CODEWITHSUNDEEP
rdataframetransposena
Tom H
Tom H

Reputation: 85

Conditional Transpose with NAs in R

I've used the dplyr package to summarize some data. The dataframe I produced looks something like this:

Iteration   Degree   Proportion
    1          0         .5
    1          30        .7
    1          60        .8
    2          0         .6
    2          30        .9
    3          0         .3
    3          30        .8
    3          60        .8

I would like to transform my dataframe into a new dataframe where the each of the 3 degree conditions are their own column, and the corresponding proportion values are filled in. MOST IMPORTANTLY I need to impute 'NA' values whenever an iteration does not have a degree value.

The dataframe I am thinking of would look something like this:

Iteration  0_Degree  30_Degree  60_Degree
     1        .5         .7         .8
     2        .6         .9         NA
     3        .3         .8         .8

Identifying where NAs need to be filled in is the major challenge I am running at the moment.

Does anyone have an idea for how I might accomplish this?

Thank you!

Upvotes: 2

Views: 248

Answers (2)

Boris
Boris

Reputation: 36

This can easily be achieved with the spread function from the tidyr package. tidyr is part of the the tidyverse.

Simply use:

library(tidyverse)
df %>% 
  spread(key = Degree, value = Proportion)

The default option for filling missing observations is fill = 'NA'.

Upvotes: 2

G. Grothendieck
G. Grothendieck

Reputation: 269501

Omit as.data.frame if you don't need it as a data frame. No packages are used.

as.data.frame(tapply(dd[[3]], dd[-3], c))

giving:

    0  30  60
1 0.5 0.7 0.8
2 0.6 0.9  NA
3 0.3 0.8 0.8

Note

The input in reproducible form is:

dd <- structure(list(Iteration = c(1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L), 
    Degree = c(0L, 30L, 60L, 0L, 30L, 0L, 30L, 60L), Proportion = c(0.5, 
    0.7, 0.8, 0.6, 0.9, 0.3, 0.8, 0.8)), class = "data.frame", row.names = c(NA, 
-8L))

Upvotes: 1

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