How can I force the dimensions of a table to be equal in R?
For example:
a <- c(0,1,2,3,4,5,1,3,4,5,3,4,5)
b <- c(1,2,3,3,3,3,3,3,3,3,5,5,6)
c <- table(a,b)
print(c)
# b
#a 1 2 3 5 6
# 0 1 0 0 0 0
# 1 0 1 1 0 0
# 2 0 0 1 0 0
# 3 0 0 2 1 0
# 4 0 0 2 1 0
# 5 0 0 2 0 1
However, I am looking for the following result:
print(c)
# b
#a 0 1 2 3 4 5 6
# 0 0 1 0 0 0 0 0
# 1 0 0 1 1 0 0 0
# 2 0 0 0 1 0 0 0
# 3 0 0 0 2 0 1 0
# 4 0 0 0 2 0 1 0
# 5 0 0 0 2 0 0 1
# 6 0 0 0 0 0 0 0
By using factors. table doesn't know the levels of your variable unless you tell it in some way!
a <- c(0,1,2,3,4,5,1,3,4,5,3,4,5)
b <- c(1,2,3,3,3,3,3,3,3,3,5,5,6)
a <- factor(a, levels = 0:6)
b <- factor(b, levels = 0:6)
table(a,b)
# b
#a 0 1 2 3 4 5 6
# 0 0 1 0 0 0 0 0
# 1 0 0 1 1 0 0 0
# 2 0 0 0 1 0 0 0
# 3 0 0 0 2 0 1 0
# 4 0 0 0 2 0 1 0
# 5 0 0 0 2 0 0 1
# 6 0 0 0 0 0 0 0
Edit The general way to force a square cross-tabulation is to do something like
x <- factor(a, levels = union(a, b))
y <- factor(b, levels = union(a, b))
table(x, y)
Related
I have created a prediction matrix for large dataset as follows:
library(mice)
dfpredm <- quickpred(df, mincor=.3)
A B C D E F G H I J
A 0 1 1 1 0 1 0 1 1 0
B 1 0 0 0 1 0 1 0 0 1
C 0 0 0 1 1 0 0 0 0 0
D 1 0 1 0 0 1 0 1 0 1
E 0 1 0 1 0 1 1 0 1 0
**F 0 0 1 0 0 0 1 0 0 0**
G 0 1 0 1 0 0 0 0 0 0
H 1 0 1 0 0 1 0 0 0 1
I 0 1 0 1 1 0 1 0 0 0
J 1 0 1 0 0 1 0 1 0 0
I would like to create a subset of the original df on the basis on dfpredm.
More specifically I would like to do the following:
Let's assume that my dependent variable is F.
According to the prediction matrix F is correlated with C and G.
In addition, C and G are best predicted by D,E and B,D respectively.
The idea is now to create a subset of df based on the dependent variable F,for which in the F row the value is 1.
Fpredictors <- df[,(dfpredm["F",]) == 1]
But also do the same for the variables where the rows in F are 1. I am thinking of first getting the column names like this:
Fpredcol <-colnames(dfpredm[,(dfpredm["c241",]) == 1])
And then doing a for loop with these column names?
For the specific example I would like to end up with the subset.
dfsub <- df[,c("F","C","G","B","E","D")]
I would however like to automate this process. Could anyone show me how to do this?
Here is one strategy that seems like it would work for you:
first_preds <- function(dat, predictor) {
cols <- which(dat[predictor, ] == 1)
names(dat)[cols]
}
# wrap first_preds() for getting best and second best predictors
first_and_second_preds <- function(dat, predictor) {
matches <- first_preds(dat, predictor)
matches <- c(matches, unlist(lapply(matches, function(x) first_preds(dat, x))))
c(predictor, matches) %>% unique()
}
dat[first_and_second_preds(dat, "F")] # order is not exactly the same as your output
F C G D E B
A 1 1 0 1 0 1
B 0 0 1 0 1 0
C 0 0 0 1 1 0
D 1 1 0 0 0 0
E 1 0 1 1 0 1
F 0 1 1 0 0 0
G 0 0 0 1 0 1
H 1 1 0 0 0 0
I 0 0 1 1 1 1
J 1 1 0 0 0 0
Not sure if the ordering in the result is important, but you could add the logic if it is.
