创建一个函数以迭代列，并在r中创建新列的每条迭代

Question

偶尔，我将获得带有李克特秤的字符串项的调查数据，我需要更改为数字以计算基本的描述性统计数据。为了做到这一点，我通常使用case_时函数为每个项目创建一个新列，并为每个数据点分配一个数字值。我正在尝试编写一个可以一次为许多不同列执行此操作的函数，以便我不必保留复制和粘贴代码。我对此是相对较新的，因此任何帮助都将不胜感激:)

这是我以前在R中所做的：

#create data frame
df <- data.frame(v1 = c("Definitely True", "Somewhat True","Somewhat False","Definitely False"),
                 v2 = c("Definitely False","Somewhat False","Somewhat True","Definitely True"))

#Use case_when to add numeric columns to dataframe
df$v1n <- case_when((df$v1 == "Definitely True")==TRUE ~ "1",
                         (df$v1 == "Somewhat True")==TRUE ~ "2",
                         (df$v1 == "Somewhat False")==TRUE ~ "3",
                         (df$v1 == "Definitely False")==TRUE ~ "4")
df$v2n <- case_when((df$v2 == "Definitely True")==TRUE ~ "1",
                         (df$v2 == "Somewhat True")==TRUE ~ "2",
                         (df$v2 == "Somewhat False")==TRUE ~ "3",
                         (df$v2 == "Definitely False")==TRUE ~ "4")

如果我想用一个数字值替换每个字符串值并在现有列中覆盖数据：

for(i in colnames(data_x)) {
  data_x[[i]] <- case_when((data_x[,i] == "Definitely True")==TRUE ~ "1",
                         (data_x[,i] == "Somewhat True")==TRUE ~ "2",
                         (data_x[,i] == "Somewhat False")==TRUE ~ "3",
                         (data_x[,i] == "Definitely False")==TRUE ~ "4")
}

但是我想要找到一种方法来为每次迭代创建新列，就像我对复制和粘贴版本一样。这是我尝试过的事情，但我没有任何成功。对此的任何帮助将不胜感激。

for(i in colnames(df)) {
  df[[var[i]]] <- case_when((df[,i] == "Definitely True")==TRUE ~ "1",
                         (df[,i] == "Somewhat True")==TRUE ~ "2",
                         (df[,i] == "Somewhat False")==TRUE ~ "3",
                         (df[,i] == "Definitely False")==TRUE ~ "4")
}

Answer 1

跨的dplyr

df %>%
  mutate(across(v1:v2, ~ case_when(
    . == "Definitely True" ~ "1", 
    . == "Somewhat True" ~ "2", 
    . == "Somewhat False" ~ "3", 
    TRUE ~ "4"
    ), .names = "{.col}n")
  )
#                 v1               v2 v1n v2n
# 1  Definitely True Definitely False   1   4
# 2    Somewhat True   Somewhat False   2   3
# 3   Somewhat False    Somewhat True   3   2
# 4 Definitely False  Definitely True   4   1

• 使我们能够在多个列中做一件事情。我们可以使用v1：v2 -syntax，或其他dplyr选择器函数之一匹配， starts_ptatch_with ，等等等。
• 的第二个参数此处是tilde-function（rlang式），在其中。在每个列中替换为每个列数据迭代。例如，第一次评估此Tilde功能时，。参考vector df $ v1。
• 因为突变的默认操作（跨（...））将是替换 ，我添加.names =以控制命名结果数据。该符号使用胶水 -syntax，其中{。col}被每次迭代中要评估的列的名称所代替。

基础r

我将添加查找图的可选使用。

lookup <- c("Definitely True" = "1", "Somewhat True" = "2", "Somewhat False" = "3", "Definitely False" = "4")
df <- cbind(df, setNames(lapply(df[,1:2], function(z) lookup[z]), paste0(names(df[,1:2]), "n")))
rownames(df) <- NULL
df
#                 v1               v2 v1n v2n
# 1  Definitely True Definitely False   1   4
# 2    Somewhat True   Somewhat False   2   3
# 3   Somewhat False    Somewhat True   3   2
# 4 Definitely False  Definitely True   4   1

Answer 2

我倾向于做不同的事情。如果将李克特秤列转换为factor，级别的级别正确，则可以使用as.integer（...）直接在没有数字级别的情况下，没有所有这些case_when（...）业务。

这是一个使用data.table的示例

library(data.table)
likertScale <- c("Definitely True", "Somewhat True","Somewhat False","Definitely False")
cols        <- names(df)
setDT(df)[, c(cols):=lapply(.SD, factor, levels=likertScale)]
df[, paste0(cols, 'n'):=lapply(.SD, as.integer), .SDcols=cols]
df
##                  v1               v2 v1n v2n
## 1:  Definitely True Definitely False   1   4
## 2:    Somewhat True   Somewhat False   2   3
## 3:   Somewhat False    Somewhat True   3   2
## 4: Definitely False  Definitely True   4   1