I have a data frame and I want to create a new variable applying a function that works within rows. See the example below.
library(tidyverse)
n <- 100
The functions is as follows
z0 %>% apply(1, function(x) any("y" == x)) -> z0$new
The desired output is z0$new, and I would like to do it using mutate rather than apply.
Thanks for any sugestion
1 Like
cderv
April 18, 2018, 8:57pm
2
You can use mutate and purrr::pmap to iterate in a data.frame over rows.
There is a webinar of RStudio on this topic, it can give you ideas and exampleshttps://www.rstudio.com/resources/webinars/thinking-inside-the-box-you-can-do-that-inside-a-data-frame/
EDIT : /!\ incorrect code here, left for learning purposes. Function in pmap needs to deal with the three elements not just the first one. See others answers. (and help about pmap) /!\
library(tidyverse)
n <- 100
z0 <- data.frame(A = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)),
B = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)),
C = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)))
z0 %>%
mutate(new = pmap(., ~any("y" == .x)))
#> Warning: le package 'bindrcpp' a été compilé avec la version R 3.4.4
#> A B C new
#> 1 n n y FALSE
#> 2 y y y FALSE
#> 3 n y n FALSE
#> 4 n n y FALSE
#> 5 n y n FALSE
#> 6 y y n FALSE
#> 7 n n y FALSE
#> 8 n n n FALSE
#> 9 <NA> <NA> n NA
#> 10 n y y FALSE
#> 11 y y n FALSE
#> 12 y n n FALSE
#> 13 <NA> <NA> n NA
#> 14 n n n FALSE
#> 15 y y n FALSE
#> 16 n n n FALSE
#> 17 y <NA> n FALSE
#> 18 <NA> n y NA
#> 19 n y n FALSE
#> 20 y n n FALSE
#> 21 y n n FALSE
#> 22 y n y FALSE
#> 23 y y n FALSE
#> 24 n <NA> n FALSE
#> 25 <NA> n n NA
#> 26 <NA> n n NA
#> 27 n n <NA> FALSE
#> 28 n y y FALSE
#> 29 y <NA> y FALSE
#> 30 n y n FALSE
#> 31 y y <NA> FALSE
#> 32 n n y FALSE
#> 33 y n n FALSE
#> 34 y y <NA> FALSE
#> 35 y y y FALSE
#> 36 y y n FALSE
#> 37 y y n FALSE
#> 38 y n <NA> FALSE
#> 39 n y n FALSE
#> 40 n n n FALSE
#> 41 n <NA> n FALSE
#> 42 n <NA> y FALSE
#> 43 n n y FALSE
#> 44 y y y FALSE
#> 45 y n y FALSE
#> 46 n y n FALSE
#> 47 n y n FALSE
#> 48 y n n FALSE
#> 49 <NA> n n NA
#> 50 y y n FALSE
#> 51 y y y FALSE
#> 52 n <NA> y FALSE
#> 53 y n n FALSE
#> 54 n <NA> n FALSE
#> 55 y <NA> y FALSE
#> 56 n n y FALSE
#> 57 n n y FALSE
#> 58 y <NA> n FALSE
#> 59 y n y FALSE
#> 60 y y n FALSE
#> 61 n <NA> n FALSE
#> 62 y n y FALSE
#> 63 y y n FALSE
#> 64 y y n FALSE
#> 65 n y n FALSE
#> 66 y y y FALSE
#> 67 <NA> y <NA> NA
#> 68 y n y FALSE
#> 69 n <NA> <NA> FALSE
#> 70 y n n FALSE
#> 71 n y y FALSE
#> 72 n y <NA> FALSE
#> 73 n y n FALSE
#> 74 n y y FALSE
#> 75 <NA> y n NA
#> 76 <NA> y y NA
#> 77 y n y FALSE
#> 78 n y n FALSE
#> 79 n y y FALSE
#> 80 <NA> y y NA
#> 81 n <NA> n FALSE
#> 82 y n <NA> FALSE
#> 83 n y n FALSE
#> 84 <NA> n y NA
#> 85 y y y FALSE
#> 86 y y n FALSE
#> 87 n y y FALSE
#> 88 y y y FALSE
#> 89 y y y FALSE
#> 90 <NA> n n NA
#> 91 n n y FALSE
#> 92 <NA> n n NA
#> 93 n <NA> n FALSE
#> 94 n n <NA> FALSE
#> 95 n y y FALSE
#> 96 n y y FALSE
#> 97 n n y FALSE
#> 98 n n <NA> FALSE
#> 99 n y n FALSE
#> 100 y y y FALSE
Created on 2018-04-18 by the reprex package (v0.2.0).
