我正在寻找一种在R中编码密码生成器功能的智能方法:
generate.password (length, capitals, numbers)
示例:
generate.password(8)
[1] "hqbfpozr"
generate.password(length=8, capitals=c(2,4))
[1] "hYbFpozr"
generate.password(length=8, capitals=c(2,4), numbers=c(7:8))
[1] "hYbFpo49"
答案 0 :(得分:11)
这是一种方法
generate.password <- function(length,
capitals = integer(0),
numbers = integer(0)) {
stopifnot(is.numeric(length), length > 0L,
is.numeric(capitals), capitals > 0L, capitals <= length,
is.numeric(numbers), numbers > 0L, numbers <= length,
length(intersect(capitals, numbers)) == 0L)
lc <- sample(letters, length, replace = TRUE)
uc <- sample(LETTERS, length(capitals), replace = TRUE)
num <- sample(0:9, length(numbers), replace = TRUE)
pass <- lc
pass[capitals] <- uc
pass[numbers] <- num
paste0(pass, collapse = "")
}
## Examples
set.seed(1)
generate.password(8)
# [1] "gjoxfxyr"
set.seed(1)
generate.password(length=8, capitals=c(2,4))
# [1] "gQoBfxyr"
set.seed(1)
generate.password(length=8, capitals=c(2,4), numbers=c(7:8))
# [1] "gQoBfx21"
您还可以以相同的方式添加其他特殊字符。如果您想要重复字母和数字的值,请在replace =TRUE
函数中添加sample
。
答案 1 :(得分:11)
有一个函数可以在stringi(版本&gt; = 0.2-3)包中生成随机字符串:
require(stringi)
stri_rand_strings(n=2, length=8, pattern="[A-Za-z0-9]")
## [1] "90i6RdzU" "UAkSVCEa"
因此,使用不同的模式,您可以为所需的密码生成部分,然后将其粘贴如下:
x <- stri_rand_strings(n=4, length=c(2,1,2,3), pattern=c("[a-z]","[A-Z]","[0-9]","[a-z]"))
x
## [1] "ex" "N" "81" "tsy"
stri_flatten(x)
## [1] "exN81tsy"
答案 2 :(得分:0)
我喜欢@Hadd E. Nuff给出的解决方案......我所做的是包含0到9之间的数字,随机...这里是修改后的解决方案......
generate.password <- function(LENGTH){
punct <- c("!", "#", "$", "%", "&", "(", ")", "*", "+", "-", "/", ":",
";", "<", "=", ">", "?", "@", "[", "^", "_", "{", "|", "}", "~")
nums <- c(0:9)
chars <- c(letters, LETTERS, punct, nums)
p <- c(rep(0.0105, 52), rep(0.0102, 25), rep(0.02, 10))
pword <- paste0(sample(chars, LENGTH, TRUE, prob = p), collapse = "")
return(pword)
}
这将生成非常强大的密码,如:
"C2~mD20U" # 8 alpha-numeric-specialchar
"+J5Gi3" # 6 alpha-numeric-specialchar
"77{h6RsGQJ66if5" # 15 alpha-numeric-specialchar