Automatically expanding an R factor into a collection of 1/0 indicator variables for every factor level
Use the model.matrix function:
model.matrix( ~ Species - 1, data=iris )
R: Expanding an R factor into dummy columns for every factor level
This worked for me perfectly:
library(reshape2)
m <- acast(data = d, User ~ Code)
The only thing was that it produced NAs, instead of 0s, but this can be easily changed with this:
m[is.na(m)] <- 0
Convert a factor to indicator variables?
One way is to use model.matrix():
model.matrix(~Species, iris)
(Intercept) Speciesversicolor Speciesvirginica
1 1 0 0
2 1 0 0
3 1 0 0
....
148 1 0 1
149 1 0 1
150 1 0 1
attr(,"assign")
[1] 0 1 1
attr(,"contrasts")
attr(,"contrasts")$Species
[1] "contr.treatment"
Split variable into multiple multiple factor variables
A fast and easy way is to use fastDummies::dummy_cols:
fastDummies::dummy_cols(df, "x")
An alternative with tidyverse functions:
library(tidyverse)
df %>%
left_join(., df %>% mutate(value = 1) %>%
pivot_wider(names_from = x, values_from = value, values_fill = 0) %>%
relocate(n, sort(colnames(.)[-1])))
output
> dummmy <- fastDummies::dummy_cols(df, "x")
> colnames(dummy)[-c(1,2)] <- LETTERS
> dummy
n x A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
1 1 Z 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
2 2 Q 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0
3 3 E 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
4 4 H 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
5 5 T 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0
6 6 X 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0
7 7 R 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0 0
8 8 F 0 0 0 0 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0
9 9 Z 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1
10 10 S 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0 1 0 0 0 0 0 0 0
Benchmark
Since there are many solutions and the question involves a large dataset, a benchmark might help. The nnet solution is the fastest according to the benchmark.
set.seed(1)
df <- data.frame(n = seq(1:1000000), x = sample(LETTERS, 1000000, replace = T))
library(microbenchmark)
bm <- microbenchmark(
fModel.matrix(),
fContrasts(),
fnnet(),
fdata.table(),
fFastDummies(),
fDplyr(),
times = 10L,
setup = gc(FALSE)
)
autoplot(bm)
converting a DocumentTermMatrix to factor
If you have a DocumentTermMatrix as defined in the tm package, you can just set the count of each word to one, by replacing all values in "v" by 1 as so:
dtm[["v"]] <- rep(1, length(dtm[["v"]]))
Full reprex:
library(tm)
#> Loading required package: NLP
data("crude")
dtm <- DocumentTermMatrix(crude)
head(inspect(dtm))
#> <<DocumentTermMatrix (documents: 20, terms: 1266)>>
#> Non-/sparse entries: 2255/23065
#> Sparsity : 91%
#> Maximal term length: 17
#> Weighting : term frequency (tf)
#> Sample :
#> Terms
#> Docs and for its mln oil opec prices said that the
#> 144 9 5 6 4 11 10 3 9 10 17
#> 236 7 4 8 4 7 6 2 6 4 15
#> 237 11 3 3 1 3 1 0 0 1 30
#> 242 3 1 0 0 3 2 1 3 0 6
#> 246 9 6 3 0 4 1 0 4 2 18
#> 248 6 2 2 3 9 6 7 5 2 27
#> 273 5 4 0 9 5 5 4 5 0 21
#> 489 5 4 2 2 4 0 2 2 1 8
#> 502 6 5 2 2 4 0 2 2 1 13
#> 704 5 3 1 0 3 0 2 3 3 21
#> Terms
#> Docs and for its mln oil opec prices said that the
#> 144 9 5 6 4 11 10 3 9 10 17
#> 236 7 4 8 4 7 6 2 6 4 15
#> 237 11 3 3 1 3 1 0 0 1 30
#> 242 3 1 0 0 3 2 1 3 0 6
#> 246 9 6 3 0 4 1 0 4 2 18
#> 248 6 2 2 3 9 6 7 5 2 27
dtm[["v"]] <- rep(1, length(dtm[["v"]]))
head(inspect(dtm))
#> <<DocumentTermMatrix (documents: 20, terms: 1266)>>
#> Non-/sparse entries: 2255/23065
#> Sparsity : 91%
#> Maximal term length: 17
#> Weighting : term frequency (tf)
#> Sample :
#> Terms
#> Docs and for its last oil prices reuter said the was
#> 144 1 1 1 1 1 1 1 1 1 1
#> 236 1 1 1 1 1 1 1 1 1 1
#> 237 1 1 1 1 1 0 1 0 1 1
#> 242 1 1 0 0 1 1 1 1 1 1
#> 246 1 1 1 1 1 0 1 1 1 1
#> 248 1 1 1 1 1 1 1 1 1 1
#> 273 1 1 0 1 1 1 1 1 1 1
#> 489 1 1 1 0 1 1 1 1 1 0
#> 502 1 1 1 0 1 1 1 1 1 0
#> 704 1 1 1 0 1 1 1 1 1 0
#> Terms
#> Docs and for its last oil prices reuter said the was
#> 144 1 1 1 1 1 1 1 1 1 1
#> 236 1 1 1 1 1 1 1 1 1 1
#> 237 1 1 1 1 1 0 1 0 1 1
#> 242 1 1 0 0 1 1 1 1 1 1
#> 246 1 1 1 1 1 0 1 1 1 1
#> 248 1 1 1 1 1 1 1 1 1 1
Created on 2022-06-26 by the reprex package (v2.0.1)
fit an `lm` model for every level of a factor
You can nest the dataframe and use map to apply lm for each factor_gear.
library(dplyr)
mtcars %>%
group_by(factor_gear) %>%
tidyr::nest() %>%
mutate(model = map(data, ~lm(mpg ~ cyl, data = .x)))
# factor_gear data model
# <fct> <list> <list>
#1 4 <tibble [12 × 11]> <lm>
#2 3 <tibble [15 × 11]> <lm>
#3 5 <tibble [5 × 11]> <lm>
In the new dplyr you can use cur_data to refer to current data in group which avoids the need of nest and map.
mtcars %>%
group_by(factor_gear) %>%
summarise(model = list(lm(mpg ~ cyl, data = cur_data())))
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