How to add a row to a data frame in R?

Question

In R, how do you add a new row to a data frame once the data frame has already been initialized?

So far I have this:

df <- data.frame("hi", "bye")
names(df) <- c("hello", "goodbye")

#I am trying to add "hola" and "ciao" as a new row
de <- data.frame("hola", "ciao")

merge(df, de) # Adds to the same row as new columns

# Unfortunately, I couldn't find an rbind() solution that wouldn't give me an error

Any help would be appreciated

assign names to `de` too. `names(de) <- c("hello","goodbye")` and `rbind` — Khashaa, Feb 12 '15 at 00:14
This really is an area which base R fails miserably at, and has for a long time: http://stackoverflow.com/questions/13599197/rbind-data-frames-without-names — thelatemail, Feb 12 '15 at 00:49
@thelatemail disagree. data frames are a special structure in r. a list of lists with common dimnames and attributes and methods. I think it is very expected that one cannot `rbind(data.frame(a = 1), data.frame(b = 2))`.. why would you want to? I would hope that would throw an error regardless. It's like `merge`'ing with a random `by` variable. And this is 2015, doesn't everyone set `options(stringsAsFactors = FALSE)`? — rawr, Feb 12 '15 at 15:34
@rawr - sure, different names shouldn't be bound, but R can't handle binding no names to no names, binding names to no names with the same dimensions, or binding new data to incorporate new factor levels. I think that's a weakness. Particularly when it can handle binding repeated names and all NA names. And setting `stringsAsFactors=FALSE` can be a quick fix, but changing the defaults that other people are going to have set differently can really ruin a day. — thelatemail, Feb 12 '15 at 22:21
@thelatemail again, rbinding is akin to cbinding. you *would not*, or I should say *should not*, be cbinding data frames. merging explicitly cbinds by a common by variable. it is dangerous to cbind blindly as it should be to rbind blindly. And IMO adding factor levels is not straight-forward, as it should not be, because it is an unusual operation. How often do you need to add a level to sex or race? It's just a level of error checking to me. And I'm not sure what you mean by no names to no names--arent we still talking about data frames here? — rawr, Feb 12 '15 at 22:38
Just use the "split" command: [split](https://i.stack.imgur.com/A6tJe.png) — AngelaR, Apr 14 '18 at 05:03

score 200 · Answer 1 · edited Aug 01 '19 at 22:53

200

Let's make it simple:

df[nrow(df) + 1,] = c("v1","v2")

edited Aug 01 '19 at 22:53

M--

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answered May 24 '17 at 06:39

Matheus Araujo

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This causes problems when trying to add a new row with mixed data types (some string, some numeric). In such a case, even the numeric values are converted to string. One workaround is to add the values separately, something like the following (assuming there are 3 columns): `df[nrow(df) + 1, 1:2] = c("v1", "v2")` and `df[nrow(df), 3] = 100` But still it's a good point about adding new row. So, +1 – The Student Soul Aug 03 '17 at 08:05
29

Or use "list" instead of "c". – Ytsen de Boer Aug 03 '17 at 09:13
nice idea, but how can I do if I want to insert or add a new row a the first position? – Darwin PC May 08 '18 at 14:55
1

@Matheus Araujo: Is this the most efficient way to add row to a df? I have 100k+ rows to be added in a loop. Feel like nrow would get slower as number of rows increase. – Chintan Pathak Jun 14 '18 at 19:51
@ChintanPathak, to be honest, I don't know if using nrow() is very naive (I'm open to suggestion). However, I know that if you know the size of your final data.frame and if you allocate it in memory before populating it, the action will be faster. – Matheus Araujo Jun 15 '18 at 09:12
1

Tried this with data.table but tells with nrow+1 is out of range. – Herman Toothrot Oct 17 '18 at 13:24
1

@Arani there's already an answer with `list()`. I reverted your edit. – M-- Aug 01 '19 at 22:54

score 173 · Accepted Answer · answered Feb 12 '15 at 03:52

173

Like @Khashaa and @Richard Scriven point out in comments, you have to set consistent column names for all the data frames you want to append.

