英文:
Replace values in one column with randomly generated values using group_by and mutate, while making sure every set of values is unique in R
问题
ifelse(!duplicated(data.new$ID), generateRandomString(), data.new$ID)
英文:
I am writing a function to replace ID's in one column with randomly generated ones. Suppose I have a dataset:
df <- data.frame(ID = c(1, 1, 2, 2, 1, 3), Name = c("Joseph", "Joseph", "Leo", "Leo", "Joseph", "David"))
ID Name1 1 Joseph2 1 Joseph3 2 Leo4 2 Leo5 1 Joseph6 3 David
The goal of the function is to group the dataset by ID and replace all unique ID's with randomly generated ones, like this:
library(tidyverse)generateRandomString <- function() {sample(LETTERS, 1)}replaceUniqueID <- function(data, column_id) {name.id <- data[grep(column_id, colnames(data))]data.new <- data %>%group_by(data[grep(column_id, colnames(data))]) %>%mutate_at(column_id, funs(generateRandomString()))data.new}replaceUniqueID(df, "ID")# A tibble: 6 × 2# Groups: ID [3]ID Name<chr> <chr>1 D Joseph2 D Joseph3 R Leo4 R Leo5 D Joseph6 Q David
My problem is, I want the code to make sure every ID gets replaced with unique string, e.g. David and Leo can't both have D as a replacement for their respective ID's.
I tried to come up with an ifelse statement within mutate but was unable to figure out how to do that.
答案1
得分: 0
I will only translate the code part for you:
我认为你可以使用翻译词典来相对容易地解决这个问题。genRandomIDs <- function(ID, min_length = 2) {ID <- unique(ID)len <- max(min_length, length(ID) %/% 26 + 1)ltrs <- do.call(paste0,do.call(expand.grid, replicate(len, LETTERS, simplify=FALSE)))sample(ltrs, length(ID))}set.seed(42)IDdict <- df %>%distinct(ID) %>%mutate(newID = genRandomIDs(ID))IDdict# ID newID# 1 1 OV# 2 2 IM# 3 3 WFdf %>%left_join(IDdict, by = "ID")# ID Name newID# 1 1 Joseph OV# 2 1 Joseph OV# 3 2 Leo IM# 4 2 Leo IM# 5 1 Joseph OV# 6 3 David WF
这是代码的翻译部分。
英文:
I think you can use a translation dictionary to solve this relatively easily.
genRandomIDs <- function(ID, min_length = 2) {ID <- unique(ID)len <- max(min_length, length(ID) %/% 26 + 1)ltrs <- do.call(paste0,do.call(expand.grid, replicate(len, LETTERS, simplify=FALSE)))sample(ltrs, length(ID))}set.seed(42)IDdict <- df %>%distinct(ID) %>%mutate(newID = genRandomIDs(ID))IDdict# ID newID# 1 1 OV# 2 2 IM# 3 3 WFdf %>%left_join(IDdict, by = "ID")# ID Name newID# 1 1 Joseph OV# 2 1 Joseph OV# 3 2 Leo IM# 4 2 Leo IM# 5 1 Joseph OV# 6 3 David WF
Walk-through:
genRandomIDsis just a helper function that internally produces a vector of alln-long letter permutations (combined withpaste0) and samples from them;- the
do.call(expand.grid, ...)gives us a frame that expands on eachlengrouping of letters; that is,expand.grid(LETTERS[1:3],LETTERS[1:3],LETTERS[1:3])gives us3^3permutations of three letters - the
do.call(paste0, ...)takes that frame fromexpand.grid(which is really just a glorifiedlist) and produces one string per "row".
- the
distinct(ID)reduces yourdfto just one row perID;- since we produce one
newIDfor each uniqueID, we now have a 1-to-1 mapping from old-to-new; - the
left_joinassigns thenewIDfor each row (if you aren't familiar with merges/joins, see https://stackoverflow.com/q/1299871/3358272, https://stackoverflow.com/q/5706437/3358272)
Note: this does not really scale well: since we explode the possible combinations with expand.grid, for a min-length of 2 letters, we produce 676 (26^2) permutations, not a problem. 26^3 produces 17576 possible combinations, whether or not we have that many IDs to uniquify. 26^4 (4 letters) produces 456976, and its delay is "palpable". Five letters is over 11 million, which becomes "stupid" to try to scale to that length (assuming you have that many unique IDs or choose a string of that long.
However ... while inefficient, this method is guaranteed to give you unique newIDs. There are other ways that may be guaranteed at the expense of a (however small) increase in complexity).
Okay, the "increased complexity" here for a more efficient process:
num2alpha <- function(num, chr = letters, zero = "", sep = "") {len <- length(chr)stopifnot(len > 1)signs <- ifelse(!is.na(num) & sign(num) < 0, "-", "")num <- as.integer(abs(num))is0 <- !is.na(num) & num < 1e-9# num[num < 1] <- NAout <- character(length(num))mult <- 0while (any(!is.na(num) & num > 0)) {not0 <- !is.na(num) & num > 0out[not0] <- paste0(chr[(num[not0] - 1) %% len + 1], sep, out[not0])num[not0] <- (num[not0] - 1) %/% len}if (nzchar(sep)) out <- sub(paste0(sep, "$"), "", out)out[is0] <- zeroout[is.na(num)] <- NAout[!is.na(out)] <- paste0(signs[!is.na(out)], out[!is.na(out)])out}IDdict <- df %>%distinct(ID) %>%mutate(newID = num2alpha(row_number()))IDdict# ID newID# 1 1 a# 2 2 b# 3 3 cdf %>%left_join(IDdict, by = "ID")# ID Name newID# 1 1 Joseph a# 2 1 Joseph a# 3 2 Leo b# 4 2 Leo b# 5 1 Joseph a# 6 3 David c
The num2alpha works more efficiently (using lower-case here, easily changed by using num2alpha(.., chr=LETTERS)), though it is deterministic here. If you are at all concerned about that, then
IDdict <- df %>%distinct(ID) %>%mutate(newID = sample(num2alpha(row_number())))
will randomize them for you.
Note that this produces single-letter strings up through 26, then cycles through 2-digit and 3-digit. It also recognized negatives, and while the defatul
num2alpha(c(-5, 0, NA, 1, 25:27, 51:53, 999999), zero="0")# [1] "-e" "0" NA "a" "y" "z" "aa" "ay" "az" "ba" "bdwgm"
(Note that this is not a simple base-converter, since we're ignoring "0"-values. Try num2alpha(14:16, c(1:9, LETTERS[1:6]), zero="0"). Perhaps it can be made to be more general.)
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