Using dat from here (a kinder way to share small R data on SO):
dat <- read.table(
text = "A B C D E F G H I J
A 0 1 1 1 0 1 0 1 1 0
B 1 0 0 0 1 0 1 0 0 1
C 0 0 0 1 1 0 0 0 0 0
D 1 0 1 0 0 1 0 1 0 1
E 0 1 0 1 0 1 1 0 1 0
F 0 0 1 0 0 0 1 0 0 0
G 0 1 0 1 0 0 0 0 0 0
H 1 0 1 0 0 1 0 0 0 1
I 0 1 0 1 1 0 1 0 0 0
J 1 0 1 0 0 1 0 1 0 0",
header = TRUE
)
Something a little more general that would let you use self_select predictors directly:
all_preds <- function(dat, predictors) {
unlist(lapply(predictors, function(x) names(dat)[which(dat[x, ] == 1 )]))
}
dat[all_preds(dat, c("A", "B"))]
B C D F H I A E G J
A 1 1 1 1 1 1 0 0 0 0
B 0 0 0 0 0 0 1 1 1 1
C 0 0 1 0 0 0 0 1 0 0
D 0 1 0 1 1 0 1 0 0 1
E 1 0 1 1 0 1 0 0 1 0
F 0 1 0 0 0 0 0 0 1 0
G 1 0 1 0 0 0 0 0 0 0
H 0 1 0 1 0 0 1 0 0 1
I 1 0 1 0 0 0 0 1 1 0
data:
set.seed(1337)
m <- matrix(sample(c(0,0,0,1),size = 50,replace=T),ncol=5) %>% as.data.frame
colnames(m)<-LETTERS[1:5]
code:
m %<>%
mutate(newcol = ifelse(A==1&(B==1|C==1)&(D==1|E==1),1,
ifelse(any(A,B,C,D,E),0,NA)),
desiredResult= ifelse(A==1&(B==1|C==1)&(D==1|E==1),1,
ifelse(!(A==0&B==0&C==0&D==0&E==0),0,NA)))
looks like:
A B C D E newcol desiredResult
1 0 1 1 1 0 0 0
2 0 1 0 0 1 0 0
3 0 1 0 0 0 0 0
4 0 0 0 0 0 0 NA
5 0 1 0 1 0 0 0
6 0 0 1 0 0 0 0
7 1 1 1 1 0 1 1
8 0 1 1 0 0 0 0
9 0 0 0 0 0 0 NA
10 0 0 1 0 0 0 0
question
I want newcol to be the same as desiredResult.
Why can't I use any in that "stratified" manner of ifelse. Is there a function like any that would work in that situation?
possible workaround
I could define a function
any_vec <- function(...) {apply(cbind(...),1,any)} but this does not make me smile too much.
like suggested in the answer
using pmax works exactly like a vectorized any.
m %>%
mutate(pmaxResult = ifelse(A==1& pmax(B,C) & pmax(D,E),1,
ifelse(pmax(A,B,C,D,E),0,NA)),
desiredResult= ifelse(A==1&(B==1|C==1)&(D==1|E==1),1,
ifelse(!(A==0&B==0&C==0&D==0&E==0),0,NA)))
Here's an alternative approach. I converted to logical at the beginning and back to integer at the end:
m %>%
mutate_all(as.logical) %>%
mutate(newcol = A & pmax(B,C) & pmax(D, E) ,
newcol = replace(newcol, !newcol & !pmax(A,B,C,D,E), NA)) %>%
mutate_all(as.integer)
# A B C D E newcol
# 1 0 1 1 1 0 0
# 2 0 1 0 0 1 0
# 3 0 1 0 0 0 0
# 4 0 0 0 0 0 NA
# 5 0 1 0 1 0 0
# 6 0 0 1 0 0 0
# 7 1 1 1 1 0 1
# 8 0 1 1 0 0 0
# 9 0 0 0 0 0 NA
# 10 0 0 1 0 0 0
I basically replaced the any with pmax.