Hence, you need to explicitly declare the columns names for the second data frame, de, then use rbind(). You only set column names for the first data frame, df:

df<-data.frame("hi","bye")
names(df)<-c("hello","goodbye")

de<-data.frame("hola","ciao")
names(de)<-c("hello","goodbye")

newdf <- rbind(df, de)

answered Feb 12 '15 at 03:52

Parfait

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Thanks! Any idea how to fix this if I dont have a second dataframe declared, but instead have each value I want to add to a new row stored as a variable? – Rilcon42 Feb 12 '15 at 03:57
12

Try: `newdf<-rbind(df, data.frame(hello="hola", goodbye="ciao"))` OR with variable: `newdf<-rbind(df, data.frame(hello=var1, goodbye=var2))` – Parfait Feb 12 '15 at 04:04

score 83 · Answer 3 · answered Dec 22 '18 at 20:37

83

There's now add_row() from the tibble or tidyverse packages.

library(tidyverse)
df %>% add_row(hello = "hola", goodbye = "ciao")

Unspecified columns get an NA.

answered Dec 22 '18 at 20:37

Joe

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I liked this approach if you stick to the tidyverse philosophy. Otherwise basic R syntax is a survival skill that comes in handy when you are in an environment where you don't have privileges to import packages. I particularly like the answer using plain R syntax with `rbind` and `as.matrix` [below](https://stackoverflow.com/a/28467834/1585486) – Pablo Adames Mar 21 '20 at 18:08
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Just I'd like to mention that the library is dplyr. – Ariel Dec 13 '22 at 17:51
@Ariel To be specific, yes. But it's usually going to just be "tidyverse" to load anything else that you might want. – CubicInfinity May 28 '23 at 22:23

score 59 · Answer 4 · edited Sep 14 '17 at 23:19

59

Or, as inspired by @MatheusAraujo:

df[nrow(df) + 1,] = list("v1","v2")

This would allow for mixed data types.

edited Sep 14 '17 at 23:19

Mabyn

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answered Aug 03 '17 at 09:15

Ytsen de Boer

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score 24 · Answer 5 · edited Jun 24 '18 at 19:06

I like list instead of c because it handles mixed data types better. Adding an additional column to the original poster's question:

#Create an empty data frame
df <- data.frame(hello=character(), goodbye=character(), volume=double())
de <- list(hello="hi", goodbye="bye", volume=3.0)
df = rbind(df,de, stringsAsFactors=FALSE)
de <- list(hello="hola", goodbye="ciao", volume=13.1)
df = rbind(df,de, stringsAsFactors=FALSE)

Note that some additional control is required if the string/factor conversion is important.

Or using the original variables with the solution from MatheusAraujo/Ytsen de Boer:

df[nrow(df) + 1,] = list(hello="hallo",goodbye="auf wiedersehen", volume=20.2)

Note that this solution doesn't work well with the strings unless there is existing data in the dataframe.

If `hello` and `goodbye` are in character in `df`, you can do the following. You do not necessarily use names in a list. `df <- data.frame(hello = "hi", goodbye = "bye", volume = 1,stringsAsFactors = FALSE); rbind(df, list("hola", "ciao", 100))`. — jazzurro, Jun 17 '19 at 07:47

score 13 · Answer 6 · answered Feb 12 '15 at 01:37

13

Not terribly elegant, but:

data.frame(rbind(as.matrix(df), as.matrix(de)))

From documentation of the rbind function:

For rbind column names are taken from the first argument with appropriate names: colnames for a matrix...

answered Feb 12 '15 at 01:37

J. Win.

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This solution works without needing to specify the columns to add, which is much better for applications on large datasets – Phil_T Sep 01 '17 at 23:50

score 3 · Answer 7 · answered Apr 28 '20 at 16:22

If you want to make an empty data frame and add contents in a loop, the following may help:

# Number of students in class
student.count <- 36

# Gather data about the students
student.age <- sample(14:17, size = student.count, replace = TRUE)
student.gender <- sample(c('male', 'female'), size = student.count, replace = TRUE)
student.marks <- sample(46:97, size = student.count, replace = TRUE)

# Create empty data frame
student.data <- data.frame()

# Populate the data frame using a for loop
for (i in 1 : student.count) {
    # Get the row data
    age <- student.age[i]
    gender <- student.gender[i]
    marks <- student.marks[i]

    # Populate the row
    new.row <- data.frame(age = age, gender = gender, marks = marks)

    # Add the row
    student.data <- rbind(student.data, new.row)
}

# Print the data frame
student.data

Hope it helps :)

score 3 · Answer 8 · answered Sep 20 '21 at 12:01

3

To build a data.frame in a loop:

df <- data.frame()
for(i in 1:10){
  df <- rbind(df, data.frame(str="hello", x=i, y=i*10))
}

answered Sep 20 '21 at 12:01

Stuart Ball

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score 1 · Answer 9 · edited Aug 16 '18 at 17:42

