如何将调查响应的数据框转换为频率表？

Question

我有一个R数据框，包含调查结果。每一列都是调查问题的回答。它可以取1到10的值和NA。我想把它转换成频率表。

这是我拥有的数据示例。我假装值从1到3而不是1到10。

data.frame(
  "Person" = c(1,2,3),
  "Question1" = c(NA, "1", "1"),
  "Question2" = c("1", "2", "3")
)

我想要的：

data.frame(
  "Question" = c("Question1", "Question2"),
  "Frequency of 1" = c(2, 1),
  "Frequency of 2" = c(0 , 1),
  "Frequency of 3" = c(0, 1)
)

我尝试使用likert包中的likert()函数，但是得到了分数结果，这显然不正确。有没有简单的解决方案？

- Eli

---

问题2中，为什么频率1被设置为0而频率3被设置为3？ - NelsonGon

我犯了一个错误。我编辑了这个问题。 - Eli

Answer 1

这里提供一个使用dplyr和purrr包的解决方案

library(dplyr)
library(purrr)

data.frame(
  "Person" = c(1,2,3),
  "Question1" = c(NA, "1", "1"),
  "Question2" = c("1", "2", "3")
)

df %>% 
  select(-Person) %>% 
  mutate_all(~ factor(.x, levels =  as.character(1:10) ) %>% addNA() ) %>% 
  map(table) %>% 
  transpose() %>% 
  map(as.integer) %>% 
  set_names( ~ paste0("Frequency of ",ifelse(is.na(.), "NA", .))) %>% 
  as_tibble() %>% 
  mutate(Question = setdiff(names(df),"Person")) %>% 
  select(Question,everything(), "Frequency of NA" = `Frequency of ` ) 

- Adam Gruer

Answer 2

一个 data.table 的解决方案：

require(data.table)
setDT(df)    

# Melt data:
df <- melt(df, id.vars = "Person", value.name = "Question")

# Cast data to required structure:
df <- data.frame(dcast(df, variable ~ Question))

# Rename variables and remove NA count (as per Ops question):
names(df)[1] <- "Question"
names(df)[-1] <- gsub("X", "Frequency of ", names(df)[-1])
df$NA. <- NULL

df
#   Question Frequency of 1 Frequency of 2 Frequency of 3
#1 Question1              2              0              0
#2 Question2              1              1              1

最初的回答：

或者简单地回答：

dcast(melt(setDT(df), id.vars="Person", value.name="Question")[!Question %in% NA][, Question := paste0("Frequency of ", Question)], variable ~ Question)

- Khaynes

Answer 3

一种不同的 `tidyverse` 可能性是：

df %>%
 gather(Question, val, -Person, na.rm = TRUE) %>%
 group_by(Question, val) %>%
 summarise(res = length(val)) %>%
 ungroup() %>%
 mutate(val = paste0("Frequency.of.", val)) %>%
 spread(val, res, fill = NA)

  Question  Frequency.of.1 Frequency.of.2 Frequency.of.3
  <chr>              <int>          <int>          <int>
1 Question1              2             NA             NA
2 Question2              1              1              1

这里首先将数据从宽格式转换为长格式，其次按照问题计算频率。最后，创建“Frequency.of.”变量并将数据返回到所需的形式。

或者，如果您还想计算每个问题的NA值：

df %>%
 gather(Question, val, -Person) %>%
 group_by(Question, val) %>%
 summarise(res = length(val)) %>%
 ungroup() %>%
 mutate(val = paste0("Frequency.of.", val)) %>%
 spread(val, res, fill = NA)

  Question  Frequency.of.1 Frequency.of.2 Frequency.of.3 Frequency.of.NA
  <chr>              <int>          <int>          <int>           <int>
1 Question1              2             NA             NA               1
2 Question2              1              1              1              NA

- tmfmnk

Answer 4

这不是最优雅的方法，但可能会有所帮助：df2是您的数据集。 数据：

   df2<-data.frame(
  "Person" = c(1,2,3),
  "Question1" = c(NA, "1", "1"),
  "Question2" = c("1", "2", "3"),stringsAsFactors = F
)

目标： 编辑：您可以按以下方式“自动化”

df2[is.na(df2)]<-0 #To allow numeric manipulation
values<-c("1","2","3")
    Final_df<-sapply(values,function(val) apply(df2[,-1],2,function(x) sum(x==val)))
    Final_df<-as.data.frame(Final_df)
    names(Final_df)<-paste0("Frequency of_",1:ncol(Final_df))

这将产生：

             Frequency of_1          Frequency of_2          Frequency of_3
Question1              2                0                    0
Question2              1                1                    1

- NelsonGon