R 匹配列表中的元素

刚开始在R。。。在这件事上被难住了，也许是因为我不知道从哪里开始

定义一个随机变量，使其等于匹配前的试验次数。如果你有一个数字列表，4,5,7,11,3,11,12,8,8,1…，随机变量的第一个值是6，因为那时有两个11。4,5,7,11,3,11第二个值是3，因为那时你有两个8…，12,8,8。 下面的代码通过模拟均匀分布创建数字列表u

谢谢你的帮助和指点。如果有人有兴趣尝试通过编写统计文本来学习，我将在下面完整描述我正在解决的问题

set.seed(1); u = matrix(runif(1000), nrow=1000)
u[u > 0    & u <= 1/12]   <- 1
u[u > 1/12 & u <= 2/12]   <- 2
u[u > 2/12 & u <= 3/12]   <- 3
u[u > 3/12 & u <= 4/12]   <- 4
u[u > 4/12 & u <= 5/12]   <- 5
u[u > 5/12 & u <= 6/12]   <- 6
u[u > 6/12 & u <= 7/12]   <- 7
u[u > 7/12 & u <= 8/12]   <- 8
u[u > 8/12 & u <= 9/12]   <- 9
u[u > 9/12 & u <= 10/12]  <- 10
u[u > 10/12 & u <= 11/12] <- 11
u[u > 11/12 & u < 12/12] <- 12
table(u); u[1:10,]

示例2.6-3 Higgins概率和随机建模中的概念

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假设我们随机询问人们出生的月份。让随机变量X表示我们在发现两个同月出生的人之前需要询问的人数。X的可能值为2,3，…13。也就是说，至少需要询问两个人才能进行比赛，并且询问的人数不超过13人。通过简化假设，每个月都是一个同样可能的响应候选，使用计算机模拟来估计X的概率质量函数。模拟生成出生月，直到找到匹配。在1000次重复实验的基础上，获得了以下经验分布和样本统计…

R具有陡峭的初始学习曲线。我认为假设这是你的家庭作业是不公平的，是的，如果你知道你在寻找什么，就有可能找到解决办法。然而，我记得有时很难在网上研究问题，因为我不知道搜索什么，也不太熟悉术语

下面是解决R中问题的一种方法的解释。阅读注释代码，并尝试准确地了解它在做什么。尽管如此，我还是建议使用一个好的初学者资源。从记忆中，一个很好的开始和运行是，但有很多在那里

# set the number of simulations
nsim <- 10000

# Create a matrix, with nsim columns, and fill it with something. 
#  The something with which you'll populate it is a random sample, 
#  with replacement, of month names (held in a built-in vector called
#  'month.abb'). We're telling the sample function that it should take 
#  13*nsim samples, and these will be used to fill the matrix, which 
#  has nsim columns (and hence 13 rows). We've chosen to take samples 
#  of length 13, because as your textbook states, 13 is the maximum
#  number of month names necessary for a month name to be duplicated.
mat <- matrix(sample(month.abb, 13*nsim, replace=TRUE), ncol=nsim)

# If you like, take a look at the first 10 columns
mat[, 1:10]

# We want to find the position of the first duplicated value for each column. 
#  Here's one way to do this, but it might be a bit confusing if you're just 
#  starting out. The 'apply' family of functions is very useful for
#  repeatedly applying a function to columns/rows/elements of an object.
#  Here, 'apply(mat, 2, foo)' means that for each column (2 represents columns,
#  1 would apply to rows, and 1:2 would apply to every cell), do 'foo' to that
#  column. Our function below extends this a little with a custom function. It
#  says: for each column of mat in turn, call that column 'x' and perform 
#  'match(1, duplicated(x))'. This match function will return the position
#  of the first '1' in the vector 'duplicated(x)'. The vector 'duplicated(x)'
#  is a logical (boolean) vector that indicates, for each element of x,
#  whether that element has already occurred earlier in the vector (i.e. if 
#  the month name has already occurred earlier in x, the corresponding element
#  of duplicated(x) will be TRUE (which equals 1), else it will be false (0).
#  So the match function returns the position of the first duplicated month 
#  name (well, actually the second instance of that month name). e.g. if 
#  x consists of 'Jan', 'Feb', 'Jan', 'Mar', then duplicated(x) will be 
#  FALSE, FALSE, TRUE, FALSE, and match(1, duplicated(x)) will return 3. 
#  Referring back to your textbook problem, this is x, a realisation of the 
#  random variable X.
# Because we've used the apply function, the object 'res' will end up with
#  nsim realisations of X, and these can be plotted as a histogram.
res <- apply(mat, 2, function(x) match(1, duplicated(x)))
hist(res, breaks=seq(0.5, 13.5, 1))

