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I have following R matrix:
> matrix
8 4 3 5 7 2 1 6 ...
8 0 0 1 0 0 0 0 0
4 1 0 1 1 0 2 0 0
3 5 0 0 1 0 0 0 0
5 0 0 1 0 0 3 0 0
7 0 0 0 0 0 0 0 0
2 3 4 1 0 0 7 0 0
1 8 0 4 0 0 0 8 0
6 9 0 1 0 0 0 0 0
...
[ reached getOption("max.print") -- omitted 23 rows ]
Question: Is it possible to sort the matrix rows and columns, so that
1 2 3 4 5 6 7 8 ...
1 ...
2
3
4
5
6
7
8
...
?
I only found this here and wondered if there is not better native option.
Thanks!
Assuming you are refering to the row and column names of the matrix such as in this example matrix
m<-matrix(scan(text="
0 0 1 0 0 0 0 0
1 0 1 1 0 2 0 0
5 0 0 1 0 0 0 0
0 0 1 0 0 3 0 0
0 0 0 0 0 0 0 0
3 4 1 0 0 7 0 0
8 0 4 0 0 0 8 0
9 0 1 0 0 0 0 0"), ncol=8)
colnames(m)<-c(8,4,3,5,7,2,1,6)
rownames(m)<-c(8,4,3,5,7,2,1,6)
You could sort the rows and columns by name with
m[, sort(colnames(m))][sort(rownames(m)), ]
Row and column names are always treated as strings. So if you have larger numbers, you may want to convert to numeric before sorting: sort(as.numeric(colnames(m)))
You can also use order() function and pick up rows and columns by positions:
mat[order(rownames(mat)),order(colnames(mat))]
# 1 2 3 4 5 6 7 8
#1 8 0 4 0 0 0 0 8
#2 0 7 1 4 0 0 0 3
#3 0 0 0 0 1 0 0 5
#4 0 2 1 0 1 0 0 1
#5 0 3 1 0 0 0 0 0
#6 0 0 1 0 0 0 0 9
#7 0 0 0 0 0 0 0 0
#8 0 0 1 0 0 0 0 0
Related
I would like to do transform Gender and Country using One-Hot-Encoding.
With the code below I can not create the new dataset including the ID
library(caret)
ID<-1:10
Gender<-c("F","F","F","M","M","F","M","M","F","M")
Country<-c("Mali","France","France","Guinea","Senegal",
"Mali","France","Mali","Senegal","France")
data<-data.frame(ID,Gender,Country)
#One hot encoding
dmy <- dummyVars(" ~Gender+Country", data = data, fullRank = T)
dat_transformed <- data.frame(predict(dmy, newdata = data))
dat_transformed
Gender.M Country.Guinea Country.Mali Country.Senegal
1 0 0 1 0
2 0 0 0 0
3 0 0 0 0
4 1 1 0 0
5 1 0 0 1
6 0 0 1 0
7 1 0 0 0
8 1 0 1 0
9 0 0 0 1
10 1 0 0 0
I want to get a dataset that include the ID without enconding it.
ID Gender.M Country.Guinea Country.Mali Country.Senegal
1 1 0 0 1 0
2 2 0 0 0 0
3 3 0 0 0 0
4 4 1 1 0 0
5 5 1 0 0 1
6 6 0 0 1 0
7 7 1 0 0 0
8 8 1 0 1 0
9 9 0 0 0 1
10 10 1 0 0 0
dat_transformed <- cbind(ID,dat_transformed)
dat_transformed
ID Gender.M Country.Guinea Country.Mali Country.Senegal
1 0 0 1 0
2 0 0 0 0
3 0 0 0 0
4 1 1 0 0
5 1 0 0 1
6 0 0 1 0
7 1 0 0 0
8 1 0 1 0
9 0 0 0 1
10 1 0 0 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
This question already has an answer here:
Include levels of zero count in result of table()
(1 answer)
Closed 8 years ago.
I need to extend a table in R language.
result 3 4 5 6 7 8
5 6 29 295 104 6 0
6 1 9 112 238 66 5
7 0 0 5 29 40 6
Should be extended to
result 1 2 3 4 5 6 7 8 9 10
1 0 0 0 0 0 0 0 0 0 0
2 0 0 0 0 0 0 0 0 0 0
3 0 0 0 0 0 0 0 0 0 0
4 0 0 0 0 0 0 0 0 0 0
5 0 0 6 29 295 104 6 0 0 0
6 0 0 1 9 112 238 66 5 0 0
7 0 0 0 0 5 29 40 6 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
So I need add zeros in missing values. Also, in alternative scenario an output as a matrix (10x10) with the same data would be satisfying.
EDIT:
table(factor(x, levels = 1:10), factor(y, levels = 1:10)) worked perfectly.
As the guys in the comments mentioned. Factoring works perfectly.
table(factor(x, levels = 1:10), factor(y, levels = 1:10))
I have a table created using the table() command in R:
y
x 0 1 2 3 4 5 6 7 8 9
0 23 0 0 0 0 1 0 0 0 0
1 0 23 1 0 1 0 1 2 0 2
2 1 1 28 0 0 0 1 0 2 2
3 0 1 0 24 0 1 0 0 0 1
4 1 1 0 0 34 0 3 0 0 0
5 0 0 0 0 0 33 0 0 0 0
6 0 0 0 0 0 2 32 0 0 0
7 0 1 0 1 0 0 0 36 0 1
8 1 1 1 1 0 0 0 1 20 1
9 1 3 0 1 0 1 0 1 0 24
This table shows the results of a classification, and I want to sum the leading diagonal of it (the diagonal with the large numbers - like 23, 23, 28 etc). Is there a sensible/easy way to do this in R?
How about sum(diag(tbl)), where tbl is your table?
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