带标签的李克特数据的 R 频率 Table

Question

我尝试从共享相同答案类别（李克特类型）的多个变量中创建一个频率 table。具有三个变量（问题 1-3）和 5 个答案类别（-- 到 ++）的结果应该如下所示：

 |           |  -- |  -  |  ~  |  +  |  ++ |
 | --------- | --- | --- | --- | --- | --- |
 | Question1 |  5% | 20% | 25% | 30% | 20% |
 | Question2 | 15% | 10% | 20% | 25% | 30% |
 | Question3 | 10% | 30% | 10% | 30% | 20% |

我在找到了一个有效的解决方案，其中包含包 expss 中的函数，这对创建加权和标记频率 table 非常有帮助。但是我在使用标签时遇到了一些麻烦，因为当变量被标记时这个解决方案似乎不起作用：

1) @GregoryDemin 的 expss 解决方案来自：

# The data we'll also use in the examples below.
q1<-c(2,2,3,3,3,4,4,4,5,5)
q2<-c(2,3,3,4,4,4,4,5,5,5)
q3<-c(2,2,2,3,4,4,4,5,5,5)
df<-data.frame(q1,q2,q3)

library(expss)
# add value lables for preserving empty categories
val_lab(df) = autonum(1:5)
res = df
for(each in colnames(df)){
    res = res %>% 
        tab_cells(list(each)) %>% 
        tab_cols(vars(each)) %>% 
        tab_stat_rpct(total_row_position = "none")
}


res = res %>% tab_pivot() 
# add percentage sign
recode(res[,-1]) = other ~ function(x) ifelse(is.na(x), NA, paste0(round(x, 0), "%"))
res

输出：

 |    |  1 |   2 |   3 |   4 |   5 |
 | -- | -- | --- | --- | --- | --- |
 | q1 |    | 20% | 30% | 30% | 20% |
 | q2 |    | 10% | 20% | 40% | 30% |
 | q3 |    | 30% | 10% | 30% | 30% |

看起来不错 - 尽管 NA 应该是零，不是吗？。我们如何确保未使用的类别显示 0% 而不是 NA？

2) 现在我们添加一些 variable/value 标签：

q1<-c(2,2,3,3,3,4,4,4,5,5)
q2<-c(2,3,3,4,4,4,4,5,5,5)
q3<-c(2,2,2,3,4,4,4,5,5,5)
df<-data.frame(q1,q2,q3)

library(expss)

# Label variables and categories
df %<>% apply_labels(q1 = "Question 1",
                     q2 = "Question 2",
                     q3 = "Question 3",
                     q1 = c("strongly agree" = 5, "agree" = 4, "neutral" = 3, "disagree" = 2, "strongly disagree" = 1),
                     q2 = c("strongly agree" = 5, "agree" = 4, "neutral" = 3, "disagree" = 2, "strongly disagree" = 1),
                     q3 = c("strongly agree" = 5, "agree" = 4, "neutral" = 3, "disagree" = 2, "strongly disagree" = 1))

# add value lables for preserving empty categories
#val_lab(df) = autonum(1:5)  # we labelled before, so no need for that anymore

# Now for the table
res = df
for(each in colnames(df)){
    res = res %>% 
        tab_cells(list(each)) %>% 
        tab_cols(vars(each)) %>% 
        tab_stat_rpct(total_row_position = "none")
}
res = res %>% tab_pivot() 
# add percentage sign
recode(res[,-1]) = other ~ function(x) ifelse(is.na(x), NA, paste0(round(x, 0), "%"))
res

输出：

 |    |        Question 1 |          |         |       |                |        Question 2 |          |         |       |
 |    | strongly disagree | disagree | neutral | agree | strongly agree | strongly disagree | disagree | neutral | agree |
 | -- | ----------------- | -------- | ------- | ----- | -------------- | ----------------- | -------- | ------- | ----- |
 | q1 |                   |      20% |     30% |   30% |            20% |                   |          |         |       |
 | q2 |                   |          |         |       |                |                   |      10% |     20% |   40% |
 | q3 |                   |          |         |       |                |                   |          |         |       |

                |        Question 3 |          |         |       |                |
 strongly agree | strongly disagree | disagree | neutral | agree | strongly agree |
 -------------- | ----------------- | -------- | ------- | ----- | -------------- |
                |                   |          |         |       |                |
            30% |                   |          |         |       |                |
                |                   |      30% |     10% |   30% |            30% |

变量不再堆叠，而是并排站立。如果我们添加变量标签，这个可行的解决方案似乎会中断。您知道如何防止这种情况发生吗？

Answer 1

中的方法仅适用于没有变量标签的变量。零百分比的通用代码：

q1<-c(2,2,3,3,3,4,4,4,5,5)
q2<-c(2,3,3,4,4,4,4,5,5,5)
q3<-c(2,2,2,3,4,4,4,5,5,5)
df<-data.frame(q1,q2,q3)

library(expss)

# Label variables and categories
df %<>% apply_labels(q1 = "Question 1",
                     q2 = "Question 2",
                     q3 = "Question 3",
                     q1 = c("strongly agree" = 5, "agree" = 4, "neutral" = 3, "disagree" = 2, "strongly disagree" = 1),
                     q2 = c("strongly agree" = 5, "agree" = 4, "neutral" = 3, "disagree" = 2, "strongly disagree" = 1),
                     q3 = c("strongly agree" = 5, "agree" = 4, "neutral" = 3, "disagree" = 2, "strongly disagree" = 1))

# add value lables for preserving empty categories
#val_lab(df) = autonum(1:5)  # we labelled before, so no need for that anymore

# Now for the table
res = df
for(each in colnames(df)){
    res = res %>% 
        tab_cells(total(label = "|")) %>% # suppress total label 
        tab_cols(unvr(vars(each))) %>%  # remove variable label 
        tab_stat_rpct(total_row_position = "none", label = var_lab(vars(each))) # use variable label as statistic label
}
res = res %>% tab_pivot() 
# add percentage sign
recode(res[,-1]) = other ~ function(x) ifelse(is.na(x), "0%", paste0(round(x, 0), "%"))
res

# |            | strongly disagree | disagree | neutral | agree | strongly agree |
# | ---------- | ----------------- | -------- | ------- | ----- | -------------- |
# | Question 1 |                0% |      20% |     30% |   30% |            20% |
# | Question 2 |                0% |      10% |     20% |   40% |            30% |
# | Question 3 |                0% |      30% |     10% |   30% |            30% |

带标签的李克特数据的 R 频率 Table

R Frequency Table of Likert Data with Labels

r

r-markdown

expss