1 Like
If you compare your solution with my apply solution they differ. indeed, first row of your example should be TRUE since there is an "y" in column C
I think part of the problem was in setting up z0 as a data.frame coerces A, B, and C into factors that then get coerced into ints in pmap.
Here's one solution, but you have to be explicit about how you use columns A, B, and C with ..1,..2,..3. Check out docs and discussions of pmap for more details obviously.
library(tidyverse)
n <- 100
z0 <- tibble(
A = sample(c("y","n",NA), n, replace = TRUE, prob = c(.4,.4,.1)),
B = sample(c("y","n",NA), n, replace = TRUE, prob = c(.4,.4,.1)),
C = sample(c("y","n",NA), n, replace = TRUE, prob = c(.4,.4,.1))
)
z0 %>%
mutate(new = pmap(., .f=~any( "y" %in% c(..1, ..2, ..3))) %>% unlist)
#> # A tibble: 100 x 4
#> A B C new
#> <chr> <chr> <chr> <lgl>
#> 1 y n y T
#> 2 n n y T
#> 3 n <NA> n F
#> 4 y n y T
#> 5 <NA> y y T
#> 6 n n y T
#> 7 y <NA> y T
#> 8 n n y T
#> 9 n n n F
#> 10 n y y T
#> # ... with 90 more rows
Created on 2018-04-18 by the reprex package (v0.2.0).
cderv
April 19, 2018, 6:14am
6
Sorry I was in a hurry when I answered and yet willing to point you to mutate + pmap.
Thank you @EconomiCurtis for correcting my answer. This is how to use pmap here.
To complete, it is possible to name your arguments' function and use the column name.pmap_lgl to flatten the result.
library(tidyverse)
n <- 100
z0 <- data_frame(A = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)),
B = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)),
C = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)))
z0 %>%
mutate(new = pmap_lgl(., function(A, B, C) any("y" %in% c(A, B, C))))
#> Warning: le package 'bindrcpp' a été compilé avec la version R 3.4.4
#> # A tibble: 100 x 4
#> A B C new
#> <chr> <chr> <chr> <lgl>
#> 1 y n n TRUE
#> 2 n n n FALSE
#> 3 <NA> y n TRUE
#> 4 y y <NA> TRUE
#> 5 <NA> n y TRUE
#> 6 y y n TRUE
#> 7 <NA> n n FALSE
#> 8 n <NA> n FALSE
#> 9 n n y TRUE
#> 10 n y y TRUE
#> # ... with 90 more rows
Created on 2018-04-19 by the reprex package (v0.2.0).
2 Likes
Thank you for your comments. However, behavior of %in% is different from ==. See a complete discussion of this issue here.
r, apply, dplyr
It is worht to note that working with factors is trublesome.
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1 Like
krose
April 19, 2018, 6:56pm
9
You can also use the dplyr rowwise function. It works like a group_by but for rows instead of grouping by a variable.
library(tidyverse)
n <- 100
z0 <- data.frame(A = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)),
B = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)),
C = sample(c("y","n",NA), n, replace = T, prob = c(.4,.4,.1)))
z0$new <- z0 %>% apply(1, function(x) any("y" == x))
identical(
as.tibble(z0),
z0 %>%
dplyr::rowwise() %>%
mutate(new = any("y" == A, "y" == B, "y" == C)) %>%
ungroup()
)
1 Like
Sorry about that. I will adhere to the policies.