Replace values greater than zero with 1 in r

Question

I want to use tidyverse to take a dataframe df and replace all non-zero values to a value of 1.

So greater than zero or non-zero? And please provide a clear [reproducible example](https://stackoverflow.com/q/5963269/1320535) without screenshots. — Julius Vainora, Jan 16 '19 at 23:40

score 15 · Accepted Answer · answered Jan 16 '19 at 23:43

15

The following will convert all non-zero numeric values to 1:

df.richness %>% mutate_if(is.numeric, ~1 * (. != 0))

while

df.richness %>% mutate_if(is.numeric, ~1 * (. > 0))

will do that with those greater than zero.

answered Jan 16 '19 at 23:43

Julius Vainora

47,421
9
90
102

score 9 · Answer 2 · answered Aug 14 '19 at 15:29

9

Alternatively, if you had only numeric data in the dataframe e.g. with sites as the rownames, this would be a simple way without the tidyverse.

df.richness[df.richness > 0] <- 1

answered Aug 14 '19 at 15:29

A.Elsy

128
1
7

score 0 · Answer 3 · answered Sep 17 '21 at 11:31

We can also convert to logical, then back to numeric:

library(dplyr)

df %>% mutate(across(where(is.numeric), ~+as.logical(.x)))

 num1 logical1 num2 char1
1    1     TRUE    0     a
2    1     TRUE    1     b
3    1     TRUE    1     c
4    0     TRUE    1     d
5    1    FALSE    1     e
6    1    FALSE    1     f
7    0    FALSE    0     g
8    1    FALSE    1     h

data

structure(list(num1 = c(1, 2, 3, 0, 1, 99, 0, 2), logical1 = c(TRUE, 
TRUE, TRUE, TRUE, FALSE, FALSE, FALSE, FALSE), num2 = c(0, 6, 
7, 8, 9, 10, 0, 1), char1 = c("a", "b", "c", "d", "e", "f", "g", 
"h")), class = "data.frame", row.names = c(NA, -8L))

  num1 logical1 num2 char1
1    1     TRUE    0     a
2    2     TRUE    6     b
3    3     TRUE    7     c
4    0     TRUE    8     d
5    1    FALSE    9     e
6   99    FALSE   10     f
7    0    FALSE    0     g
8    2    FALSE    1     h

Replace values greater than zero with 1 in r

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