总结R数据框中的因子分布

时间:2016-08-14 20:57:32

标签: r dataframe summarization

假设我有一个像这样的data.frame:

  X1   X2   X3
1 A    B    A
2 A    C    B
3 B    A    B
4 A    A    C

我想计算每列中A,B,C等的出现次数,并将结果返回为

    A_count B_count C_count
X1  3       1       0       
X2  2       1       1
X3  1       2       1

我确定这个问题有一千个重复,但我似乎无法找到适合我的答案:(

运行

apply(mydata, 2, table)

我得到像

这样的东西
$X1
   B     A
   1     3
$X2
   A     C     B
   2     1     1

但它并不完全是我想要的,如果我尝试将其重新构建到数据框中,它就无法工作,因为我不会为每一行获得相同数量的列(比如上面没有C&#39的$ X1)。

我错过了什么?

非常感谢!

2 个答案:

答案 0 :(得分:3)

您可以重构以包含每列共有的因子级别,然后制表。我还建议使用lapply()代替apply(),因为apply()代表矩阵。

df <- read.table(text = "X1   X2   X3
1 A    B    A
2 A    C    B
3 B    A    B
4 A    A    C", h=T)

do.call(
    rbind, 
    lapply(df, function(x) table(factor(x, levels=levels(unlist(df)))))
)
#    A B C
# X1 3 1 0
# X2 2 1 1
# X3 1 2 1

答案 1 :(得分:3)

假设您的数据框为x,我只会这样做:

do.call(rbind, tapply(unlist(x, use.names = FALSE),
                      rep(1:ncol(x), each = nrow(x)),
                      table))

#  A B C
#1 3 1 0
#2 2 1 1
#3 1 2 1

<强>基准

# a function to generate toy data
# `k` factor levels
# `n` row
# `p` columns
datsim <- function(n, p, k) {
  as.data.frame(replicate(p, sample(LETTERS[1:k], n, TRUE), simplify = FALSE),
                col.names = paste0("X",1:p), stringsAsFactors = TRUE)
  }

# try `n = 100`, `p = 500` and `k = 3`
x <- datsim(100, 500, 3)

## DirtySockSniffer's answer
system.time(do.call(rbind, lapply(x, function(u) table(factor(u, levels=levels(unlist(x)))))))
#   user  system elapsed 
# 21.240   0.068  21.365 

## my answer
system.time(do.call(rbind, tapply(unlist(x, use.names = FALSE), rep(1:ncol(x), each = nrow(x)), table)))
#   user  system elapsed 
#  0.108   0.000   0.111 

通过以下方式可以改善肮脏的答案:

## improved DirtySockSniffer's answer
system.time({clevels <- levels(unlist(x, use.names = FALSE));
             do.call(rbind, lapply(x, function(u) table(factor(u, levels=clevels))))})
#   user  system elapsed 
#  0.108   0.000   0.108

另请考虑user20650的回答

## Let's try a large `n`, `p`, `k`
x <- datsim(200, 5000, 5)

system.time(t(table(stack(lapply(x, as.character)))))
#   user  system elapsed 
#  0.592   0.052   0.646 

虽然我的回答是:

system.time(do.call(rbind, tapply(unlist(x, use.names = FALSE), rep(1:ncol(x), each = nrow(x)), table)))
#   user  system elapsed 
#  1.844   0.056   1.904 

改进了Dirty的回答:

system.time({clevels <- levels(unlist(x, use.names = FALSE));
             do.call(rbind, lapply(x, function(u) table(factor(u, levels=clevels))))})
#   user  system elapsed 
#  1.240   0.012   1.263