计算级别内的值

Question

我在R中使用cut生成一组级别，例如：说0到1之间的小数值，分为0.1个分档：

> frac <- cut(c(0, 1), breaks=10)
> levels(frac)
[1] "(-0.001,0.1]" "(0.1,0.2]"    "(0.2,0.3]"    "(0.3,0.4]"    "(0.4,0.5]"
[6] "(0.5,0.6]"    "(0.6,0.7]"    "(0.7,0.8]"    "(0.8,0.9]"    "(0.9,1]"

给定v包含[0.0, 1.0]之间连续值的向量v，如何计算levels(frac)中每个级别内v中元素的频率？

我可以自定义中断的数量和/或我创建级别的时间间隔，所以我正在寻找一种方法来使用标准R命令，这样我就可以构建一个两列数据框：一列用于作为因子的水平，以及

> table(frac)
frac
(-0.001,0.1]    (0.1,0.2]    (0.2,0.3]    (0.3,0.4]    (0.4,0.5]    (0.5,0.6]
           1            0            0            0            0            0
   (0.6,0.7]    (0.7,0.8]    (0.8,0.9]      (0.9,1]
           0            0            0            1

中该元素的总元素的分数或百分比值的第二列。

注意：以下内容不起作用：

cut

如果我直接在v上使用cut，那么当我在不同的向量上运行

frac = seq(0,1,by=0.1)

ranges = paste(head(frac,-1), frac[-1], sep=" - ")
freq   = hist(v, breaks=frac, include.lowest=TRUE, plot=FALSE)

data.frame(range = ranges, frequency = freq$counts)

时，我得不到相同的水平，因为值的范围 - 它们的最小值和最大值 - 在任意向量之间会有所不同，所以尽管我可能有相同数量的休息时间，等级间隔将不相同。

我的目标是采用不同的向量并将它们分成同一组级别。希望这有助于澄清我的问题。谢谢你的帮助。

Answer 1

frac

Answer 2

修改table实际代表您所需的间隔，然后使用

x = runif(100) # For example.
frac = cut(x, breaks = seq(0, 1, 0.1))
table(frac)

函数：

frac
  (0,0.1] (0.1,0.2] (0.2,0.3] (0.3,0.4] (0.4,0.5] (0.5,0.6] (0.6,0.7] (0.7,0.8]
       14         9         8        10         8        12         7         7
(0.8,0.9]   (0.9,1]
       16         9

结果：

c(0, 1)

Answer 3

将极端v引入cut，然后使用相同的

library(dplyr)

#dummy data
set.seed(1)
v <- round(runif(7), 2)

#result
data.frame(v,
           vFrac = cut(c(0, 1, v), breaks = 10)[-c(1, 2)]) %>% 
  group_by(vFrac) %>% 
  mutate(vFreq = n())

# Source: local data frame [10 x 3]
# Groups: vFrac [8]
# 
#        v        vFrac vFreq
#    <dbl>       <fctr> <int>
# 1   0.27    (0.2,0.3]     1
# 2   0.37    (0.3,0.4]     1
# 3   0.57    (0.5,0.6]     1
# 4   0.91      (0.9,1]     2
# 5   0.20    (0.1,0.2]     1
# 6   0.90    (0.8,0.9]     1
# 7   0.94      (0.9,1]     2

：

v<-data.frame(v=runif(100,0,1))

library(plyr)
v$x<-findInterval(v$v,seq(0,1,by=0.1))*0.1
ddply(v, .(x), summarize, n=length(x))

Answer 4

使用findInterval而不是cut：

frac = seq(0, 1, 0.1)
set.seed(42); v = rnorm(10, 0.5, 0.2)
sapply(1:(length(frac)-1), function(i) sum(frac[i]<v & frac[i+1]>=v))
#[1] 0 0 0 1 3 2 1 1 1 1

Answer 5

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