r - 在 R 中使用均值、总和、长度和 sd 实现频率计数的更简单方法

Question

我的任务是创建包含统计摘要的频率表。我的目标是创建一个可以简单地导出到 excel 的数据框。其中大部分可能在使用存储过程的 sql 中，但我决定在 R 中执行此操作。我正在学习 R，所以我可能会做很长的路要走。这是来自 getting-r-frequency-counts-for-all-possible-answers 的后续问题

给定

    Id <- c(1,2,3,4,5,6,7,8,9,10)
    ClassA <- c(1,NA,3,1,1,2,1,4,5,3)
    ClassB <- c(2,1,1,3,3,2,1,1,3,3)
    R <- c(1,2,3,NA,9,2,4,5,6,7)
    S <- c(3,7,NA,9,5,8,7,NA,7,6)
    df <- data.frame(Id,ClassA,ClassB,R,S)

    ZeroTenNAScale <- c(0:10,NA);

    R.freq <- setNames(nm=c('answer','value'),data.frame(table(factor(df$R,levels=ZeroTenNAScale,exclude=NULL))));
    R.freq[, 1] <- as.numeric(as.character( R.freq[, 1] ))
    R.freq <- cbind(question='R',R.freq)

    S.freq <- setNames(nm=c('answer','value'),data.frame(table(factor(df$S,levels=ZeroTenNAScale,exclude=NULL))));
    S.freq[, 1] <- as.numeric(as.character( S.freq[, 1] ))
    S.freq <- cbind(question='S',S.freq)

    R.mean = mean(df$R, na.rm = TRUE) 
    R.length = sum(!is.na(df$R)) 
    R.sd = sd(df$R, na.rm = TRUE) 
    R.sum = sum(df$R, na.rm = TRUE)

    S.mean = mean(df$S, na.rm = TRUE) 
    S.length = sum(!is.na(df$S)) 
    S.sd = sd(df$S, na.rm = TRUE) 
    S.sum = sum(df$S, na.rm = TRUE)

    S.row <- cbind('S','sum',as.numeric(S.sum))
    S.row <- setNames(nm=c('question','answer','value'),data.frame(S.row))
    S.freq = rbind(S.freq, S.row )

    S.row <- cbind('S','length',as.numeric(S.length))
    S.row <- setNames(nm=c('question','answer','value'),data.frame(S.row))
    S.freq = rbind(S.freq, S.row )

    S.row <- cbind('S','mean',as.numeric(S.mean))
    S.row <- setNames(nm=c('question','answer','value'),data.frame(S.row))
    S.freq = rbind(S.freq, S.row )

    S.row <- cbind('S','sd',as.numeric(S.sd))
    S.row <- setNames(nm=c('question','answer','value'),data.frame(S.row))
    S.freq = rbind(S.freq, S.row )

    R.row <- cbind('R','sum',as.numeric(R.sum))
    R.row <- setNames(nm=c('question','answer','value'),data.frame(R.row))
    R.freq = rbind(R.freq, R.row )

    R.row <- cbind('R','length',as.numeric(R.length))
    R.row <- setNames(nm=c('question','answer','value'),data.frame(R.row))
    R.freq = rbind(R.freq, R.row )

    R.row <- cbind('R','mean',as.numeric(R.mean))
    R.row <- setNames(nm=c('question','answer','value'),data.frame(R.row))
    R.freq = rbind(R.freq, R.row )

    R.row <- cbind('R','sd',as.numeric(R.sd))
    R.row <- setNames(nm=c('question','answer','value'),data.frame(R.row))
    R.freq = rbind(R.freq, R.row )

    result <- rbind(R.freq,S.freq)
    result <- cbind(filter='None',result)
    result  

我明白了

   filter question answer            value
1    None        R      0                0
2    None        R      1                1
3    None        R      2                2
4    None        R      3                1
5    None        R      4                1
6    None        R      5                1
7    None        R      6                1
8    None        R      7                1
9    None        R      8                0
10   None        R      9                1
11   None        R     10                0
12   None        R   <NA>                1
13   None        R    sum               39
14   None        R length                9
15   None        R   mean 4.33333333333333
16   None        R     sd 2.64575131106459
17   None        S      0                0
18   None        S      1                0
19   None        S      2                0
20   None        S      3                1
21   None        S      4                0
22   None        S      5                1
23   None        S      6                1
24   None        S      7                3
25   None        S      8                1
26   None        S      9                1
27   None        S     10                0
28   None        S   <NA>                2
29   None        S    sum               52
30   None        S length                8
31   None        S   mean              6.5
32   None        S     sd  1.8516401995451

这正是我要找的。我认为下一步是开始包装一些函数以简化代码，然后再开始添加来自 ClassA=1、ClassA=n+1 ... ClassA=NA、ClassB=1、ClassB= 的类似结果集2 ... ClassB=NA。有更简单的方法吗？

学习了Ernest A的答案后的新代码和 Imo是

    # https://stackoverflow.com/questions/36790376/a-simpler-way-to-achieve-a-frequency-count-with-mean-sum-length-and-sd-in-r/36794422#36794422

    # create the summary function
    summaryStatistics <- function(x) {
        xx <- na.omit(x)
        c(table(factor(x, levels=0:10), useNA='always', exclude=NULL),
          sum=sum(xx), length=length(x), mean=mean(xx), sd=sqrt(var(xx)))
    }

    # create the test data frame
    Id <- c(1,2,3,4,5,6,7,8,9,10)
    ClassA <- c(1,NA,3,1,1,2,1,4,5,3)
    ClassB <- c(2,1,1,3,3,2,1,1,3,3)
    R <- c(1,2,3,NA,9,2,4,5,6,7)
    S <- c(3,7,NA,9,5,8,7,NA,7,6)
    df <- data.frame(Id,ClassA,ClassB,R,S)

    # create the result
    result <- setNames(
        nm=c('answer','question','value'),
        as.data.frame(
            as.table(
                simplify2array(
                    lapply(
                        df[c('R', 'S')], 
                        summaryStatistics
                    )
                )
            )
        )
    )

    # change the order to question, answer, value
    result <- result[, c(2, 1, 3)]

    # add the filter
    result <- cbind(filter='None',result)

    # return the result
    result 

这要简单得多，也使我培训团队的其他任务变得更加简单。感谢Ernest A和 Imo .

关于我对 R 的理解的下一个问题是 Using vectors in R to change the output of a function

Answer 1

减少代码大小的一个方法是将摘要统计信息包装在一个函数中:

myStats <- function(x) {
  answer <- c("sum"=sum(x, na.rm = TRUE), "length"=sum(!is.na(x)), 
              "mean"=mean(x, na.rm = TRUE), "sd"=sd(x, na.rm = TRUE))

  return(answer)
}

这将返回一个命名的摘要统计向量，按照您在输出中的顺序排序。然后，您可以rbind 返回的值以及您的频率表的名称:

R.stats <- myStats(df$R)
rbind(R.freq, data.frame("question"='R', "answer"=names(R.stats),
                         "value"=R.stats))