我正在用R(和tidyverse)工作，数据来自一份由11个问题组成的问卷，每个问题的答案是4分李克特量表:
- 低于正常值
- 不超过正常
- 高于正常值
- 大于正常
数据在数据框架中，参与者作为行，每个问题的回答作为有序因子存储在单独的列中。
以下代码复制当前存储的5行数据:
library(tidyverse)
df <- tibble(id = c(1, 2, 3, 4, 5), q1 = c(3, 4, 2, 3, 3),
q2 = c(4, 4, 2, 3, 2), q3 = c(3, 3, 2, 2, 3),
q4 = c(2, 2, 3, 2, 1), q5 = c(3, 3, 3, 3, 3),
q6 = c(4, 3, 2, 2, 2), q7 = c(1, 2, 2, 2, 2),
q8 = c(3, 3, 3, 2, 1), q9 = c(3, 4, 4, 2, 1),
q10 = c(2, 4, 3, 2, 1), q11 = c(2, 3, 2, 2, 1)) %>%
mutate(across(q1:q11, ~factor(.x,
levels = c(1, 2, 3, 4),
labels = c("Less than usual",
"No more than usual",
"More than usual",
"Much more than usual"),
ordered = TRUE)))
str(df)
# tibble [5 × 12] (S3: tbl_df/tbl/data.frame)
# $ id : num [1:5] 1 2 3 4 5
# $ q1 : Ord.factor w/ 4 levels "Less than usual"<..: 3 4 2 3 3
# $ q2 : Ord.factor w/ 4 levels "Less than usual"<..: 4 4 2 3 2
# $ q3 : Ord.factor w/ 4 levels "Less than usual"<..: 3 3 2 2 3
# $ q4 : Ord.factor w/ 4 levels "Less than usual"<..: 2 2 3 2 1
# $ q5 : Ord.factor w/ 4 levels "Less than usual"<..: 3 3 3 3 3
# $ q6 : Ord.factor w/ 4 levels "Less than usual"<..: 4 3 2 2 2
# $ q7 : Ord.factor w/ 4 levels "Less than usual"<..: 1 2 2 2 2
# $ q8 : Ord.factor w/ 4 levels "Less than usual"<..: 3 3 3 2 1
# $ q9 : Ord.factor w/ 4 levels "Less than usual"<..: 3 4 4 2 1
# $ q10: Ord.factor w/ 4 levels "Less than usual"<..: 2 4 3 2 1
# $ q11: Ord.factor w/ 4 levels "Less than usual"<..: 2 3 2 2 1
我需要为整个问卷以及选择问题的两个子量表使用两种不同的评分系统来计算总数。第一个子量表包括问题1-7，第二个子量表包括问题8-11。
- 第一个评分系统(Likert)分别将值0、1、2和3分配给因子水平。
- 第二个评分系统(Binary)分别赋值0、0、1和1。
我如何使用两个评分系统来计算这些总数以获得6(子)总数:total_likert,total_binary,total_ss1_likert,total_ss1_binary,total_ss2_likert和total_ss2_binary?
您可以首先使用across和recode根据评分系统更新您的值(您可能也想选择replace)，然后使用rowwise计算每个id的总和分数:
df %>%
mutate(across(starts_with("q"), ~ recode(.x, "Less than usual" = 0,
"No more than usual" = 1,
"More than usual" = 2,
"Much more than usual" = 3),
.names = "likert_{.col}")) %>%
mutate(across(starts_with("q"), ~ recode(.x, "Less than usual" = 0,
"No more than usual" = 0,
"More than usual" = 1,
"Much more than usual" = 1),
.names = "binary_{.col}")) %>%
rowwise(id) %>% mutate(total_likert = sum(c_across(likert_q1:likert_q11)),
total_ss1_likert = sum(c_across(likert_q1:likert_q7)),
total_ss2_likert = sum(c_across(likert_q8:likert_q11)),
total_binary = sum(c_across(binary_q1:binary_q11)),
total_ss1_binary = sum(c_across(binary_q1:binary_q7)),
total_ss2_binary = sum(c_across(binary_q8:binary_q11))) %>%
select(id, total_likert, total_binary, total_ss1_likert, total_ss1_binary, total_ss2_likert, total_ss2_binary)