I have a priorly unknown number of variables, and for each variable I need to define a for loop and perform a series of operations. For each subsequent variable, I need to define a nested loop inside the previous one, performing the same operations. I guess there must be a way of doing this recursively, but I am struggling with it.
Consider for instance the following easy example:
results = c()
index = 0
for(i in 1:5)
{
a = i*2
for(j in 1:5)
{
b = a*2 + j
for(k in 1:5)
{
index = index + 1
c = b*2 + k
results[index] = c
}
}
}
In this example, I would have 3 variables. The loop on j requires information from the loop i, and the loop on k requires information from the loop j. This is a simplified example of my problem and the operations here are pretty simple. I am not interested on another way of getting the "results" vector, what I would like to know is if there is a way to recursevily do this operations for an unknown number of variables, lets say 10 variables, so that I do not need to nest manually 10 loops.
Here is one approach that you might be able to modify for your situation...
results <- 0 #initialise
for(level in 1:3){ #3 nested loops - change as required
results <- c( #converts output to a vector
outer(results, #results so far
1:5, #as in your loops
FUN = function(x,y) {x*2+y} #as in your loops
)
)
}
The two problems with this are
a) that your formula is different in the first (outer) loop, and
b) the order of results is different from yours
However, you might be able to find workarounds for these depending on your actual problem.
I have tried to change the code so that it is a function that allows to define how many iterations need to happen.
library(tidyverse)
fc <- function(i_end, j_end, k_end){
i <- 1:i_end
j <- 1:j_end
k <- 1:k_end
df <- crossing(i, j, k) %>%
mutate(
a = i*2,
b = a*2 + j,
c = b*2 + k,
index = row_number())
df
}
fc(5,5,5)
While trying to learn R, I want to implement the algorithm below in R. Consider the two lists below:
List 1: "crashed", "red", "car"
List 2: "crashed", "blue", "bus"
I want to find out how many actions it would take to transform 'list1' into 'list2'.
As you can see I need only two actions:
1. Replace "red" with "blue".
2. Replace "car" with "bus".
But, how we can find the number of actions like this automatically.
We can have several actions to transform the sentences: ADD, REMOVE, or REPLACE the words in the list.
Now, I will try my best to explain how the algorithm should work:
At the first step: I will create a table like this:
rows: i= 0,1,2,3,
columns: j = 0,1,2,3
(example: value[0,0] = 0 , value[0, 1] = 1 ...)
crashed red car
0 1 2 3
crashed 1
blue 2
bus 3
Now, I will try to fill the table. Please, note that each cell in the table shows the number of actions we need to do to reformat the sentence (ADD, remove, or replace).
Consider the interaction between "crashed" and "crashed" (value[1,1]), obviously we don't need to change it so the value will be '0'. Since they are the same words. Basically, we got the diagonal value = value[0,0]
crashed red car
0 1 2 3
crashed 1 0
blue 2
bus 3
Now, consider "crashed" and the second part of the sentence which is "red". Since they are not the same word we can use calculate the number of changes like this :
min{value[0,1] , value[0,2] and value[1,1]} + 1
min{ 1, 2, 0} + 1 = 1
Therefore, we need to just remove "red".
So, the table will look like this:
crashed red car
0 1 2 3
crashed 1 0 1
blue 2
bus 3
And we will continue like this :
"crashed" and "car" will be :
min{value[0,3], value[0,2] and value[1,2]} + 1
min{3, 2, 1} +1 = 2
and the table will be:
crashed red car
0 1 2 3
crashed 1 0 1 2
blue 2
bus 3
And we will continue to do so. the final result will be :
crashed red car
0 1 2 3
crashed 1 0 1 2
blue 2 1 1 2
bus 3 2 2 2
As you can see the last number in the table shows the distance between two sentences: value[3,3] = 2
Basically, the algorithm should look like this:
if (characters_in_header_of_matrix[i]==characters_in_column_of_matrix [j] &
value[i,j] == value[i+1][j-1] )
then {get the 'DIAGONAL VALUE' #diagonal value= value[i, j-1]}
else{
value[i,j] = min(value[i-1, j], value[i-1, j-1], value[i, j-1]) + 1
}
endif
for finding the difference between the elements of two lists that you can see in the header and the column of the matrix, I have used the strcmp() function which will give us a boolean value(TRUE or FALSE) while comparing the words. But, I fail at implementing this.
