用R中的均值，总和，长度和sd实现频率计数的更简单方法 [英] A simpler way to achieve a frequency count with mean, sum, length and sd in R
问题描述
我的任务是创建带有统计摘要的频率表.我的目标是创建一个可以简单导出为excel的数据框. 大部分可能是在使用存储过程的sql中，但是我决定在R中这样做.我正在学习R，所以我可能会做得很长.这是来自 getting-r-frequency-counts-for -all-possible-answers
I've been tasked with creating frequency tables with statistical summaries. My goal is to create a data frame that can be exported simply to excel. Most of this could be in sql using stored procedures but I decided to do this in R. I'm learning R so I might be doing it the long way. This is a follow on question from getting-r-frequency-counts-for-all-possible-answers
给出
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
我知道
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
几乎是我要寻找的东西.我看到的下一步是开始包装一些函数以简化代码，然后再开始从ClassA = 1，ClassA = n + 1 ... ClassA = NA，然后ClassB = 1，ClassB = 2 ... ClassB = NA.有更简单的方法吗?
Which is pretty much what I'm looking for. The next step as I see it is to start wrapping in some functions to simplify the code before I start adding in similar result sets from ClassA=1, ClassA=n+1 ... ClassA=NA, then ClassB=1, ClassB=2 ... ClassB=NA. Is there a much simpler way of doing this?
研究了 Ernest A 和这要简单得多，而使我训练团队的其他任务也要简单得多.感谢 Ernest A 和 Imo .
Which is much simpler and make my other task of training our team much simpler. Thanks to Ernest A and Imo.
The next question in relation to my understanding of R is Using vectors in R to change the output of a function
推荐答案
是的，绝对可以简化.通常，您会使用汇总功能，例如
Yes, it definitely can be simplified. Typically you would use a summary function such as
smry <- function(x, levels) {
xx <- na.omit(x)
c(table(factor(x, levels=levels), useNA='always', exclude=NULL),
sum=sum(xx), length=length(x), mean=mean(xx), sd=sqrt(var(xx)))
}
然后将其应用于数据的不同子集
then apply it to the different subsets of the data
> lapply(df[c('R', 'S')], smry, 0:10)
$R
0 1 2 3 4 5 6 7
0.000000 1.000000 2.000000 1.000000 1.000000 1.000000 1.000000 1.000000
8 9 10 <NA> sum length mean sd
0.000000 1.000000 0.000000 1.000000 39.000000 10.000000 4.333333 2.645751
$S
0 1 2 3 4 5 6 7
0.00000 0.00000 0.00000 1.00000 0.00000 1.00000 1.00000 3.00000
8 9 10 <NA> sum length mean sd
1.00000 1.00000 0.00000 2.00000 52.00000 10.00000 6.50000 1.85164
如果您绝对必须将所有内容都放在数据框中
If you absolutely have to put everything in a data frame
> as.data.frame(as.table(simplify2array(lapply(df[c('R', 'S')], smry, 0:10))))
Var1 Var2 Freq
1 0 R 0.000000
2 1 R 1.000000
3 2 R 2.000000
4 3 R 1.000000
5 4 R 1.000000
6 5 R 1.000000
7 6 R 1.000000
8 7 R 1.000000
9 8 R 0.000000
10 9 R 1.000000
11 10 R 0.000000
12 <NA> R 1.000000
13 sum R 39.000000
14 length R 10.000000
15 mean R 4.333333
16 sd R 2.645751
17 0 S 0.000000
18 1 S 0.000000
19 2 S 0.000000
20 3 S 1.000000
21 4 S 0.000000
22 5 S 1.000000
23 6 S 1.000000
24 7 S 3.000000
25 8 S 1.000000
26 9 S 1.000000
27 10 S 0.000000
28 <NA> S 2.000000
29 sum S 52.000000
30 length S 10.000000
31 mean S 6.500000
32 sd S 1.851640
，然后只需更改列名称/根据需要添加列即可.
and then simply change the column names / add columns as you need.
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