There is a simpler way to append a record from one dataframe to another IF you know that the two dataframes share the same columns and types. To append one row from xx to yy just do the following where i is the i'th row in xx.

yy[nrow(yy)+1,] <- xx[i,]

Simple as that. No messy binds. If you need to append all of xx to yy, then either call a loop or take advantage of R's sequence abilities and do this:

zz[(nrow(zz)+1):(nrow(zz)+nrow(yy)),] <- yy[1:nrow(yy),]

score 1 · Answer 10 · answered Aug 19 '18 at 00:20

I need to add stringsAsFactors=FALSE when creating the dataframe.

> df <- data.frame("hello"= character(0), "goodbye"=character(0))
> df
[1] hello   goodbye
<0 rows> (or 0-length row.names)
> df[nrow(df) + 1,] = list("hi","bye")
Warning messages:
1: In `[<-.factor`(`*tmp*`, iseq, value = "hi") :
  invalid factor level, NA generated
2: In `[<-.factor`(`*tmp*`, iseq, value = "bye") :
  invalid factor level, NA generated
> df
  hello goodbye
1  <NA>    <NA>
>

.

> df <- data.frame("hello"= character(0), "goodbye"=character(0), stringsAsFactors=FALSE)
> df
[1] hello   goodbye
<0 rows> (or 0-length row.names)
> df[nrow(df) + 1,] = list("hi","bye")
> df[nrow(df) + 1,] = list("hola","ciao")
> df[nrow(df) + 1,] = list(hello="hallo",goodbye="auf wiedersehen")
> df
  hello         goodbye
1    hi             bye
2  hola            ciao
3 hallo auf wiedersehen
>

score 1 · Answer 11 · answered Sep 16 '18 at 18:33

Make certain to specify stringsAsFactors=FALSE when creating the dataframe:

> rm(list=ls())
> trigonometry <- data.frame(character(0), numeric(0), stringsAsFactors=FALSE)
> colnames(trigonometry) <- c("theta", "sin.theta")
> trigonometry
[1] theta     sin.theta
<0 rows> (or 0-length row.names)
> trigonometry[nrow(trigonometry) + 1, ] <- c("0", sin(0))
> trigonometry[nrow(trigonometry) + 1, ] <- c("pi/2", sin(pi/2))
> trigonometry
  theta sin.theta
1     0         0
2  pi/2         1
> typeof(trigonometry)
[1] "list"
> class(trigonometry)
[1] "data.frame"

Failing to use stringsAsFactors=FALSE when creating the dataframe will result in the following error when attempting to add the new row:

> trigonometry[nrow(trigonometry) + 1, ] <- c("0", sin(0))
Warning message:
In `[<-.factor`(`*tmp*`, iseq, value = "0") :
  invalid factor level, NA generated

score 1 · Answer 12 · edited Jan 19 '22 at 19:54

1

I think,

rbind.data.frame(df, de)

should do the trick

edited Jan 19 '22 at 19:54

buddemat

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answered Jan 13 '22 at 12:09

user17924177

11
1

Your answer could be improved with additional supporting information. Please [edit] to add further details, such as citations or documentation, so that others can confirm that your answer is correct. You can find more information on how to write good answers [in the help center](/help/how-to-answer). – Community Jan 13 '22 at 12:43

score 1 · Answer 13 · answered May 11 '22 at 17:23

In dplyr >= 1.0.0 you could use row_insert:

df1 <- data.frame(hello = "hi", goodbye = "bye")
df2 <- data.frame(hello = "hola", goodbye = "ciao")

library(dplyr)

df1 %>% 
  rows_insert(df2)
Matching, by = "hello"
  hello goodbye
1    hi     bye
2  hola    ciao

Note: all columns in df2 must exist in df1, but not all columns in df1 have to be in df2.

