Hello,
Here is a comparative test of 3 options.
cumstats::Mode returns a list with two members,
Values: all the modes.
Frequency: their frequency
The value of the mode must be extracted after. cumstats::Mode is by far
the slowest but returns more information.
The function below is in this StackOverflow post [1]. It's the fastest
but only returns one mode, the first found.
set.seed(2020)
V <- LETTERS
df <- replicate(100, sample(V, 1000, replace = TRUE))
df <- as.data.frame(t(df))
Mode <- function(x) {
ux <- unique(x)
ux[which.max(tabulate(match(x, ux)))]
}
res1 <- apply(df, 1, prettyR::Mode)
res2 <- apply(df, 1, cumstats::Mode)
res3 <- apply(df, 1, Mode)
head(res1)
res2vals <- lapply(res2, '[[', 1)
head(res2vals)
head(res3)
library(microbenchmark)
mb <- microbenchmark(
pre = apply(df, 1, prettyR::Mode),
cum = cumstats::Mode(x),
so = apply(x, 1, Mode),
times = 10
)
print(mb, unit = "relative", order = "median")
[1] https://stackoverflow.com/a/8189441/8245406
Hope this helps,
Rui Barradas
Às 17:12 de 31/10/20, Luigi Marongiu escreveu:
Thank you. The problem was not finding the mode but applying it the R
way (I have the tendency to loop into each line of the dataframes,
which I believe is NOT the R way).
I'll try them.
Best regards
Luigi
On Sat, Oct 31, 2020 at 5:40 PM Bert Gunter <bgunter.4...@gmail.com> wrote:
As usual, a web search ("find statistical mode in R") brought up something that
is possibly useful -- Did you try this before posting? If not, please do so in future and
let us know what your results were if you subsequently post here.
Here's what SO suggested:
Mode <- function(x) {
ux <- unique(x)
ux[which.max(tabulate(match(x, ux)))]
}
# ergo:
apply(as.matrix(df),1,Mode)
Note that all the functionality in Mode is via .Internal functions. So you can
determine whether this is faster than Jim's code for your use case, but I'm
pretty sure it will be faster than yours. However, note that this gives only
the value of the *first* mode if there is more than one, while Jim's code
alerts you to multiple modes.
Bert Gunter
"The trouble with having an open mind is that people keep coming along and sticking
things into it."
-- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
On Sat, Oct 31, 2020 at 2:29 AM Jim Lemon <drjimle...@gmail.com> wrote:
Hi Luigi,
If I understand your request:
library(prettyR)
apply(as.matrix(df),1,Mode)
[1] "C" "B" "D" ">1 mode" ">1 mode" ">1 mode" "D"
[8] "C" "B" ">1 mode"
Jim
On Sat, Oct 31, 2020 at 7:56 PM Luigi Marongiu <marongiu.lu...@gmail.com>
wrote:
Hello,
I have a large dataframe (1 000 000 rows, 1000 columns) where the
columns contain a character. I would like to determine the most common
character for each row.
In the example below, I can parse one row at the time and find the
most common character (apart for ties...). But I think this will be
very slow and memory consuming.
Is there a way to run it more efficiently?
Thank you
```
V = c("A", "B", "C", "D")
df = data.frame(n = 1:10,
col_01 = sample(V, 10, replace = TRUE, prob = NULL),
col_02 = sample(V, 10, replace = TRUE, prob = NULL),
col_03 = sample(V, 10, replace = TRUE, prob = NULL),
col_04 = sample(V, 10, replace = TRUE, prob = NULL),
col_05 = sample(V, 10, replace = TRUE, prob = NULL),
stringsAsFactors = FALSE)
q = vector()
for(i in 1:nrow(df)) {
x = as.vector(t(df[i,2:ncol(df)]))
q[i] = names(which.max(table(x)))
}
df$most = q
```
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