Sample random rows in dataframe

I am struggling to find the appropriate function that would return a specified number of rows picked up randomly without replacement from a data frame in R language? Can anyone help me out?

13 Answers

First make some data:

> df = data.frame(matrix(rnorm(20), nrow=10)) > df X1 X2 1 0.7091409 -1.4061361 2 -1.1334614 -0.1973846 3 2.3343391 -0.4385071 4 -0.9040278 -0.6593677 5 0.4180331 -1.2592415 6 0.7572246 -0.5463655 7 -0.8996483 0.4231117 8 -1.0356774 -0.1640883 9 -0.3983045 0.7157506 10 -0.9060305 2.3234110 

Then select some rows at random:

> df[sample(nrow(df), 3), ] X1 X2 9 -0.3983045 0.7157506 2 -1.1334614 -0.1973846 10 -0.9060305 2.3234110 

The answer John Colby gives is the right answer. However if you are a dplyr user there is also the answer sample_n:

sample_n(df, 10) 

randomly samples 10 rows from the dataframe. It calls sample.int, so really is the same answer with less typing (and simplifies use in the context of magrittr since the dataframe is the first argument).

The data.table package provides the function DT[sample(.N, M)], sampling M random rows from the data table DT.

library(data.table) set.seed(10) mtcars <- data.table(mtcars) mtcars[sample(.N, 6)] mpg cyl disp hp drat wt qsec vs am gear carb 1: 14.7 8 440.0 230 3.23 5.345 17.42 0 0 3 4 2: 19.2 6 167.6 123 3.92 3.440 18.30 1 0 4 4 3: 17.3 8 275.8 180 3.07 3.730 17.60 0 0 3 3 4: 21.5 4 120.1 97 3.70 2.465 20.01 1 0 3 1 5: 22.8 4 108.0 93 3.85 2.320 18.61 1 1 4 1 6: 15.5 8 318.0 150 2.76 3.520 16.87 0 0 3 2 

Write one! Wrapping JC's answer gives me:

randomRows = function(df,n){ return(df[sample(nrow(df),n),]) } 

Now make it better by checking first if n<=nrow(df) and stopping with an error.

Just for completeness sake:

dplyr also offers to draw a proportion or fraction of the sample by

df %>% sample_frac(0.33) 

This is very convenient e.g. in machine learning when you have to do a certain split ratio like 80%:20%

EDIT: This answer is now outdated, see the updated version.

In my R package I have enhanced sample so that it now behaves as expected also for data frames:

library(devtools); install_github('kimisc', 'krlmlr') library(kimisc) example(sample.data.frame) smpl..> set.seed(42) smpl..> sample(data.frame(a=c(1,2,3), b=c(4,5,6), row.names=c('a', 'b', 'c')), 10, replace=TRUE) a b c 3 6 c.1 3 6 a 1 4 c.2 3 6 b 2 5 b.1 2 5 c.3 3 6 a.1 1 4 b.2 2 5 c.4 3 6 

This is achieved by making sample an S3 generic method and providing the necessary (trivial) functionality in a function. A call to setMethod fixes everything. The original implementation still can be accessed through base::sample.

Outdated answer. Please use dplyr::sample_frac() or dplyr::sample_n() instead.

In my R package there is a function sample.rows just for this purpose:

install.packages('kimisc') library(kimisc) example(sample.rows) smpl..> set.seed(42) smpl..> sample.rows(data.frame(a=c(1,2,3), b=c(4,5,6), row.names=c('a', 'b', 'c')), 10, replace=TRUE) a b c 3 6 c.1 3 6 a 1 4 c.2 3 6 b 2 5 b.1 2 5 c.3 3 6 a.1 1 4 b.2 2 5 c.4 3 6 

Enhancing sample by making it a generic S3 function was a bad idea, according to comments by Joris Meys to a previous answer.

You could do this:

library(dplyr) cols <- paste0("a", 1:10) tab <- matrix(1:1000, nrow = 100) %>% as.tibble() %>% set_names(cols) tab # A tibble: 100 x 10 a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> 1 1 101 201 301 401 501 601 701 801 901 2 2 102 202 302 402 502 602 702 802 902 3 3 103 203 303 403 503 603 703 803 903 4 4 104 204 304 404 504 604 704 804 904 5 5 105 205 305 405 505 605 705 805 905 6 6 106 206 306 406 506 606 706 806 906 7 7 107 207 307 407 507 607 707 807 907 8 8 108 208 308 408 508 608 708 808 908 9 9 109 209 309 409 509 609 709 809 909 10 10 110 210 310 410 510 610 710 810 910 # ... with 90 more rows 

Above I just made a dataframe with 10 columns and 100 rows, ok?

Now you can sample it with sample_n:

sample_n(tab, size = 800, replace = T) # A tibble: 800 x 10 a1 a2 a3 a4 a5 a6 a7 a8 a9 a10 <int> <int> <int> <int> <int> <int> <int> <int> <int> <int> 1 53 153 253 353 453 553 653 753 853 953 2 14 114 214 314 414 514 614 714 814 914 3 10 110 210 310 410 510 610 710 810 910 4 70 170 270 370 470 570 670 770 870 970 5 36 136 236 336 436 536 636 736 836 936 6 77 177 277 377 477 577 677 777 877 977 7 13 113 213 313 413 513 613 713 813 913 8 58 158 258 358 458 558 658 758 858 958 9 29 129 229 329 429 529 629 729 829 929 10 3 103 203 303 403 503 603 703 803 903 # ... with 790 more rows 

Select a Random sample from a tibble type in R:

library("tibble") a <- your_tibble[sample(1:nrow(your_tibble), 150),] 

nrow takes a tibble and returns the number of rows. The first parameter passed to sample is a range from 1 to the end of your tibble. The second parameter passed to sample, 150, is how many random samplings you want. The square bracket slicing specifies the rows of the indices returned. Variable 'a' gets the value of the random sampling.

As @matt_b indicates, sample_n() & sample_frac() have been soft deprecated in favour of slice_sample(). See the dplyr docs.

Example from docstring:

# slice_sample() allows you to random select with or without replacement mtcars %>% slice_sample(n = 5) mtcars %>% slice_sample(n = 5, replace = TRUE) 

You could do this:

sample_data = data[sample(nrow(data), sample_size, replace = FALSE), ] 

I'm new in R, but I was using this easy method that works for me:

sample_of_diamonds <- diamonds[sample(nrow(diamonds),100),] 

PS: Feel free to note if it has some drawback I'm not thinking about.

The 2021 way of doing this in the tidyverse is:

library(tidyverse) df = data.frame( A = letters[1:10], B = 1:10 ) df #> A B #> 1 a 1 #> 2 b 2 #> 3 c 3 #> 4 d 4 #> 5 e 5 #> 6 f 6 #> 7 g 7 #> 8 h 8 #> 9 i 9 #> 10 j 10 df %>% sample_n(5) #> A B #> 1 e 5 #> 2 g 7 #> 3 h 8 #> 4 b 2 #> 5 j 10 df %>% sample_frac(0.5) #> A B #> 1 i 9 #> 2 g 7 #> 3 j 10 #> 4 c 3 #> 5 b 2 

^{Created on 2021-10-05 by the reprex package (v2.0.0.9000)}