For additional behavior, there are other row_* options. For example, you could use row_upsert which will overwrite the values if they exist already, otherwise it will insert them:

df2 <- data.frame(hello = c("hi", "hola"), goodbye = c("goodbye", "ciao"))

library(dplyr)

df1 %>% 
  rows_upsert(df2)
Matching, by = "hello"
  hello goodbye
1    hi goodbye # bye updated to goodbye since "hi" was already in data frame
2  hola    ciao # inserted because "hola" was not in the data frame

These functions work by matching key columns. If the by argument is not specified then the default behavior is to match the first column in the second data frame (df2 in this example) to the first data frame (df1 in this example).

score 0 · Answer 14 · answered Jan 20 '21 at 15:03

To formalize what someone else used setNames for:

add_row <- function(original_data, new_vals_list){ 
  # appends row to dataset while assuming new vals are ordered and classed appropriately. 
  # new_vals must be a list not a single vector. 
  rbind(
    original_data,
    setNames(data.frame(new_vals_list), colnames(original_data))
    )
  }

It preserves class when legal and passes errors elsewhere.

m <- mtcars[ ,1:3]
m$cyl <- as.factor(m$cyl)
str(m)

#'data.frame':  32 obs. of  3 variables:
# $ mpg : num  21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
# $ cyl : Factor w/ 3 levels "4","6","8": 2 2 1 2 3 2 3 1 1 2 ...
# $ disp: num  160 160 108 258 360 ...

Factor preserved when adding 4, even though it was passed as a numeric.

str(add_row(m, list(20,4,160)))
#'data.frame':  33 obs. of  3 variables:
# $ mpg : num  21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
# $ cyl : Factor w/ 3 levels "4","6","8": 2 2 1 2 3 2 3 1 1 2 ... 
# $ disp: num  160 160 108 258 360 ...

Attempting to pass a non- 4,6,8 would return an error that factor level is invalid.

str(add_row(m, list(20,3,160)))
# 'data.frame': 33 obs. of  3 variables:
# $ mpg : num  21 21 22.8 21.4 18.7 18.1 14.3 24.4 22.8 19.2 ...
# $ cyl : Factor w/ 3 levels "4","6","8": 2 2 1 2 3 2 3 1 1 2 ...
# $ disp: num  160 160 108 258 360 ...
Warning message:
In `[<-.factor`(`*tmp*`, ri, value = 3) :
  invalid factor level, NA generated

score 0 · Answer 15 · edited Dec 19 '21 at 02:41

I will add to the other suggestions. I use the base r code to create a dataframe:

data_set_name <- data.frame(data_set)

Now I always suggest making a duplicate of the original data frame just in case you need to go back or test something out. I listed that below:

data_set_name_copy <- data_set_name

Now if you wanted to add a new column the code would look like the following:

data_set_name_copy$Name_of_New_Column <- Data_for_New_Column

The $ signifies that you are adding a new column and right after as outlined you insert the nomenclature/name for your new entry.

score 0 · Answer 16 · answered Jan 05 '23 at 23:28

Building on prior answers, new rows can be added to a data frame using replacement functions. Replacement functions can encapsulate code complexity, which is advantageous when row additions are occurring multiple times in the same code.

Multiple versions of such a function are presented in order of increasing complexity.

Version 1: This version is like the answers by @MatheusAraujo or @YtsendeBoer. It is compact and useful if all column data for the new row is present in a fixed order.

'new_row<-'<- function(x, value){x[nrow(x) + 1,] <- value; x}

 df <- data.frame(A = 1,  B = 2,  C = 3)
 new_row(df) <- c(4,  5,  6)
 new_row(df) <- list(7,  8,  9)

Version 2: Though slightly longer, this version improves traceability by keying new data to the column name. All named columns must be present, though not necessarily in order, when adding a new row.

'new_row<-'<- function(x, value){
    x[nrow(x) + 1,] <- sapply(names(x), function(y){value[y]}); x
 }

 df <- data.frame(A = 1,  B = 2,  C = 3)
 new_row(df) <- c(B = 1, C = 2,  A = 3)     
 new_row(df) <- list(C = 1,  A = 2,  B = 3)
 new_row(df) <- data.frame(A = 3,  B = 4,  C = 5)

Version 3: This bulkier version will work when columns are missing and when new named columns are included. This is advantageous when new rows need adding while column data is still incomplete or when new rows only partially fit the data frame.

'new_row<-'<- function(x, value){
  x[names(value)[!is.element(names(value), names(x))]] <- numeric()
  x[nrow(x) + 1,] <- sapply(names(x), function(y){
    if(is.element(y,names(value))){return(value[y])}else{return(NA)}
  }); x}  

df <- data.frame(A = 1,  B = 2,  C = 3)

new_row(df) <- NA
new_row(df) <- c(A = 5)
new_row(df) <- list(C = 1,  A = 2, B = 1)
new_row(df) <- data.frame(Z = 1000)

How to add a row to a data frame in R?

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