I'd appreciate your help on this one, thanks.
The question
After some clarification in a previous post, and after the update of the post, my understanding is that Zero is asking: 'how one can iteratively count the number of word differences in two strings'.
I am unaware of any implementation in R, although i would be surprised if i doesn't already exists. I took a bit of time out to create a simple implementation, altering the algorithm slightly for simplicity (For anyone not interested scroll down for 2 implementations, 1 in pure R, one using the smallest amount of Rcpp). The general idea of the implementation:
Initialize with string_1 and string_2 of length n_1 and n_2
Calculate the cumulative difference between the first min(n_1, n_2) elements,
Use this cumulative difference as the diagonal in the matrix
Set the first off-diagonal element to the very first element + 1
Calculate the remaining off diagonal elements as: diag(i) - diag(i-1) + full_matrix(i-1,j)
In the previous step i iterates over diagonals, j iterates over rows/columns (either one works), and we start in the third diagonal, as the first 2x2 matrix is filled in step 1 to 4
Calculate the remaining abs(n_1 - n_2) elements as full_matrix[,min(n_1 - n_2)] + 1:abs(n_1 - n_2), applying the latter over each value in the prior, and bind them appropriately to the full_matrix.
The output is a matrix with dimensions row and column names of the corresponding strings, which has been formatted for some easier reading.
Implementation in R
Dist_between_strings <- function(x, y,
split = " ",
split_x = split, split_y = split,
case_sensitive = TRUE){
#Safety checks
if(!is.character(x) || !is.character(y) ||
nchar(x) == 0 || nchar(y) == 0)
stop("x, y needs to be none empty character strings.")
if(length(x) != 1 || length(y) != 1)
stop("Currency the function is not vectorized, please provide the strings individually or use lapply.")
if(!is.logical(case_sensitive))
stop("case_sensitivity needs to be logical")
#Extract variable names of our variables
# used for the dimension names later on
x_name <- deparse(substitute(x))
y_name <- deparse(substitute(y))
#Expression which when evaluated will name our output
dimname_expression <-
parse(text = paste0("dimnames(output) <- list(",make.names(x_name, unique = TRUE)," = x_names,",
make.names(y_name, unique = TRUE)," = y_names)"))
#split the strings into words
x_names <- str_split(x, split_x, simplify = TRUE)
y_names <- str_split(y, split_y, simplify = TRUE)
#are we case_sensitive?
if(isTRUE(case_sensitive)){
x_split <- str_split(tolower(x), split_x, simplify = TRUE)
y_split <- str_split(tolower(y), split_y, simplify = TRUE)
}else{
x_split <- x_names
y_split <- y_names
}
#Create an index in case the two are of different length
idx <- seq(1, (n_min <- min((nx <- length(x_split)),
(ny <- length(y_split)))))
n_max <- max(nx, ny)
#If we have one string that has length 1, the output is simplified
if(n_min == 1){
distances <- seq(1, n_max) - (x_split[idx] == y_split[idx])
output <- matrix(distances, nrow = nx)
eval(dimname_expression)
return(output)
}
#If not we will have to do a bit of work
output <- diag(cumsum(ifelse(x_split[idx] == y_split[idx], 0, 1)))
#The loop will fill in the off_diagonal
output[2, 1] <- output[1, 2] <- output[1, 1] + 1
if(n_max > 2)
for(i in 3:n_min){
for(j in 1:(i - 1)){
output[i,j] <- output[j,i] <- output[i,i] - output[i - 1, i - 1] + #are the words different?
output[i - 1, j] #How many words were different before?
}
}
#comparison if the list is not of the same size
if(nx != ny){
#Add the remaining words to the side that does not contain this
additional_words <- seq(1, n_max - n_min)
additional_words <- sapply(additional_words, function(x) x + output[,n_min])
#merge the additional words
if(nx > ny)
output <- rbind(output, t(additional_words))
else
output <- cbind(output, additional_words)
}
#set the dimension names,
# I would like the original variable names to be displayed, as such i create an expression and evaluate it
eval(dimname_expression)
output
}
Note that the implementation is not vectorized, and as such can only take single string inputs!
