I have a vector based on col names which looks like
x <- c("C", "A", "T")
my dataframe looks like with rownames and colnames defined.
names A B C D T
Dan 1 0 1 0 1
Joe 0 1 0 1 0
I want to order the dataframe so the columns in the vector appear first followed by columns not in the vector
names C A T B D
Dan 1 1 1 0 0
Joe 0 0 0 1 1
Thanks
The following will rearrange your data to set the columns specified in the vector x at the beginning, and the remaining columns in their original order afterwards.
x <- c("C", "A", "T")
mydata <- mydata[, c(x, setdiff(names(mydata), x))]
If the names column should stay at the first position and is not specified within x, use (Thanks #StevenBeaupré for pointing it out and providing the code):
mydata <- mydata[, c(names(mydata)[1], x, setdiff(names(mydata)[-1], x))]
Small data example:
mydata <- data.frame(names = c("Dan", "Joe"), A = c(1, 0), B = c(0,1),
C = c(1, 0), D = c(0,1), T = c(1, 0))
> mydata
names A B C D T
1 Dan 1 0 1 0 1
2 Joe 0 1 0 1 0
mydata <- mydata[, c(names(mydata)[1], x, setdiff(names(mydata)[-1], x))]
> mydata
names C A T B D
1 Dan 1 1 1 0 0
2 Joe 0 0 0 1 1
Related
I am wondering what an efficient approach to the following question would be:
Suppose I have three characters in group 1 and two characters in group 2:
group_1 = c("X", "Y", "Z")
group_2 = c("A", "B")
Clearly, the "all" possible combinations for group_1 and group_2 are given by:
group_1_combs = data.frame(X = c(0,1,0,0,1,1,0,1),
Y = c(0,0,1,0,1,0,1,1),
Z = c(0,0,0,1,0,1,1,1))
group_2_combs = data.frame(A = c(0,1,0,1),
B = c(0,0,1,1))
My question is the following:
(1) How do I go from group_1 to group_1_combs efficiently (given that the character vector might be large).
(2) How do I do an "all possible" combinations of each row of group_1_combs and group_2_combs? Specifically, I want a "final" data.frame where each row of group_1_combs is "permuted" with every row of group_2_combs. This means that the final data.frame would have 8 x 4 rows (since there are 8 rows in group_1_combs and 4 rows in group_2_combs) and 5 columns (X,Y,Z,A,B).
Thanks!
You want expand.grid and merge:
Question 1:
group_1_combs <- expand.grid(setNames(rep(list(c(0, 1)), length(group_1)), group_1))
group_2_combs <- expand.grid(setNames(rep(list(c(0, 1)), length(group_2)), group_2))
Question 2:
> merge(group_1_combs, group_2_combs)
X Y Z A B
1 0 0 0 0 0
2 1 0 0 0 0
3 0 1 0 0 0
4 1 1 0 0 0
5 0 0 1 0 0
6 1 0 1 0 0
7 0 1 1 0 0
...
Or you can go directly to the merged data.frame:
group_12 <- c(group_1, group_2)
expand.grid(setNames(rep(list(c(0, 1)), length(group_12)), group_12))
I have a data frame of the following way
dat <- data.frame(A=c("D", "A", "D", "B"), B=c("B", "B", "D", "R"), C=c("A", "D", "C", ""), D=c("D", "C", "A", "A"))
My idea is to create a matrix with this information, based on the number of occasions that each column variable refers to the other columns (and ignore when referring to other things that are not in one of the columns (e.g. "R")). So I want to fill the following matrix:
n <- ncol(dat)
names_d <- colnames(dat)
mat <- matrix(0, nrow=n, ncol=n)
rownames(mat) <- names_d
colnames(mat) <- names_d
So in the end, I would have something like this:
A B C D
A 1 1 0 2
B 0 2 0 1
C 1 0 1 1
D 2 0 1 1
Which would be the most efficient way of doing this in R?
