Using R, I have generated several strings of letters that range from 6-25 characters. I'd like for each one to generate an output that consists of all the combinations of these strings with every "I" substituted for a "L" and vice versa, the order of the characters should stay the same.
For example:
Input
"IVGLWEA"
OUTPUT
"IVGLWEA"
"LVGLWEA"
"LVGIWEA"
'IVGIWEA"
"LVGLWEA"
many thanks
rob
Edit: Thanks to #Skaqqs for the dynamic solution!
string <- "IVGLWEA"
# find the number of I's and L's in the string
n <- length(unlist(gregexpr("I|L", string)))
# make a grid of all possible combinations with this amount of I's and L's
df <- expand.grid(rep(list(c("I", "L")), n))
# replace I's and L's with %s
string_ <- gsub("I|L", "\\%s", string)
# replace %s with letters in grid
do.call(sprintf, as.list(c(string_, df)))
Result:
[1] "IVGIWEA" "LVGIWEA" "IVGLWEA" "LVGLWEA"
Here's an extremely inefficient (but concise!) approach:
Create all potential combinations of your input characters and use regex to extract the desired pattern.
pattern <- "(I|L)VG(I|L)WEA"
b <- c("I", "V", "G", "L", "W", "E", "A")
strings <- apply(expand.grid(rep(list(b), 7)), 1, paste0, collapse = "")
grep(pattern, strings, value = TRUE)
[1] "IVGIWEA" "LVGIWEA" "IVGLWEA" "LVGLWEA"
I'm trying to make a table in R with formattable but some rows are returning NA and I don't know how to remove them.
In Excel the table looks like this, and this is how I would like it to be:
In R looks like this:
How can I make this table in R looks like in Excel?
Code:
library(readxl)
library(formattable)
indicadores <- read_excel("Dados/Tabela Indicadores.xlsx")
tabela_indicadores <- formattable(indicadores,
align = c("c", "c", "c", "c"))
File that I'm using: https://drive.google.com/file/d/1CDhMkW7l78WEj45mYZ86LTY6PjCyhsSd/view?usp=sharing
The easiest way to do this is to replace the missing values with a character, "-" or "":
indicadores$Tema <- ifelse(is.na(indicadores$Tema), "-", indicadores$Tema)
tabela_indicadores <- formattable(indicadores, align = c("c", "c", "c", "c"))
I am assuming you do not want to exclude the lines completely. If you do want to remove them, use
tabela_indicadores <- formattable(na.omit(indicadores), align = c("c", "c", "c", "c"))
I want to generate a random password for employees with the function below. This is my first attempt with functions in R. So I need a bit help.
genPsw <- function(num, len=8) {
# Vorgaben für die Passwortkonventionen festlegen
myArr <- c("", 2, 3, 4, 5, 6, 7, 8, 9, "A", "B",
"C", "D", "E", "F", "G", "H", "J", "K", "L", "M",
"N", "P", "Q", "R", "S", "T", "U", "V", "W", "X", "Y", "Z",
"a", "b", "c", "d", "e", "f", "g", "h", "i", "j", "k", "l", "m", "n", "o",
"p", "q", "r", "s", "t", "u", "v", "w", "x", "y", "z",
"!", "§", "$", "%", "&", "(", ")", "*")
# replicate is a wrapper for the common use of sapply for repeated evaluation of an expression
# (which will usually involve random number generation).
replicate(num, paste(sample(myArr, size=len, replace=T), collapse=""))
# nrow of dataframe mitarbeiter
dim_mitarbeiter <- nrow(mitarbeiter)
for(i in 1:dim_mitarbeiter) {
# Random Number Generation with i
set.seed(i)
# Generate Passwort for new variable password
mitarbeiter$passwort <- genPsw(i)
}
}
With the answer form Floo0 I've changed the function to somthing like that, but it doesn't work:
genPsw <- function(num, len=8) {
# Vorgaben für die Passwortkonventionen festlegen
sam<-list()
sam[[1]]<-1:9
sam[[2]]<-letters
sam[[3]]<-LETTERS
sam[[4]]<-c("!", "§", "$", "%", "&", "(", ")", "*")
# nrow of dataframe mitarbeiter
dim_mitarbeiter <- nrow(mitarbeiter)
for(i in 1:dim_mitarbeiter) {
# Random Number Generation with i
tmp<-mapply(sample,sam,c(2,2,2,2))
# Generate Passwort for new variable password
mitarbeiter$passwort <- paste(sample(tmp),collapse="")
}
}
What about
samp<-c(2:9,letters,LETTERS,"!", "§", "$", "%", "&", "(", ")", "*")
paste(sample(samp,8),collapse="")
result is something like this
"HKF§VvnD"
Caution:
This approch does not enforce having capitals, numbers, and non alpha numeric symbols
EDIT:
If you want to enforce a certain number of capitals, numbers, and non alpha numeric symbols you could go with this:
sam<-list()
sam[[1]]<-1:9
sam[[2]]<-letters
sam[[3]]<-LETTERS
sam[[4]]<-c("!", "§", "$", "%", "&", "(", ")", "*")
tmp<-mapply(sample,sam,c(2,2,2,2))
paste(sample(tmp),collapse="")
Where c(2,2,2,2) specifies the number of numbers, letters, capital letters and symbild (in this order). Result:
[1] "j$bP%5R3"
EDIT2:
To produce an new column in you table mitarbeiter just use
passwort<-replicate(nrow(mitarbeiter),paste(mapply(sample,sam,c(2,2,2,2)),collapse=""))
mitarbeiter$passwort<-passwort
There is function which generates random strings in stringi package:
require(stringi)
stri_rand_strings(n=2, length=8, pattern="[A-Za-z0-9]")
## [1] "90i6RdzU" "UAkSVCEa"
This might work, one might want to alter ASCII to avoid unwanted symbols:
generatePwd <- function(plength=8, ASCII=c(33:126)) paste(sapply(sample(ASCII, plength), function(x) rawToChar(as.raw(x))), collapse="")
The below script creates a password of specified length from a combination of upper and lowercase letters, digits and 32 symbols (punctuation, etc.).