Testing the implementation
To test the implementation, one could use the strings given. As they were said to be contained in lists, we will have to convert them to strings. Note that the function lets one split each string differently, however it assumes space separated strings. So first I'll show how one could achieve a conversion to the correct format:
list_1 <- list("crashed","red","car")
list_2 <- list("crashed","blue","bus")
string_1 <- paste(list_1,collapse = " ")
string_2 <- paste(list_2,collapse = " ")
Dist_between_strings(string_1, string_2)
output
#Strings in the given example
string_2
string_1 crashed blue bus
crashed 0 1 2
red 1 1 2
car 2 2 2
This is not exactly the output, but it yields the same information, as the words are ordered as they were given in the string.
More examples
Now i stated it worked for other strings as well and this is indeed the fact, so lets try some random user-made strings:
#More complicated strings
string_3 <- "I am not a blue whale"
string_4 <- "I am a cat"
string_5 <- "I am a beautiful flower power girl with monster wings"
string_6 <- "Hello"
Dist_between_strings(string_3, string_4, case_sensitive = TRUE)
Dist_between_strings(string_3, string_5, case_sensitive = TRUE)
Dist_between_strings(string_4, string_5, case_sensitive = TRUE)
Dist_between_strings(string_6, string_5)
Running these show that these do yield the correct answers. Note that if either string is of size 1, the comparison is a lot faster.
Benchmarking the implementation
Now as the implementation is accepted, as correct, we would like to know how well it performs (For the uninterested reader, one can scroll past this section, to where a faster implementation is given). For this purpose, i will use much larger strings. For a complete benchmark i should test various string sizes, but for the purposes i will only use 2 rather large strings of size 1000 and 2500. For this purpose i use the microbenchmark package in R, which contains a microbenchmark function, which claims to be accurate down to nanoseconds. The function itself executes the code 100 (or a user defined) number of times, returning the mean and quartiles of the run times. Due to other parts of R such as the Garbage Cleaner, the median is mostly considered a good estimate of the actual average run-time of the function.
The execution and results are shown below:
#Benchmarks for larger strings
set.seed(1)
string_7 <- paste(sample(LETTERS,1000,replace = TRUE), collapse = " ")
string_8 <- paste(sample(LETTERS,2500,replace = TRUE), collapse = " ")
microbenchmark::microbenchmark(String_Comparison = Dist_between_strings(string_7, string_8, case_sensitive = FALSE))
# Unit: milliseconds
# expr min lq mean median uq max neval
# String_Comparison 716.5703 729.4458 816.1161 763.5452 888.1231 1106.959 100
Profiling
Now i find the run-times very slow. One use case for the implementation could be an initial check of student hand-ins to check for plagiarism, in which case a low difference count very likely shows plagiarism. These can be very long and there may be hundreds of handins, an as such i would like the run to be very fast.
To figure out how to improve my implementation i used the profvis package with the corrosponding profvis function. To profile the function i exported it in another R script, that i sourced, running the code 1 once prior to profiling to compile the code and avoid profiling noise (important). The code to run the profiling can be seen below, and the most important part of the output is visualized in an image below it.
library(profvis)
profvis(Dist_between_strings(string_7, string_8, case_sensitive = FALSE))
Now, despite the colour, here i can see a clear problem. The loop filling the off-diagonal by far is responsible for most of the runtime. R (like python and other not compiled languages) loops are notoriously slow.
Using Rcpp to improve performance
To improve the implementation, we could implement the loop in c++ using the Rcpp package. This is rather simple. The code is not unlike the one we would use in R, if we avoid iterators. A c++ script can be made in file -> new file -> c++ File. The following c++ code would be pasted into the corrosponding file and sourced using the source button.
//Rcpp Code
#include <Rcpp.h>
using namespace Rcpp;
// [[Rcpp::export]]
NumericMatrix Cpp_String_difference_outer_diag(NumericMatrix output){
long nrow = output.nrow();
for(long i = 2; i < nrow; i++){ // note the
for(long j = 0; j < i; j++){
output(i, j) = output(i, i) - output(i - 1, i - 1) + //are the words different?
output(i - 1, j);
output(j, i) = output(i, j);
}
}
return output;
}
The corresponding R function needs to be altered to use this function instead of looping. The code is similar to the first function, only switching the loop for a call to the c++ function.
