Passing a column name to R tidyr spread

Question

I am trying to pass an object with the column name to the spread function, but instead of reading the value inside the object it just tries to use the object name itself

Here just a toy example

library(tidyr)
d = (1:4)
n = c("a"," a", "b","b") 
s = c(1, 2,5,7) 

df = data.frame(d,n, s) 

Value <- n
data_wide <- spread(df, Value , s)

Error: Key column 'Value' does not exist in input.

whilst below works fine:

data_wide <- spread(df, n, s)
d  a  a  b
1 1 NA  1 NA
2 2  2 NA NA
3 3 NA NA  5
4 4 NA NA  7

Was that space in 2nd "a" intentional? `n = c("a"," a", "b","b")` — zx8754, Jun 06 '16 at 11:31

zx8754 · Accepted Answer · 2016-06-06T11:29:39.353

13

We can use spread_() to pass variable names as strings:

library(tidyr)
# dummy data
df1 <- data.frame(d = (1:4),
                  n = c("a", "a", "b", "b") ,
                  s = c(1, 2, 5, 7)) 

myKey <- "n"
myValue <- "s"
spread_(data = df1, key_col = myKey , value_col = myValue)

edited Jun 06 '16 at 11:29

answered Jun 06 '16 at 11:22

zx8754

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score 3 · Answer 2 · edited May 23 '17 at 11:46

3

Using data.table

library(data.table)
 dcast(setDT(df), eval(as.name(myValue))~ eval(as.name(myKey)), value.var=myValue)

Regarding passing names in tidyr functions, this answer could also help (which was posted a couple of hours back).

edited May 23 '17 at 11:46

Community

1
1

answered Jun 06 '16 at 11:26

akrun

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score 0 · Answer 3 · answered Jul 27 '18 at 14:12

looks like tidyr recognize it automatically now - it works both ways: type in the column names or put them in strings

stocks <- data.frame(
    time = as.Date('2009-01-01') + 0:9,
    X = rnorm(10, 0, 1),
    Y = rnorm(10, 0, 2),
    Z = rnorm(10, 0, 4)
)
stocksm <- stocks %>% gather(stock, price, -time) # make it to skinny table

# make it wide table
stocksm %>% spread(stock, price)

# time          X           Y          Z
# 1  2009-01-01  0.7444343 -0.07030219  0.9140019
# 2  2009-01-02  1.1988507  2.98659296  5.3044361
# 3  2009-01-03 -0.4344259 -0.11526884 -3.8380602
# 4  2009-01-04  0.8154400  2.08313458 -0.1152524
# 5  2009-01-05  1.1965647 -0.59055846  3.5647410
# ...



# it's same if put column names in strings
stocksm %>% spread('stock', 'price')

# time          X           Y          Z
# 1  2009-01-01  0.7444343 -0.07030219  0.9140019
# 2  2009-01-02  1.1988507  2.98659296  5.3044361
# 3  2009-01-03 -0.4344259 -0.11526884 -3.8380602
# 4  2009-01-04  0.8154400  2.08313458 -0.1152524
# 5  2009-01-05  1.1965647 -0.59055846  3.5647410



# or put in string variables
col1 = 'stock'
col2 = 'price'
stocksm %>% spread(col1, col2)

# time          X           Y          Z
# 1  2009-01-01  0.7444343 -0.07030219  0.9140019
# 2  2009-01-02  1.1988507  2.98659296  5.3044361
# 3  2009-01-03 -0.4344259 -0.11526884 -3.8380602
# 4  2009-01-04  0.8154400  2.08313458 -0.1152524
# 5  2009-01-05  1.1965647 -0.59055846  3.5647410

Passing a column name to R tidyr spread

3 Answers3