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R具有陡峭的初始学习曲线。我认为假设这是你的家庭作业是不公平的，是的，如果你知道你在寻找什么，就有可能找到解决办法。然而，我记得有时很难在网上研究问题，因为我不知道搜索什么，也不太熟悉术语

下面是解决R中问题的一种方法的解释。阅读注释代码，并尝试准确地了解它在做什么。尽管如此，我还是建议使用一个好的初学者资源。从记忆中，一个很好的开始和运行是，但有很多在那里

# set the number of simulations
nsim <- 10000

# Create a matrix, with nsim columns, and fill it with something. 
#  The something with which you'll populate it is a random sample, 
#  with replacement, of month names (held in a built-in vector called
#  'month.abb'). We're telling the sample function that it should take 
#  13*nsim samples, and these will be used to fill the matrix, which 
#  has nsim columns (and hence 13 rows). We've chosen to take samples 
#  of length 13, because as your textbook states, 13 is the maximum
#  number of month names necessary for a month name to be duplicated.
mat <- matrix(sample(month.abb, 13*nsim, replace=TRUE), ncol=nsim)

# If you like, take a look at the first 10 columns
mat[, 1:10]

# We want to find the position of the first duplicated value for each column. 
#  Here's one way to do this, but it might be a bit confusing if you're just 
#  starting out. The 'apply' family of functions is very useful for
#  repeatedly applying a function to columns/rows/elements of an object.
#  Here, 'apply(mat, 2, foo)' means that for each column (2 represents columns,
#  1 would apply to rows, and 1:2 would apply to every cell), do 'foo' to that
#  column. Our function below extends this a little with a custom function. It
#  says: for each column of mat in turn, call that column 'x' and perform 
#  'match(1, duplicated(x))'. This match function will return the position
#  of the first '1' in the vector 'duplicated(x)'. The vector 'duplicated(x)'
#  is a logical (boolean) vector that indicates, for each element of x,
#  whether that element has already occurred earlier in the vector (i.e. if 
#  the month name has already occurred earlier in x, the corresponding element
#  of duplicated(x) will be TRUE (which equals 1), else it will be false (0).
#  So the match function returns the position of the first duplicated month 
#  name (well, actually the second instance of that month name). e.g. if 
#  x consists of 'Jan', 'Feb', 'Jan', 'Mar', then duplicated(x) will be 
#  FALSE, FALSE, TRUE, FALSE, and match(1, duplicated(x)) will return 3. 
#  Referring back to your textbook problem, this is x, a realisation of the 
#  random variable X.
# Because we've used the apply function, the object 'res' will end up with
#  nsim realisations of X, and these can be plotted as a histogram.
res <- apply(mat, 2, function(x) match(1, duplicated(x)))
hist(res, breaks=seq(0.5, 13.5, 1))

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所有这些代码都可以更容易地用u-GoogleforR来代替。生日问题带来了一些可能被证明非常有用的匹配项。如果这是家庭作业，那么值得阅读社区指南不，这不是家庭作业。我试图通过将R编码应用于旧的统计文本来学习它。谢谢你所有的评论，它们被证明是非常有用的。找到Bruce Trumbo的pdf，这导致了Suess和Trumbo的一本书，非常适合我的目的。介绍概率模拟和Gibbs采样与RAll，代码可以更容易地替换为u Google for r Birth问题提出了一些匹配，可能会被证明非常有用。如果这是家庭作业，值得阅读社区指南不，这不是家庭作业。我试图通过将R编码应用于旧的统计文本来学习它。谢谢你所有的评论，它们被证明是非常有用的。找到Bruce Trumbo的pdf，这导致了Suess和Trumbo的一本书，非常适合我的目的。概率模拟与R-Gibbs抽样简介