Dist_between_strings_cpp <- function(x, y,
split = " ",
split_x = split, split_y = split,
case_sensitive = TRUE){
#Safety checks
if(!is.character(x) || !is.character(y) ||
nchar(x) == 0 || nchar(y) == 0)
stop("x, y needs to be none empty character strings.")
if(length(x) != 1 || length(y) != 1)
stop("Currency the function is not vectorized, please provide the strings individually or use lapply.")
if(!is.logical(case_sensitive))
stop("case_sensitivity needs to be logical")
#Extract variable names of our variables
# used for the dimension names later on
x_name <- deparse(substitute(x))
y_name <- deparse(substitute(y))
#Expression which when evaluated will name our output
dimname_expression <-
parse(text = paste0("dimnames(output) <- list(", make.names(x_name, unique = TRUE)," = x_names,",
make.names(y_name, unique = TRUE)," = y_names)"))
#split the strings into words
x_names <- str_split(x, split_x, simplify = TRUE)
y_names <- str_split(y, split_y, simplify = TRUE)
#are we case_sensitive?
if(isTRUE(case_sensitive)){
x_split <- str_split(tolower(x), split_x, simplify = TRUE)
y_split <- str_split(tolower(y), split_y, simplify = TRUE)
}else{
x_split <- x_names
y_split <- y_names
}
#Create an index in case the two are of different length
idx <- seq(1, (n_min <- min((nx <- length(x_split)),
(ny <- length(y_split)))))
n_max <- max(nx, ny)
#If we have one string that has length 1, the output is simplified
if(n_min == 1){
distances <- seq(1, n_max) - (x_split[idx] == y_split[idx])
output <- matrix(distances, nrow = nx)
eval(dimname_expression)
return(output)
}
#If not we will have to do a bit of work
output <- diag(cumsum(ifelse(x_split[idx] == y_split[idx], 0, 1)))
#The loop will fill in the off_diagonal
output[2, 1] <- output[1, 2] <- output[1, 1] + 1
if(n_max > 2)
output <- Cpp_String_difference_outer_diag(output) #Execute the c++ code
#comparison if the list is not of the same size
if(nx != ny){
#Add the remaining words to the side that does not contain this
additional_words <- seq(1, n_max - n_min)
additional_words <- sapply(additional_words, function(x) x + output[,n_min])
#merge the additional words
if(nx > ny)
output <- rbind(output, t(additional_words))
else
output <- cbind(output, additional_words)
}
#set the dimension names,
# I would like the original variable names to be displayed, as such i create an expression and evaluate it
eval(dimname_expression)
output
}
Testing the c++ implementation
To be sure the implementation is correct we check if the same output is obtained with the c++ implementation.
#Test the cpp implementation
identical(Dist_between_strings(string_3, string_4, case_sensitive = TRUE),
Dist_between_strings_cpp(string_3, string_4, case_sensitive = TRUE))
#TRUE
Final benchmarks
Now is this actually faster? To see this we could run another benchmark using the microbenchmark package. The code and results are shown below:
#Final microbenchmarking
microbenchmark::microbenchmark(R = Dist_between_strings(string_7, string_8, case_sensitive = FALSE),
Rcpp = Dist_between_strings_cpp(string_7, string_8, case_sensitive = FALSE))
# Unit: milliseconds
# expr min lq mean median uq max neval
# R 721.71899 753.6992 850.21045 787.26555 907.06919 1756.7574 100
# Rcpp 23.90164 32.9145 54.37215 37.28216 47.88256 243.6572 100
From the microbenchmark median improvement factor of roughly 21 ( = 787 / 37), which is a massive improvement from just implementing a single loop!
There is already an edit-distance function in R we can take advantage of: adist().
As it works on the character level, we'll have to assign a character to each unique word in our sentences, and stitch them together to form pseudo-words we can calculate the distance between.
s1 <- c("crashed", "red", "car")
s2 <- c("crashed", "blue", "bus")
ll <- list(s1, s2)
alnum <- c(letters, LETTERS, 0:9)
ll2 <- relist(alnum[factor(unlist(ll))], ll)
ll2 <- sapply(ll2, paste, collapse="")
adist(ll2)
# [,1] [,2]
# [1,] 0 2
# [2,] 2 0
Main limitation here, as far as I can tell, is the number of unique characters available, which in this case is 62, but can be extended quite easily, depending on your locale. E.g: intToUtf8(c(32:126, 161:300), TRUE).
