本文介绍了获取所有可能答案的R频率计数的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我从R开始,但仍在语法上找到自己的出路.我希望获得比例变量的频率,该变量的值为0到10和NA.
I've started with R and I'm still finding my way with syntax.I'm looking to get the frequencies for a scaled variable which has values of 0 through 10 and NA.
Id <- c(1,2,3,4,5)
ClassA <- c(1,NA,3,1,1)
ClassB <- c(2,1,1,3,3)
R <- c(5,5,7,NA,9)
S <- c(3,7,NA,9,5)
df <- data.frame(Id,ClassA,ClassB,R,S)
library(plyr)
count(df,'R')
我得到的结果
R freq
1 5 2
2 7 1
3 9 1
4 NA 1
我正在寻找
R freq
1 0 0
2 1 0
3 2 0
4 3 0
5 4 0
6 5 2
7 6 0
8 7 1
9 8 0
10 9 1
11 10 0
12 NA 1
如果我有向量显示可能的结果
If I have the vector showing the possible results
RAnswers <- c(0,1,2,3,4,5,6,7,8,9,10,NA)
我如何将其与数据集一起应用以获得上述结果?
How would I apply it with the data set to get the above result?
推荐答案
这是围绕table()
,match()
和replace()
构建的基本R解决方案:
Here's a base R solution built around table()
, match()
, and replace()
:
freq <- table(df$R,useNA='ifany');
freq;
##
## 5 7 9 <NA>
## 2 1 1 1
R <- c(0:10,NA);
df2 <- data.frame(R=R,freq=freq[match(R,as.integer(names(freq)))]);
df2$freq[is.na(df2$freq)] <- 0;
df2;
## R freq
## 1 0 0
## 2 1 0
## 3 2 0
## 4 3 0
## 5 4 0
## 6 5 2
## 7 6 0
## 8 7 1
## 9 8 0
## 10 9 1
## 11 10 0
## 12 NA 1
弗兰克有一个更好的答案,这是如何在因子上使用table()
来获得所需输出的方法:
Frank has a better answer, here's how you can use table()
on a factor to get the required output:
setNames(nm=c('R','freq'),data.frame(table(factor(df$R,levels=RAnswers,exclude=NULL))));
## R freq
## 1 0 0
## 2 1 0
## 3 2 0
## 4 3 0
## 5 4 0
## 6 5 2
## 7 6 0
## 8 7 1
## 9 8 0
## 10 9 1
## 11 10 0
## 12 <NA> 1
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