问题
I'm looking to permute (or combine) c("a","b","c") within six positions under the condition to have always sequences with alternate elements, e.g abcbab.
Permutations could easily get with:
abc<-c("a","b","c")
permutations(n=3,r=6,v=abc,repeats.allowed=T)
I think is not possible to do that with gtools, and I've been trying to design a function for that -even though I think it may already exist.
回答1:
Since you're looking for permutations, expand.grid can work as well as permutations. But since you don't want like-neighbors, we can shorten the dimensionality of it considerably. I think this is legitimate random-wise!
Up front:
r <- replicate(6, seq_len(length(abc)-1), simplify=FALSE)
r[[1]] <- c(r[[1]], length(abc))
m <- t(apply(do.call(expand.grid, r), 1, cumsum) %% length(abc) + 1)
m[] <- abc[m]
dim(m)
# [1] 96 6
head(as.data.frame(cbind(m, apply(m, 1, paste, collapse = ""))))
# Var1 Var2 Var3 Var4 Var5 Var6 V7
# 1 b c a b c a bcabca
# 2 c a b c a b cabcab
# 3 a b c a b c abcabc
# 4 b a b c a b babcab
# 5 c b c a b c cbcabc
# 6 a c a b c a acabca
Walk-through:
- since you want all recycled permutations of it, we can use
gtools::permutations, or we can useexpand.grid... I'll use the latter, I don't know if it's much faster, but it does a short-cut I need (more later) - when dealing with constraints like this, I like to expand on the indices of the vector of values
however, since we don't want neighbors to be the same, I thought that instead of each row of values being the straight index, we
cumsumthem; by using this, we can control the ability of the cumulative sum to re-reach the same value ... by removing0andlength(abc)from the list of possible values, we remove the possibility of (a) never staying the same, and (b) never increasing actually one vector-length (repeating the same value); as a walk-through:head(expand.grid(1:3, 1:2, 1:2, 1:2, 1:2, 1:2), n = 6) # Var1 Var2 Var3 Var4 Var5 Var6 # 1 1 1 1 1 1 1 # 2 2 1 1 1 1 1 # 3 3 1 1 1 1 1 # 4 1 2 1 1 1 1 # 5 2 2 1 1 1 1 # 6 3 2 1 1 1 1Since the first value can be all three values, it's
1:3, but each additional is intended to be 1 or 2 away from it.head(t(apply(expand.grid(1:3, 1:2, 1:2, 1:2, 1:2, 1:2), 1, cumsum)), n = 6) # Var1 Var2 Var3 Var4 Var5 Var6 # [1,] 1 2 3 4 5 6 # [2,] 2 3 4 5 6 7 # [3,] 3 4 5 6 7 8 # [4,] 1 3 4 5 6 7 # [5,] 2 4 5 6 7 8 # [6,] 3 5 6 7 8 9okay, that doesn't seem that useful (since it goes beyond the length of the vector), so we can invoke the modulus operator and a shift (since modulus returns 0-based, we want 1-based):
head(t(apply(expand.grid(1:3, 1:2, 1:2, 1:2, 1:2, 1:2), 1, cumsum) %% 3 + 1), n = 6) # Var1 Var2 Var3 Var4 Var5 Var6 # [1,] 2 3 1 2 3 1 # [2,] 3 1 2 3 1 2 # [3,] 1 2 3 1 2 3 # [4,] 2 1 2 3 1 2 # [5,] 3 2 3 1 2 3 # [6,] 1 3 1 2 3 1To verify this works, we can do a
diffacross each row and look for0:m <- t(apply(expand.grid(1:3, 1:2, 1:2, 1:2, 1:2, 1:2), 1, cumsum) %% 3 + 1) any(apply(m, 1, diff) == 0) # [1] FALSEto automate this to an arbitrary vector, we enlist the help of
replicateto generate the list of possible vectors:r <- replicate(6, seq_len(length(abc)-1), simplify=FALSE) r[[1]] <- c(r[[1]], length(abc)) str(r) # List of 6 # $ : int [1:3] 1 2 3 # $ : int [1:2] 1 2 # $ : int [1:2] 1 2 # $ : int [1:2] 1 2 # $ : int [1:2] 1 2 # $ : int [1:2] 1 2and then
do.callto expand it.one you have the matrix of indices,
head(m) # Var1 Var2 Var3 Var4 Var5 Var6 # [1,] 2 3 1 2 3 1 # [2,] 3 1 2 3 1 2 # [3,] 1 2 3 1 2 3 # [4,] 2 1 2 3 1 2 # [5,] 3 2 3 1 2 3 # [6,] 1 3 1 2 3 1and then replace each index with the vector's value:
m[] <- abc[m] head(m) # Var1 Var2 Var3 Var4 Var5 Var6 # [1,] "b" "c" "a" "b" "c" "a" # [2,] "c" "a" "b" "c" "a" "b" # [3,] "a" "b" "c" "a" "b" "c" # [4,] "b" "a" "b" "c" "a" "b" # [5,] "c" "b" "c" "a" "b" "c" # [6,] "a" "c" "a" "b" "c" "a"and then we
cbindthe united string (viaapplyandpaste)
Performance:
library(microbenchmark)
library(dplyr)
library(tidyr)
library(stringr)
microbenchmark(
tidy1 = {
gtools::permutations(n = 3, r = 6, v = abc, repeats.allowed = TRUE) %>%
data.frame() %>%
unite(united, sep = "", remove = FALSE) %>%
filter(!str_detect(united, "([a-c])\\1"))
},
tidy2 = {
filter(unite(data.frame(gtools::permutations(n = 3, r = 6, v = abc, repeats.allowed = TRUE)),
united, sep = "", remove = FALSE),
!str_detect(united, "([a-c])\\1"))
},
base = {
r <- replicate(6, seq_len(length(abc)-1), simplify=FALSE)
r[[1]] <- c(r[[1]], length(abc))
m <- t(apply(do.call(expand.grid, r), 1, cumsum) %% length(abc) + 1)
m[] <- abc[m]
},
times=10000
)
# Unit: microseconds
# expr min lq mean median uq max neval
# tidy1 1875.400 2028.8510 2446.751 2165.651 2456.051 12790.901 10000
# tidy2 1745.402 1875.5015 2284.700 2000.051 2278.101 50163.901 10000
# base 796.701 871.4015 1020.993 919.801 1021.801 7373.901 10000
I tried the infix (non-%>%) tidy2 version just for kicks, and though I was confident it would theoretically be faster, I didn't realize it would shave over 7% off the run-times. (The 50163 is likely R garbage-collecting, not "real".) The price we pay for readability/maintainability.
回答2:
There are probably cleaner methods, but here ya go:
abc <- letters[1:3]
library(tidyverse)
res <- gtools::permutations(n = 3, r = 6, v = abc, repeats.allowed = TRUE) %>%
data.frame() %>%
unite(united, sep = "", remove = FALSE) %>%
filter(!str_detect(united, "([a-c])\\1"))
head(res)
united X1 X2 X3 X4 X5 X6
1 ababab a b a b a b
2 ababac a b a b a c
3 ababca a b a b c a
4 ababcb a b a b c b
5 abacab a b a c a b
6 abacac a b a c a c
If you want a vector, you can use res$united or add %>% pull(united) as an additional step at the end of the pipes above.
来源：https://stackoverflow.com/questions/53566191/permutations-of-3-elements-within-6-positions