I was asked a programming question by a friend about how to determine all possible combinations of values from a set can be added to make a desired total. I have a solution, but it is less than elegant (its basically just a series of for-loops and if-statements). I'm sure dplyr has a solution that I can't think of, as I know how useful it is but I just haven't gotten very good at it yet. I will post the question and my script below.
Question:
There is a target with six rings on it, each ring is worth a different value. These values are either 1, 2, 3, 4, 5, or 6. How many different combinations of rings can you use to score EXACTLY 9 points?
So to consider:
Order is not important
You can use as few or as many values as you want
You can get the same value more than once (so 9 1's is a totally valid option)
I had considered first using combn() from combinat package, but combn() does not replace values.
Then I decided to use a series of nested for loops and if statements (I truncated it to where you could only use a max of 6-values, because while I may have free time, I'm not a masochist about to write a loop that allows up to 9-values). So essentially, it is running through 6 for-loops worth of possible values. I included the number 0 to the list of possible values to represent no attempt when I only need 2 values instead of 6 (so 4+5+0+0+0+0 is a valid output in this loop, but it wouldn't be able to do 4+5 as it would always try to add more non-zero values).
## Create a vector x with possible values
x = c(1,2,3,4,5,6)
## Add in value 0 because I need to be able to write this dumb loop that allows many terms to be used, but also allows smaller amounts of terms to be used
x = c(x,0);x
## Creating empty data.frame to input solutions to so that I can check for uniqueness of solution
df = data.frame("a" = as.numeric(),
"b" = as.numeric(),
"c" = as.numeric(),
"d" = as.numeric(),
"e" = as.numeric(),
"f" = as.numeric())
for (a in x){
for (b in x){
for (c in x){
for (d in x){
for (e in x){
for (f in x){
m = sum(a,b,c,d,e,f)
if(m == 9) {
p = 0
n = c(a,b,c,d,e,f)
if (nrow(df) == 0){
df[1,] = n
}
if (nrow(df) >= 1){
for (i in (1:nrow(df))){
if(setequal(n,df[i,]) == TRUE){
p = p+1
}}
if(p == 0){
df = rbind(df,n)
}
}
}
}
}
}
}
}
}
## Convert any 0 values to NA
df[df==0] = NA
## Check Solutions
df
I created an empty data.frame to store solutions, and then within the loop, I created a test to see if a new solution in the loop matches a combination of values previously found, and if so, it would not rbind() that to the data.frame.
I'm sure there is a better way to do this, one that allows a dynamic max number of values (so it could be soft coded in this case to change the max number of values in each solution to 9 instead of my hard-coded 6, or drop it to 5 if my desired total was 5 instead of 9). If you have any suggestions to make this less of a clunky, loop-filled mess, it would be appreciated!
You can maybe try this :
library(modelr)
library(dplyr)
range = 1:6
df = data.frame("a" = range,
"b" = range,
"c" = range,
"d" = range,
"e" = range,
"f" = range)
data_grid(df,a,b,c,d,e,f) %>%
mutate(sum = a+b+c+d+e+f) %>%
filter(sum == 9) %>% nrow
This is the function :
foo <- function(sum_needed, max_value){
range <- 1:max_value
df = data.frame("a" = range,
"b" = range,
"c" = range,
"d" = range,
"e" = range,
"f" = range)
result <- data_grid(df,a,b,c,d,e,f) %>%
mutate(sum = a+b+c+d+e+f) %>%
filter(sum == sum_needed) %>% nrow
return(result)
}
foo(9,6)
#[1] 56
x = 1:6
mysum = 9
#Repeat each element of x as long the sum of repetitions does not exceed mysum
temp = rep(x, floor(mysum/x))
#Calculate total unique combinations of temp that sum up to mysum
sum(sapply(1:max(floor(mysum/x)),
function(i) sum(rowSums(unique(t(combn(temp, i)))) == mysum)))
#[1] 26
Following should list all the combinations
sapply(1:max(floor(mysum/x)), function(i){
temp2 = unique(t(combn(temp, i)))
temp2[rowSums(temp2) == mysum,]
})
I have a for loop in R in which I want to store the result of each calculation (for all the values looped through). In the for loop a function is called and the output is stored in a variable r in the moment. However, this is overwritten in each successive loop. How could I store the result of each loop through the function and access it afterwards?
