I have data that looks similar to these two adjacency matrices:
data1999 <- data.frame(node1=c("A", "A", "B", "D", "B", "C", "D"),
node2=c("A", "A", "D", "B", "B", "C", "D"),
link=c(1, 1, 1, 1, 1, 1, 1),
stringsAsFactors = FALSE)
adj.m1999 <- reshape2::acast(data1999, node1 ~ node2)
> adj.m1999
A B C D
A 2 0 0 0
B 0 1 0 1
C 0 0 1 0
D 0 1 0 1
data2000 <- data.frame(node1=c("A", "A", "B", "C", "D", "C", "D"),
node2=c("A", "A", "B", "C", "D", "D", "C"),
link=c(1, 1, 1, 1, 1, 1, 1),
stringsAsFactors = FALSE)
adj.m2000 <- reshape2::acast(data2000, node1 ~ node2)
> adj.m2000
A B C D
A 2 0 0 0
B 0 1 0 0
C 0 0 1 1
D 0 0 1 1
Note that in 1999, node D and B have a link.
Note that in 2000, node D and C have a link.
Based on this information, I want to construct a new adjacency matrix (with all the nodes of my 2000 data) in which B-D and D-B have a value of 1 while the rest has a zero:
> result
A B C D
A 0 0 0 0
B 0 0 1 0
C 0 1 0 0
D 0 0 0 0
In my real-life data, data 1999 can have additional nodes that don't return in 2000 and vice versa.
Any ideas?
In graph theory, the product of two adjacency matices m1 and m2 gives in position (i,j) the number of ways to go from i to j, going first through m1 then through m2. This is related to what you want but not exactly the same as, if we do adj.m1999 %*% adj.m2000, we get:
A B C D
A 4 0 0 0
B 0 1 1 1
C 0 0 1 1
D 0 1 1 1
So for example, you can go from C to D in one way, and that would be C -> C, followed by C -> D.
In your example, you don't take into account links (or edges) that are on the diagonal, and also your graph is not directed, so, if I understand correctly what you want, you could do:
## First make sure that you have in adj.m1999 only nodes that appear in adj.m2000:
adj.m1999 = adj.m1999[row.names(adj.m1999) %in% row.names(adj.m2000),colnames(adj.m1999) %in% colnames(adj.m2000)]
## Then turn both diagonals into zeros:
diag(adj.m1999) = 0
diag(adj.m2000) = 0
## Finally, get the sum of the two products
res = adj.m1999 %*% adj.m2000 + adj.m2000 %*% adj.m1999
A B C D
A 0 0 0 0
B 0 0 1 0
C 0 1 0 0
D 0 0 0 0
Related
set0 <- data.frame(A = c("A", "B", "C", "D", "A", "B", "C", "D", "A", "B"),
B = c("E", "F", "G", "H", "I", "E", "F", "G", "H", "I"))
set0 <- table(set0)
result:
> set0
B
A E F G H I
A 1 0 0 1 1
B 1 1 0 0 1
C 0 1 1 0 0
D 0 0 1 1 0
I know that when I want to change the column order I can use the following:
set0 <- set0[, c(5, 4, 3, 2, 1)]
The above makes it possible to change the order of the column in any way I like.
I am looking for something that makes it possible to do the exact same but for the rownames.
Any ideas?
In the brackets, columns are defined to the right of the comma (that what you did). In case you didn't know yet, rows are defined to the left of the comma. So you can use one of these approaches that you like best:
set0[c(4, 3, 2, 1), ]
set0[4:1, ]
set0[rev(seq(nrow(set0))), ]
set0[c("D", "C", "B", "A"), ]
set0[rev(rownames(set0)), ]
# B
# A E F G H I
# D 0 0 1 1 0
# C 0 1 1 0 0
# B 1 1 0 0 1
# A 1 0 0 1 1
You can use factor to define the order, e.g.,
table(
transform(
set0,
A = factor(A, levels = sort(unique(A), decreasing = TRUE))
)
)
which gives
B
A E F G H I
D 0 0 1 1 0
C 0 1 1 0 0
B 1 1 0 0 1
A 1 0 0 1 1
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)
(Sorry about the poorly made title, wasn't sure how to phrase my question in a single sentence.)
I have a data.table of matches that is sorted chronologically, with P1 representing player1, P2 representing player2, and Res representing whether P1 won ("w") or drew ("d") against P2.
EDIT: a, b, and c in the P1 and P2 columns represent individual players. Think of them as Alice, Bob, and Charlie.
DT <- data.table(time = 1:10,
P1 = c("a", "a", "b", "b", "b", "a", "a", "b", "a", "c"),
Res = c("d", "w", "w", "w", "w", "w", "d", "d", "w", "w"),
P2 = c("b", "c", "c", "a", "a", "c", "c", "a", "b", "b"))
I performed the following operations to count at the time of the match, how many wins P1 had (wins1) and how many losses P2 had (loss2).
DT[, wins1 := shift(cumsum(Res == "w"), 1, fill=0L), by=P1]
DT[, loss2 := shift(cumsum(Res == "w"), 1, fill=0L), by=P2]
I am trying to create the columns for wins2, loss1, draw1, and draw2.
That is, I right now have how many wins P1 had at the time of their match against P2, but I do not know how many losses they had. Hope that makes sense. I'm sure the method to creating these columns are all similar, so if I can make one I should be able to make them all.
The final table should look like the following:
time P1 Res P2 wins1 loss1 draw1 wins2 loss2 draw2
1: 1 a d b 0 0 0 0 0 0
2: 2 a w c 0 0 1 0 0 0
3: 3 b w c 0 0 1 0 1 0
4: 4 b w a 1 0 1 1 0 1
5: 5 b w a 2 0 1 1 1 1
6: 6 a w c 1 2 1 0 2 0
7: 7 a d c 2 2 1 0 3 0
8: 8 b d a 3 0 1 2 2 2
9: 9 a w b 2 2 2 3 0 1
10: 10 c w b 0 3 1 3 1 1
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)]