You can try the code below
> t(sapply(dat, function(x) table(factor(x, levels = names(dat)))))
A B C D
A 1 1 0 2
B 0 2 0 1
C 1 0 1 1
D 2 0 1 1
or
> t(xtabs(~., subset(stack(dat), values != "")))
values
ind A B C D
A 1 1 0 2
B 0 2 0 1
C 1 0 1 1
D 2 0 1 1
Another option is stack with table
table(subset(stack(dat), nzchar(values) & values != 'R'))
My variable is as follows
variable
D
D
B
C
B
D
C
C
D
I want to make the column in the above figure below
variable
B
C
D
D
0
0
1
D
0
0
1
B
1
0
0
C
0
1
0
B
1
0
0
D
0
0
1
C
0
1
0
C
0
1
0
D
0
0
1
But I don't want a code like the one below. Because the number of factors in the variable column is too many
data = data %>% mutate(B=ifelse(variable=="B", 1,0),
C=ifelse(variable=="C", 1,0),
D=ifelse(variable=="D", 1,0))
Here is a base R approach. We can first find all unique variable values from the data frame. Then, sapply over that vector and generate a new column for each value. Finally, we can rbind this new data frame of 0/1 valued columns to the original data frame.
cols <- sort(unique(df$variable))
df2 <- sapply(cols, function(x) ifelse(df$variable == x, 1, 0))
df <- cbind(df, df2)
df
variable B C D
1 D 0 0 1
2 D 0 0 1
3 B 1 0 0
4 C 0 1 0
5 B 1 0 0
6 D 0 0 1
7 C 0 1 0
8 C 0 1 0
9 D 0 0 1
Data:
df <- data.frame(variable=c("D", "D", "B", "C", "B",
"D", "C", "C", "D"),
stringsAsFactors=FALSE)
Try this with reshaping and duplicating the original variable in order to have a reference for values. Then, you can reshape to obtain the expected output:
library(dplyr)
library(tidyr)
#Code
new <- df %>% mutate(Var=variable,Val=1,id=row_number()) %>%
pivot_wider(names_from = Var,values_from=Val,values_fill = 0) %>%
select(-id)
Output:
# A tibble: 9 x 4
variable D B C
<chr> <dbl> <dbl> <dbl>
1 D 1 0 0
2 D 1 0 0
3 B 0 1 0
4 C 0 0 1
5 B 0 1 0
6 D 1 0 0
7 C 0 0 1
8 C 0 0 1
9 D 1 0 0
Some data used:
#Data
df <- structure(list(variable = c("D", "D", "B", "C", "B", "D", "C",
"C", "D")), class = "data.frame", row.names = c(NA, -9L))
1) model.matrix
model.matrix will generate column names like variableB so the last line removes the variable part to ensure that the column names are exactly the same as in the question. Omit the last line if it is not important that the column names be exactly as shown there.
dat2 <- cbind(dat, model.matrix(~ variable - 1, dat))
names(dat2) <- sub("variable(.)", "\\1", names(dat2))
giving:
> dat2
variable B C D
1 D 0 0 1
2 D 0 0 1
3 B 1 0 0
4 C 0 1 0
5 B 1 0 0
6 D 0 0 1
7 C 0 1 0
8 C 0 1 0
9 D 0 0 1
2) outer
This can also be done using outer as shown. Each component of variable is compared to each level. We name the levels so that outer uses them as column names. The output is the same.
levs <- sort(unique(dat$variable))
names(levs) <- levs
cbind(dat, +outer(dat$variable, levs, `==`))
Note
The input in reproducible form:
Lines <- "
variable
D
D
B
C
B
D
C
C
D"
dat <- read.table(text = Lines, header = TRUE)
I have a data frame in R with binary entries for three variables (a, b and c)
set.seed(12)
library(dplyr)
df <- data.frame(a = rbinom(10, 1, 0.5), b = rbinom(10, 2, 0.3), c = rbinom(10, 1, 0.8))
df
a b c
1 0 1 1
2 1 1 0
3 0 0 1
4 1 1 1
5 0 0 1
6 1 1 1
7 1 0 1
8 0 0 1
9 0 1 1
10 1 0 1
I also have a numeric vector whose names are the same of the columns in df:
vec <- rnorm(3, 1, 0.5)
names(vec) <- colnames(df)
vec
a b c
1.3369906 2.0360179 0.7294857
Here's the thing: for each column in df and for each observation, if the variable has a value of 1, then replace the 1 by the values in vec. Otherwise, if the original value is 0, then I want to keep it. I tried to perform a loop, but it didn't work well.
for(i in 1:ncol(df)){
df[,i][df==1] <- vec[i]
}
I feel it might be due to the need to specify the matching pattern between the matrix and the vector. Is there an alternative way to do that?
