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R 将单个值与每个直方图单元的计数相结合_R_Count_Histogram_Frequency - Fatal编程技术网

R 将单个值与每个直方图单元的计数相结合

R 将单个值与每个直方图单元的计数相结合,r,count,histogram,frequency,R,Count,Histogram,Frequency,我有一个1000值的向量,然后用100个箱子做一个直方图。我想做一个向量=(Value,Freq),其中的值是单独的值(因此我将有1000个),Freq=特定值所在的容器中的值的计数 希望这能让问题变得更清楚: data <- c(6.429229, 9.300965, 11.073744, 6.527263, 8.425178, 6.821384, 6.515991, 9.131452, 6.313888, 8.866572) Myhist <- hist

我有一个1000值的向量,然后用100个箱子做一个直方图。我想做一个向量=(Value,Freq),其中的值是单独的值(因此我将有1000个),Freq=特定值所在的容器中的值的计数

希望这能让问题变得更清楚:

data <- c(6.429229, 9.300965, 11.073744, 6.527263, 8.425178, 
          6.821384, 6.515991,  9.131452, 6.313888, 8.866572) 
Myhist <- hist(data,2)
Myhist$counts
# 5 4 1

# c(5,4,1,5,4,5,5,4,5,4)
MyDF <- cbind(data,c(5,4,1,5,4,5,5,4,5,4))
# Result I want: 

# [1,]  6.429229 5
# [2,]  9.300965 4
# [3,] 11.073744 1
# [4,]  6.527263 5
# [5,]  8.425178 4
# [6,]  6.821384 5
# [7,]  6.515991 5
# [8,]  9.131452 4
# [9,]  6.313888 5
#[10,]  8.866572 4

data根据您的示例,我猜您想知道的是,在直方图中,落入同一个容器中的值(包括有问题的值)的数量。您似乎还指定了直方图分段

data <- c(6.429229, 9.300965, 11.073744, 6.527263, 8.425178, 
          6.821384, 6.515991,  9.131452, 6.313888, 8.866572)

get.num.in.bin <- function(data, hist.breaks=2){
  Myhist   <- hist(data, breaks=hist.breaks, plot=FALSE)
  cats     <- as.numeric(cut(data, breaks=Myhist$breaks, labels=1:3))
  counts   <- Myhist$counts[cats]
  new.data <- data.frame(data=data, num.in.bin=counts)
  return(new.data)
}
get.num.in.bin(data)
#         data num.in.bin
# 1   6.429229          5
# 2   9.300965          4
# 3  11.073744          1
# 4   6.527263          5
# 5   8.425178          4
# 6   6.821384          5
# 7   6.515991          5
# 8   9.131452          4
# 9   6.313888          5
# 10  8.866572          4

data根据您的示例,我猜您想知道的是,在直方图中,落入同一个容器中的值(包括有问题的值)的数量。您似乎还指定了直方图分段

data <- c(6.429229, 9.300965, 11.073744, 6.527263, 8.425178, 
          6.821384, 6.515991,  9.131452, 6.313888, 8.866572)

get.num.in.bin <- function(data, hist.breaks=2){
  Myhist   <- hist(data, breaks=hist.breaks, plot=FALSE)
  cats     <- as.numeric(cut(data, breaks=Myhist$breaks, labels=1:3))
  counts   <- Myhist$counts[cats]
  new.data <- data.frame(data=data, num.in.bin=counts)
  return(new.data)
}
get.num.in.bin(data)
#         data num.in.bin
# 1   6.429229          5
# 2   9.300965          4
# 3  11.073744          1
# 4   6.527263          5
# 5   8.425178          4
# 6   6.821384          5
# 7   6.515991          5
# 8   9.131452          4
# 9   6.313888          5
# 10  8.866572          4

数据我不太明白这一点。你能为人们提供一个工作环境吗?如果你有1000个独特的损失,那么你将需要1k个箱子&每个箱子中观察到的损失数将是1。你能给我们一个输入和预期输出的例子吗,当然减少输入的大小(例如10个损失,5个箱子)嘿,对不起,我添加了一个基本的例子,希望这能使它更清晰。我不太明白这一点。你能为人们提供一个工作环境吗?如果你有1000个独特的损失,那么你将需要1k个箱子&每个箱子中观察到的损失数将为1。你能给我们一个输入和预期输出的例子,当然减少输入的大小(例如10个损失,5个箱子)嘿,对不起,我添加了一个基本的例子,希望这能让它更清楚