# Store symbols as a vector in variable punc
R> library(magrittr) # Load this package to use the %>% (pipe) operator
R> punc_chr <- "!#$%&’()*+,-./:;<=>?#[]^_`{|}~" %>%
str_split("", simplify = TRUE) %>%
as.vector() -> punc
# Randomly generate specified number of characters from 94 characters
R> sample(c(LETTERS, letters, 0:9, punc), 8) %>%
paste(collapse = "") -> pw
R> pw # Return password
[1] "fAXVpyOs"
I like the brevity of L.R.'s solution, although I don't follow what it does 100%.
My solution allows to specify the length of the password but also ensures that at least one lower case, one upper case, one numeric, and one special character is included, and allows reproducibility. (ht to moazzem for spelling out all the special characters.)
gen_pass <- function(len=8,seeder=NULL){
set.seed(seeder) # to allow replicability of passw generation
# get all combinations of 4 nums summing to length len
all_combs <- expand.grid(1:(len-3),1:(len-3),1:(len-3),1:(len-3))
sum_combs <- all_combs[apply(all_combs, 1, function(x) sum(x)==len),]
# special character vector
punc <- unlist(strsplit("!#$%&’()*+,-./:;<=>?#[]^_`{|}~",""))
# list of all characters to sample from
chars <- list(punc,LETTERS,0:9,letters)
# retrieve the number of characters from each list element
# specified in the sampled row of sum_combs
pass_chars_l<- mapply(sample, chars,
sum_combs[sample(1:nrow(sum_combs),1),],
replace = TRUE)
# unlist sets of password characters
pass_chars <- unlist(pass_chars_l)
# jumble characters and combine into password
passw <- str_c(sample(pass_chars),collapse = "")
return(passw)
}
I am still wondering how the (1:(len-3),1:(len-3),1:(len-3),1:(len-3)) in expand.grid(1:(len-3),1:(len-3),1:(len-3),1:(len-3)) can be expressed more elegantly?
I have just tried the function proposed by Fmerhout. It seems like an excellent solution. thanks a lot. But because of the last line of code:
passw <- str_c (sample (pass_chars), collapse = "")
the function does not work.
I tried :
passw <- str (sample (pass_chars), collapse = "")
... and now it works:
example with a seed:
>gen_pass(4,2)
gives: chr [1:4] "y" "[" "O" "1"
In order to obtain a directly usable password, I changed the end of the code this way:
zz <- sample(pass_chars)
passw <- paste(zz, sep = "", collapse = "")
return(passw)
}
So, now we get for example:
> gen_pass(35, 2)
[1] "0OD}1O}8DKMqTL[JEFZBwKMJWGD’VZ=VRnD"
It's interesting; because we only have to remember the parameters passed to the function. here, in this case: 35 2.
Thank you Mr Fmerhout.
To conclude : with this little script we have a good way to create very strong and very safe passwords with good entropy without having to use a dictionary and without having to record in any place our passwords .
I have a function, where one part reads as follows:
conefor.input <- function(conefor.file, onlyoverall, probmin, index)
{
for(i in 1:l00)
{
cat(paste(conefor.file,
if(onlyoverall=="TRUE")
{onlyoverall<-"onlyoverall"},
distance,
if(probmin=="TRUE")
{probmin<-paste("-pcHeur", min)},
index, sep="\t"),file="conef_command.txt", sep="\n")}
return("conef_command.txt")
}
My objective is to generate 100 lines where each input is divided by a tab.
There is no problem when the two 'if' statements above are true. With the resulting being:
conefor.file\tonlyoverall\tdistance\tpcHeur min\tindex
However, when the two 'if' statements above are false, I am left with two tabs where 'onlyoverall' and 'pcHeur' would have been, whereas, what I really want is no action at all, and only one tab separating each argument.
Example when 'if' statements are false:
conefor.file\t\tdistance\t\tindex
What I want when 'if' statements are false:
conefor.file\tdistance\tindex
Many thanks in advance
it's hard to suggest any form of vectorization or similar for your for loop, since it's not clear (to me at least)
what differences there are between iterations.
However, as for the paste statement inside of cat, you can use the following instead
paste(ifelse(onlyoverall,
paste(conefor.file, onlyoverall<-"onlyoverall", sep="\t"),
conefor.file),
ifelse(probmin,
paste(distance, probmin<-paste("-pcHeur", min), sep="\t"),
distance),
index,
sep="\t")
Notice, this grabs the output prior to the if statement and includes in an ifelse output, where it is given for both TRUE and FALSE
eg:
x1 <- x2 <- TRUE
paste(ifelse(x1, paste("A", "B" , sep="#"), "A"), ifelse(x2, paste("C", "D", sep="#"), "C"), "E", sep="#")
# [1] "A#B#C#D#E"
x1 <- x2 <- FALSE
paste(ifelse(x1, paste("A", "B" , sep="#"), "A"), ifelse(x2, paste("C", "D", sep="#"), "C"), "E", sep="#")
# [1] "A#C#E"
x2 <- TRUE
paste(ifelse(x1, paste("A", "B" , sep="#"), "A"), ifelse(x2, paste("C", "D", sep="#"), "C"), "E", sep="#")
# [1] "A#C#D#E"