If you want to use a loop, the following should do the trick.
for(i in 1:ncol(df)){
for(j in 1:nrow(df)){
if(df[j,i]>=1){df[j,i]<-vec[colnames(df[i])]}
else{df[j,i]<-df[j,i]}
}
}
I have a data frame like this, called df:
a b c d e f
b c f a a a
d f a b c c
f e d f f d
The first row is actually the column name. Let's take an example to explain the meaning here: df[1,1] is b, which means there is relation from a to b, so the values in the column means there is relation from the 'column name' to that entry.
I want create a matrix(df1) with 6*6 dimension, column and row names are both column names of df. The (i,j) entry is 1, if there is relation from 'i' to 'j', otherwise, 0.
The output I want is:
a b c d e f
a 0 1 0 1 0 1
b 0 0 1 0 1 1
c 1 0 0 1 0 1
d 1 1 0 0 0 1
e 1 0 1 0 0 1
f 1 0 1 1 0 0
How to do this with a loop in R?
How to do this without a loop, and only use basic R?
How to do this using some fancy packages in R?
Using the reshape2 package, this is one way to go. My sample data has all columns as character. You use melt() to reshape your data in a long format. Then, you use dcast() from the same package.
library(magrittr)
library(reshape2)
melt(mydf, measure.vars = names(mydf)) %>%
dcast(variable ~ value, length)
variable a b c d e f
1 a 0 1 0 1 0 1
2 b 0 0 1 0 1 1
3 c 1 0 0 1 0 1
4 d 1 1 0 0 0 1
5 e 1 0 1 0 0 1
6 f 1 0 1 1 0 0
EDIT
As mentioned below by akrun, you can do all work using recast() in the reshape2 package.
recast(mydf, measure.var= names(mydf),variable~value, length)
DATA
mydf <- structure(list(a = c("b", "d", "f"), b = c("c", "f", "e"), c = c("f",
"a", "d"), d = c("a", "b", "f"), e = c("a", "c", "f"), f = c("a",
"c", "d")), .Names = c("a", "b", "c", "d", "e", "f"), class = "data.frame", row.names = c(NA,
-3L))
Just use table:
table(colnames(mydf)[col(mydf)], unlist(mydf) )
# a b c d e f
# a 0 1 0 1 0 1
# b 0 0 1 0 1 1
# c 1 0 0 1 0 1
# d 1 1 0 0 0 1
# e 1 0 1 0 0 1
# f 1 0 1 1 0 0
If you have multiple matches, then:
pmin(table(colnames(mydf)[col(mydf)], unlist(mydf) ), 1)
You can do this with reshaping.
library(dplyr)
library(tidyr)
data %>%
gather(from, to) %>%
distinct %>%
mutate(value = 1) %>%
spread(to, value, fill = 0)
The other solution using dplyr is really neat and smart. I recommend using that solution.
Here is an alternative solution to your problem using most basic functions in R.
Say your data frame has n columns and m rows i.e. n <- ncol(df) and m <- nrow(df).
output_matrix <- matrix(rep(0, n*n), ncol = n)
for(i in 1:n){
for(j in 1:m){
# UTF to integer conversion
# utf8ToInt("a") = 97
rowWithRelation <- utf8ToInt(df[j, i]) - 96
output_matrix[rowWithRelation, i] <- 1
}
}
rownames(output_matrix) <- letters[seq(from = 1, to = n)]
colnames(output_matrix) <- letters[seq(from = 1, to = n)]