I have a matrix that is for example like this:
rownames V1
a 1
c 3
b 2
d 4
y 2
q 4
i 1
j 1
r 3
I want to make a Symmetric binary matrix that it's dimnames of that is the same as rownames of above matrix. I want to fill these matrix by 1 & 0 in such a way that 1 indicated placing variables that has the same number in front of it and 0 for the opposite situation.This matrix would be like
dimnames
a c b d y q i j r
a 1 0 0 0 0 0 1 1 0
c 0 1 0 0 0 0 0 0 1
b 0 0 1 0 1 0 0 0 0
d 0 0 0 1 0 1 0 0 0
y 0 0 1 0 1 0 0 0 0
q 0 0 0 1 0 1 0 0 0
i 1 0 0 0 0 0 1 1 0
j 1 0 0 0 0 0 1 1 0
r 0 1 0 0 0 0 0 0 1
Anybody know how can I do that?
Use dist:
DF <- read.table(text = "rownames V1
a 1
c 3
b 2
d 4
y 2
q 4
i 1
j 1
r 3", header = TRUE)
res <- as.matrix(dist(DF$V1)) == 0L
#alternatively:
#res <- !as.matrix(dist(DF$V1))
#diag(res) <- 0L #for the first version of the question, i.e. a zero diagonal
res <- +(res) #for the second version, i.e. to coerce to an integer matrix
dimnames(res) <- list(DF$rownames, DF$rownames)
# 1 2 3 4 5 6 7 8 9
#1 1 0 0 0 0 0 1 1 0
#2 0 1 0 0 0 0 0 0 1
#3 0 0 1 0 1 0 0 0 0
#4 0 0 0 1 0 1 0 0 0
#5 0 0 1 0 1 0 0 0 0
#6 0 0 0 1 0 1 0 0 0
#7 1 0 0 0 0 0 1 1 0
#8 1 0 0 0 0 0 1 1 0
#9 0 1 0 0 0 0 0 0 1
You can do this using table and crossprod.
tcrossprod(table(DF))
# rownames
# rownames a b c d i j q r y
# a 1 0 0 0 1 1 0 0 0
# b 0 1 0 0 0 0 0 0 1
# c 0 0 1 0 0 0 0 1 0
# d 0 0 0 1 0 0 1 0 0
# i 1 0 0 0 1 1 0 0 0
# j 1 0 0 0 1 1 0 0 0
# q 0 0 0 1 0 0 1 0 0
# r 0 0 1 0 0 0 0 1 0
# y 0 1 0 0 0 0 0 0 1
If you want the row and column order as they are found in the data, rather than alphanumerically, you can subset
tcrossprod(table(DF))[DF$rownames, DF$rownames]
or use factor
tcrossprod(table(factor(DF$rownames, levels=unique(DF$rownames)), DF$V1))
If your data is large or sparse, you can use the sparse matrix algebra in xtabs, with similar ways to change the order of the resulting table as before.
Matrix::tcrossprod(xtabs(data=DF, ~ rownames + V1, sparse=TRUE))
I have data.frames of counts such as:
a <- data.frame(id=1:10,
"1"=c(rep(1,3),rep(0,7)),
"3"=c(rep(0,4),rep(1,6)))
names(a)[2:3] <- c("1","3")
a
> a
id 1 3
1 1 1 0
2 2 1 0
3 3 1 0
4 4 0 0
5 5 0 1
6 6 0 1
7 7 0 1
8 8 0 1
9 9 0 1
10 10 0 1
and a template data.frame such as
m <- data.frame(id=1:10,
"1"= rep(0,10),
"2"= rep(0,10),
"3"= rep(0,10),
"4"= rep(0,10))
names(m)[-1] <- 1:4
m
> m
id 1 2 3 4
1 1 0 0 0 0
2 2 0 0 0 0
3 3 0 0 0 0
4 4 0 0 0 0
5 5 0 0 0 0
6 6 0 0 0 0
7 7 0 0 0 0
8 8 0 0 0 0
9 9 0 0 0 0
10 10 0 0 0 0
and I want to add the values of a into the template m
in the appropraite columns, leaving the rest as 0.
