You can also use the pivot_longer to do it:
library(tidyverse)
input <- structure(list(...1 = c(92.9925354, 76.0024254, 44.99547465,
28.00536465, 120.0068103, 31.9980405, 85.0071837, 40.1532933,
19.3120917, 113.12581575, 28.45843425, 114.400074, 143.925,
46.439634, 20.7845679, 50.82874575, 36
Paul,
I have snipped away your long message and want to suggest another approach
or way of thinking to consider.
You have received other good suggestions and I likely would have used
something like that, probably within the dplyr/tidyverse but consider
something simpler.
You seem to be viewing a
Às 04:13 de 28/10/2023, Paul Bernal escreveu:
Dear friends,
I have the following dataframe:
dim(alajuela_df)
[1] 126 12
dput(alajuela_df)
structure(list(...1 = c(92.9925354, 76.0024254, 44.99547465,
28.00536465, 120.0068103, 31.9980405, 85.0071837, 40.1532933,
19.3120917, 113.12581575, 28.4584
The tidyverse idiom looks very different but does what you want and I have come
to like it.
What idiom of R one likes, for the mostly small datasets I handle, is largely a
matter
of preferenceds for "readability", itself very personal. Here's my tidyverse
way of doing
what you wanted:
### sta
Hi Iris,
Thank you so much for your valuable feedback. I wonder why your code gives
you 1512 rows, given that the original structure has 12 columns and 126
rows, so I would expect (125*12)+ 9=1,509 total rows.
Cheers,
Paul
El El vie, 27 de oct. de 2023 a la(s) 10:40 p. m., Iris Simmons <
ikwsi...
You are not getting the structure you want because the indexes are
wrong. They should be something more like this:
i <- 0
for (row in 1:nrow(alajuela_df)){
for (col in 1:ncol(alajuela_df)){
i <- i + 1
df[i,1]=alajuela_df[row,col]
}
}
but I think what you are doing can be written much
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