【发布时间】:2016-09-24 18:58:22
【问题描述】:
我有一个看起来像这样的数据框
Country <- rep(c("Austria", "Austria","Belgium", "Belgium", "Spain", "Slovenia", "France"), times=3)
Institute <- rep(c("Inst 1","Inst 2","Inst 3","Inst 4","Inst 5","Inst 6","Inst 7"), times=3)
Ans <- rep(c(1,2,3,1,NA,2,2),times=3)
Category.1 <- rep(c("Cat 1", "Cat 2", "Cat 2", "Cat 2","Cat 2", "Cat 1", "Cat 1"),times=3)
Category.2 <- rep(c("P", "L", "M", "P", "P", "L", "M"),times=3)
qs <- c(rep("Q1.a-Some Text", times=7),rep("Q1.b-Some Text", times=7), rep("Q1.c-Some Text", times=7))
df <- data.frame(Country=Country,Institute=Institute, Category.1=Category.1, Category.2=Category.2, qs=qs, Ans=Ans)
df<-df %>% spread(qs,Ans)
head(df)
Country Institute Category.1 Category.2 Q1.a-Some Text Q1.b-Some Text Q1.c-Some Text
1 Austria Inst 1 Cat 1 P 1 1 1
2 Austria Inst 2 Cat 2 L 2 2 2
3 Belgium Inst 3 Cat 2 M 3 3 3
4 Belgium Inst 4 Cat 2 P 1 1 1
5 France Inst 7 Cat 1 M 2 2 2
6 Slovenia Inst 6 Cat 1 L 2 2 2
对数据框的简短解释:有一些问题,比如 Q1,对于这个问题,有多个“子问题”,比如 a、b、c,其中每个“子问题/选项”受访者被要求使用某种比例来回答,在这个例子中从 1 到 3。我的范围是计算每个子问题的相对频率,每个响应。所以,我使用了这个函数:
multichoice<-function(data, question.prefix){
index<-grep(question.prefix, names(data)) # identifies the index for the available options in Q.12
cases<-length(index) # The number of possible options / columns
# Identify the range of possible answers for each question
# Step 1. Search for the min in each col and across each col choose the min
# step 2. Search for the max in each col and across each col choose the max
mn<-min(data[,index[1:cases]], na.rm=T)
mx<-max(data[,index[1:cases]], na.rm=T)
d = colSums(data[, index] != 0, na.rm = TRUE) # The number of elements across column vector, that are different from zero.
vec<-matrix(,nrow=length(mn:mx),ncol=cases)
for(j in 1:cases){
for(i in mn:mx){
vec[i,j]=sum(data[, index[j]] == i, na.rm = TRUE)/d[j] # This stores the relative responses for option j for the answer that is i
}
}
vec1<-as.data.frame(vec)
names(vec1)<-names(data[index])
vec1<-t(vec1)
return(vec1)
}
调用,我得到所需数据帧的函数。
q1 <- as.data.frame(multichoiceq4(df,"^Q1"))
head(q1)
V1 V2 V3
Q1.a-Some Text 0.3333333 0.5 0.1666667
Q1.b-Some Text 0.3333333 0.5 0.1666667
Q1.c-Some Text 0.3333333 0.5 0.1666667
这表明对于选项“a”,33% 的参与者回答 1,50% 回答 2 等等......
我的问题
我想计算,相同但以类别为条件。所以,我想看看基于category1, category2 的相对频率如何。有人可以建议我如何做到这一点吗?
【问题讨论】:
标签: r data-manipulation survey data-management