This is working but I would like to know
if there is a more elegant way, perhaps using plyr or data.table:
provi <- rbind.fill(a,m)
provi[is.na(provi)] <- 0
mnew <- aggregate(provi[,-1],by=list(provi$id),FUN=sum)
names(mnew)[1] <- "id"
mnew <- mnew[c(1,order(names(mnew)[-1])+1)]
mnew
> mnew
id 1 2 3 4
1 1 1 0 0 0
2 2 1 0 0 0
3 3 1 0 0 0
4 4 0 0 0 0
5 5 0 0 1 0
6 6 0 0 1 0
7 7 0 0 1 0
8 8 0 0 1 0
9 9 0 0 1 0
10 10 0 0 1 0
I guess the concise option would be:
m[names(a)] <- a
Or we match the column names ('i1'), use that to create the column index with max.col, cbind with the row index ('i2'), and a similar step can be done to create 'i3'. We change the values in 'm' corresponding to 'i2' with the 'a' values based on 'i3'.
i1 <- match(names(a)[-1], names(m)[-1])
i2 <- cbind(m$id, i1[max.col(a[-1], 'first')]+1L)
i3 <- cbind(a$id, max.col(a[-1], 'first')+1L)
m[i2] <- a[i3]
m
# id 1 2 3 4
#1 1 1 0 0 0
#2 2 1 0 0 0
#3 3 1 0 0 0
#4 4 0 0 0 0
#5 5 0 0 1 0
#6 6 0 0 1 0
#7 7 0 0 1 0
#8 8 0 0 1 0
#9 9 0 0 1 0
#10 10 0 0 1 0
A data.table option would be melt/dcast
library(data.table)
dcast(melt(setDT(a), id.var='id')[,
variable:= factor(variable, levels=1:4)],
id~variable, value.var='value', drop=FALSE, fill=0)
# id 1 2 3 4
# 1: 1 1 0 0 0
# 2: 2 1 0 0 0
# 3: 3 1 0 0 0
# 4: 4 0 0 0 0
# 5: 5 0 0 1 0
# 6: 6 0 0 1 0
# 7: 7 0 0 1 0
# 8: 8 0 0 1 0
# 9: 9 0 0 1 0
#10: 10 0 0 1 0
A similar dplyr/tidyr option would be
library(dplyr)
library(tidyr)
gather(a, Var, Val, -id) %>%
mutate(Var=factor(Var, levels=1:4)) %>%
spread(Var, Val, drop=FALSE, fill=0)
You could use merge, too:
res <- suppressWarnings(merge(a, m, by="id", suffixes = c("", "")))
(res[, which(!duplicated(names(res)))][, names(m)])
# id 1 2 3 4
# 1 1 1 0 0 0
# 2 2 1 0 0 0
# 3 3 1 0 0 0
# 4 4 0 0 0 0
# 5 5 0 0 1 0
# 6 6 0 0 1 0
# 7 7 0 0 1 0
# 8 8 0 0 1 0
# 9 9 0 0 1 0
# 10 10 0 0 1 0
I am using the following R code to produce a confusion matrix comparing the true labels of some data to the output of a neural network.
t <- table(as.factor(test.labels), as.factor(nnetpredict))
However, sometimes the neural network doesn't predict any of a certain class, so the table isn't square (as, for example, there are 5 levels in the test.labels factor, but only 3 levels in the nnetpredict factor). I want to make the table square by adding in any factor levels necessary, and setting their counts to zero.
How should I go about doing this?