Thanks,
example
for (par1 in 1:n) {
var<-function(par1,par2)
c(var,par1)->var2
print(var2)
So print returns every instance of var2 but in var2 only the value for the last n is saved..is there any way to get an array of the data or something?
initialise an empty object and then assign the value by indexing
a <- 0
for (i in 1:10) {
a[i] <- mean(rnorm(50))
}
print(a)
EDIT:
To include an example with two output variables, in the most basic case, create an empty matrix with the number of columns corresponding to your output parameters and the number of rows matching the number of iterations. Then save the output in the matrix, by indexing the row position in your for loop:
n <- 10
mat <- matrix(ncol=2, nrow=n)
for (i in 1:n) {
var1 <- function_one(i,par1)
var2 <- function_two(i,par2)
mat[i,] <- c(var1,var2)
}
print(mat)
The iteration number i corresponds to the row number in the mat object. So there is no need to explicitly keep track of it.
However, this is just to illustrate the basics. Once you understand the above, it is more efficient to use the elegant solution given by #eddi, especially if you are handling many output variables.
To get a list of results:
n = 3
lapply(1:n, function(par1) {
# your function and whatnot, e.g.
par1*par1
})
Or sapply if you want a vector instead.
A bit more complicated example:
n = 3
some_fn = function(x, y) { x + y }
par2 = 4
lapply(1:n, function(par1) {
var = some_fn(par1, par2)
return(c(var, par1)) # don't have to type return, but I chose to make it explicit here
})
#[[1]]
#[1] 5 1
#
#[[2]]
#[1] 6 2
#
#[[3]]
#[1] 7 3
I have a vector, which contains names of data vectors, named c("tom.txt", "tim.txt" , "Amy.txt"). My task is to: build a symmetric matrix that looks like the following at the end.
> m
tom.txt tim.txt amy.txt
tom.txt 0 10 5
tim.txt 10 0 7
amy.txt 5 7 0
The entries are obtained by a function called get.result(vec1, vec2), which finds the corresponding data vectors of the 2 text files in the directory and does some operations to them and then returns a value for each position in the matrix. For instance, if I want to get the entry m["tom.txt", "tim.txt"], I need to pass "tom.txt" and "tim.txt" into get.result("tom.txt", "tim.txt"). The content of this function is not important.
However, if I want to compute the value for all entries, it will be tedious to keep typing get.result("tom.txt", "amy.txt"), get.result("tim.txt", "amy.txt"), especially when I am actually working with 100 different text files.
My question: Is there an efficient way to program this such that I am always comparing one text file against the rest (never compare against itself) and at the same time, I can keep track of their positions in the matrix ?
Should I initialise the matrix with all zeros right at the beginning and set the column and row names to be the text files name ? But in this case, I am not sure how to obtain the column names and row names such that I can pass them into get.result(vec1, vec2).
Try this solution
fn <- c("tom.txt", "tim.txt" , "Amy.txt")
n <- length(fn)
m <- matrix(0, n, n)
rownames(m) <- fn
colnames(m) <- fn
for (i in 1:n) for (j in i:n) if (i!=j) {
v <- get.result(fn[i], fn[j])
m[i,j] <- v
m[j,i] <- v
}
m
fn = dir(pattern=".txt") (change the pattern if needed) will give you the text files in your target folder. You could then loop over that list like in the previous answer.
The combn function gives you distinct combinations of vector elements:
combs <-combn( c("tom.txt", "tim.txt" , "Amy.txt") , 2)
#----------------
[,1] [,2] [,3]
[1,] "tom.txt" "tom.txt" "tim.txt"
[2,] "tim.txt" "Amy.txt" "Amy.txt"
You can then : apply( combs, 2, get.result)
Given that your file names are in a vector, say
vec <- c("tom.txt", "tim.txt" , "Amy.txt")
you can use
temp <- outer(seq(vec), seq(vec), Vectorize(function(x,y) if(x>y) get.result(vec[x],vec[y]) else 0 ))
result <- temp + t(temp)
Note that this makes sure get.result() is called only once for every relevant comparison, i.e., it's not called for equal files, nor is it called for pairs that differ only by order.
The last line creates a symmetric matrix.
EDIT: to get the names, use this:
rownames(result) <- colnames(result) <- vec