2022-06-25
9 c     2022-07-01

Base solution without using dplyr

data[order(data$id, data$date),]

Selecting observations within a data frame and reversing their order

One option is to group_split by ID and do the arrange by looping over the list with map based on whether any of the values 'b', 'e', 'g' are %n% the 'ID'

library(dplyr)
library(purrr)
out <- dat %>% 
        group_split(ID) %>%
        map_dfr(~ if(any(c('b', 'e', 'g') %in% first(.x$ID)))
         .x %>%
             arrange(desc(time)) else .x)   

out %>% 
   filter(ID %in% c('a', 'b'))
# A tibble: 20 x 3
#   ID     time    var1
#   <fct> <int>   <dbl>
# 1 a         1 -0.560 
# 2 a         2 -0.230 
# 3 a         3  1.56  
# 4 a         4  0.0705
# 5 a         5  0.129 
# 6 a         6  1.72  
# 7 a         7  0.461 
# 8 a         8 -1.27  
# 9 a         9 -0.687 
#10 a        10 -0.446 
#11 b        10 -0.473 
#12 b         9  0.701 
#13 b         8 -1.97  
#14 b         7  0.498 
#15 b         6  1.79  
#16 b         5 -0.556 
#17 b         4  0.111 
#18 b         3  0.401 
#19 b         2  0.360 
#20 b         1  1.22  

---

Or we can make use of arrange in a hacky way i.e. change the time to negative based on the ID 'b', 'e', 'g' while the rest is positive

out1 <- dat %>%
     arrange(ID,  time * c(1, -1)[c(1 + (ID %in% c('b', 'e', 'g')))])

-checking

all.equal(out, out1, check.attributes = FALSE)
#[1] TRUE

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