Example:
> table(as.factor(a), as.factor(b))
1 2 3 4 5 6 7 8 9 10
1 1 0 0 0 0 0 0 1 0 0
2 0 1 0 0 0 0 0 0 1 0
3 0 0 1 0 0 0 0 0 0 1
4 0 0 0 1 0 0 0 0 0 0
5 0 0 0 0 1 0 0 0 0 0
6 0 0 0 0 0 1 0 0 0 0
7 0 0 0 0 0 0 1 0 0 0
You can see in the table above that there are 7 rows, but 10 columns, because the a factor only has 7 levels, whereas the b factor has 10 levels. What I want to do is to pad the table with zeros so that the row labels and the column labels are the same, and the matrix is square. From the example above, this would produce:
1 2 3 4 5 6 7 8 9 10
1 1 0 0 0 0 0 0 1 0 0
2 0 1 0 0 0 0 0 0 1 0
3 0 0 1 0 0 0 0 0 0 1
4 0 0 0 1 0 0 0 0 0 0
5 0 0 0 0 1 0 0 0 0 0
6 0 0 0 0 0 1 0 0 0 0
7 0 0 0 0 0 0 1 0 0 0
8 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0 0
10 0 0 0 0 0 0 0 0 0 0
The reason I need to do this is two-fold:
For display to users/in reports
So that I can use a function to calculate the Kappa statistic, which requires a table formatted like this (square, same row and col labels)
EDIT - round II to address the additional details in the question. I deleted my first answer since it wasn't relevant anymore.
This has produced the desired output for the test cases I've given it, but I definitely advise testing thoroughly with your real data. The approach here is to find the full list of levels for both inputs into the table and set that full list as the levels before generating the table.
squareTable <- function(x,y) {
x <- factor(x)
y <- factor(y)
commonLevels <- sort(unique(c(levels(x), levels(y))))
x <- factor(x, levels = commonLevels)
y <- factor(y, levels = commonLevels)
table(x,y)
}
Two test cases:
> #Test case 1
> set.seed(1)
> x <- factor(sample(0:9, 100, TRUE))
> y <- factor(sample(3:7, 100, TRUE))
>
> table(x,y)
y
x 3 4 5 6 7
0 2 1 3 1 0
1 1 0 2 3 0
2 1 0 3 4 3
3 0 3 6 3 2
4 4 4 3 2 1
5 2 2 0 1 0
6 1 2 3 2 3
7 3 3 3 4 2
8 0 4 1 2 4
9 2 1 0 0 3
> squareTable(x,y)
y
x 0 1 2 3 4 5 6 7 8 9
0 0 0 0 2 1 3 1 0 0 0
1 0 0 0 1 0 2 3 0 0 0
2 0 0 0 1 0 3 4 3 0 0
3 0 0 0 0 3 6 3 2 0 0
4 0 0 0 4 4 3 2 1 0 0
5 0 0 0 2 2 0 1 0 0 0
6 0 0 0 1 2 3 2 3 0 0
7 0 0 0 3 3 3 4 2 0 0
8 0 0 0 0 4 1 2 4 0 0
9 0 0 0 2 1 0 0 3 0 0
> squareTable(y,x)
y
x 0 1 2 3 4 5 6 7 8 9
0 0 0 0 0 0 0 0 0 0 0
1 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0
3 2 1 1 0 4 2 1 3 0 2
4 1 0 0 3 4 2 2 3 4 1
5 3 2 3 6 3 0 3 3 1 0
6 1 3 4 3 2 1 2 4 2 0
7 0 0 3 2 1 0 3 2 4 3
8 0 0 0 0 0 0 0 0 0 0
9 0 0 0 0 0 0 0 0 0 0
>
> #Test case 2
> set.seed(1)
> xx <- factor(sample(0:2, 100, TRUE))
> yy <- factor(sample(3:5, 100, TRUE))
>
> table(xx,yy)
yy
xx 3 4 5
0 4 14 9
1 14 15 9
2 11 11 13
> squareTable(xx,yy)
y
x 0 1 2 3 4 5
0 0 0 0 4 14 9
1 0 0 0 14 15 9
2 0 0 0 11 11 13
3 0 0 0 0 0 0
4 0 0 0 0 0 0
5 0 0 0 0 0 0
> squareTable(yy,xx)
y
x 0 1 2 3 4 5
0 0 0 0 0 0 0
1 0 0 0 0 0 0
2 0 0 0 0 0 0
3 4 14 11 0 0 0
4 14 15 11 0 0 0
5 9 9 13 0 0 0