用于响应的Likert图例名称

Question

我已经开发了一个脚本来绘制一个Likert比例。剧本工作正常，情节正确。我想把回应标签改为：“强烈反对”、“不同意”、“稍不同意”、“稍同意”、“同意”、“强烈同意”。我尝试过不同的解决方案，但似乎没有一种可行。

Q1 <- read_excel("C:\\Users\\users\\Desktop\\Survey Responses\\Business Survey\\BusinessLikert.xlsx")

df <- data.frame(respondent = c(Q1$Respondent), Score = c(Q1$Q1))

df1 <- likert(items=df[,2, drop = FALSE], nlevels = 6)

summary(df1)

likert.bar.plot(df1)

likert.density.plot(df1)

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Answer 1

正如likert函数(?likert::likert)的文档中所述，items中的data.frame列应该是因素。然后，级别名称指定派生的likert绘图中使用的响应标签。由于您的数据不可复制，请考虑以下人工示例：

library(likert)

set.seed(1)
df <- data.frame(Score = factor(sample(1:6, size = 100, replace = TRUE), 
                 labels = c("Strongly Disagree", "Disagree", "Slightly Disagree", "Slightly Agree", "Agree", "Strongly Agree")))
(df_likert <- likert(items = df))
#>    Item Strongly Disagree Disagree Slightly Disagree Slightly Agree Agree
#> 1 Score                19       18                12             15    15
#>   Strongly Agree
#> 1             21

likert.bar.plot(df_likert)

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编辑：用于表示data.frame中各个响应组的多个(例如数字)列，首先将这些列作为因子重新编码，然后将likert函数应用于已编码的data.frame：

## initial data.frame of integers
df <- data.frame(
  sapply(c("Q1", "Q2", "Q3"), function(x) sample(1:6, size = 100, replace = TRUE))
)

## recode each column as a factor
df_factor <- as.data.frame(
  lapply(df, function(x) factor(x, 
      labels = c("Strongly Disagree", "Disagree", "Slightly Disagree", 
                 "Slightly Agree", "Agree", "Strongly Agree"))
  )
)

(df_likert <- likert(items = df_factor))
#>   Item Strongly Disagree Disagree Slightly Disagree Slightly Agree Agree
#> 1   Q1                19       18                12             15    15
#> 2   Q2                19       16                19             18    15
#> 3   Q3                18       15                 8             21    20
#>   Strongly Agree
#> 1             21
#> 2             13
#> 3             18

likert.bar.plot(df_likert)

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