洗牌的非正态抽样
洗牌的非正态抽样
提问于 2021-06-29 23:56:39
我有一个脚本，它使用sample()将一副牌洗成四手-如下所示。
此脚本生成正态分布的手部分割。这是在一套花色中的13张牌在四手中的分布将是"4432“21.6%的时间，"5442”15.5%的时间，... -请参阅此处：https://en.wikipedia.org/wiki/Contract_bridge_probabilities#Hand%20pattern%20probabilities。
有没有办法让sample()给出非正态分布，比如展平轮廓，这样分布就会偏离预期概率？sample()可以接受一个加权参数，但看不到如何使用它来实现我的目标。或者，是否有其他采样函数可以提供此功能？
谢谢,
TC
# Set up
library(tidyverse)
set.seed(123)
# Build pack
pack <- expand.grid(rank = c("A", 2:9, "T", "J", "Q", "K"), suit = c("S", "H", "D", "C")) %>%
as_tibble(.name_repair = "minimal") %>%
mutate(card = paste(suit, rank, sep = "-"))
# Divide cards into hands
for (i in 1:4) {
temp <- sample(pack$card, 13, replace = FALSE) %>%
as_tibble(.name_repair = "minimal") %>%
separate(value, sep = "-", into = c("suit", "rank")) %>%
mutate(
suit = factor(suit, levels = c("S", "H", "D", "C")),
rank = factor(rank, levels = c("A", "K", "Q", "J", "T", 9:2, " "))
) %>%
arrange(suit, rank) %>%
unite("card", sep = "-")
assign(glue::glue("hand{i}"), temp)
pack <- pack %>%
filter(!card %in% unname(unlist(temp)))
}
# Reassemble pack
pack <- hand1 %>%
cbind(hand2) %>%
cbind(hand3) %>%
cbind(hand4) %>%
rename(N = 1, E = 2, S = 3, W = 4)回答 1
Stack Overflow用户
发布于 2021-06-30 06:54:50
您可以简化创建卡片组、洗牌和处理的过程，如下所示：
# Create the deck
Suit <- c("S", "H", "D", "C")
Rank <- c("A", 2:9, "T", "J", "Q", "K")
Deck <- data.frame(Rank=rep(Rank, 4), Suit=rep(Suit, each=13))
# Shuffle and deal
Shuffle <- Deck[sample(nrow(Deck)), ] # Shuffle the deck
Hand <- factor(rep(c("N", "E", "S", "W"), 13), levels=c("N", "E", "S", "W"))
Deal <- data.frame(Hand, Shuffle)现在发牌是一个数据框，显示了每一手中的牌。要获得按花色分配的卡片，请：
xtabs(~Suit+Hand, Even)
# Hand
# Suit N E S W
# C 3 3 3 4
# D 3 3 4 3
# H 3 4 3 3
# S 4 3 3 3要获得手，请执行以下操作：
split(Deal, Deal$Hand)
# $N
# Hand Rank Suit
# 12 N Q S
# 2 N 2 S
# 37 N J D
# 49 N T C
# 43 N 4 C
# 17 N 4 H
# 33 N 7 D
# 26 N K H
# 13 N K S
# 44 N 5 C
# 18 N 5 H
# 46 N 7 C
# 11 N J S
. . . . .既然您提到了手部模式概率，那么展示如何根据经验生成它们可能会很有帮助：
HPP <- function() {
Shuffle <- Deck[sample(nrow(Deck)), ] # Shuffle the deck
Hand <- factor(rep(c("N", "E", "S", "W"), 13), levels=c("N", "E", "S", "W"))
Deal <- data.frame(Hand, Shuffle)
tbl <- xtabs(~Suit+Hand, Deal)
return(unname(apply(tbl, 1, function(x) paste(sort(x, decreasing=TRUE), collapse="-"))))
}
X <- replicate(1000, HPP())
HP <- prop.table(sort(table(X), decreasing=TRUE))
as.matrix(HP)
# [,1]
# 4-4-3-2 0.22200
# 5-3-3-2 0.14925
# 5-4-3-1 0.12075
# 4-3-3-3 0.11225
# 5-4-2-2 0.10775
# 6-3-2-2 0.05175
# 6-4-2-1 0.04850
# . . . .为了获得更多的“均匀”分布的牌，你必须打破随机分布的随机洗牌，例如，我们可以洗牌中的排名值，但保持花色的顺序：
Even <- Deck
Even$Rank <- c(replicate(4, sample(Rank)))
Even <- data.frame(Hand, Even)现在，牌值会有所不同，但花色的分配总是相同的。
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原文链接：
https://stackoverflow.com/questions/68182103
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