mutate_if错误:无法将列表转换为函数

Question

我试图将12小时后包含单词"length"的变量的所有值更改为NA.

df_data <- cbind(
  seq(0, 15, by = 0.5),
  sample(seq(from = 100, to = 300, by = 10), size = 31, replace = TRUE),
  sample(seq(from = 1, to = 100, by = 9), size = 31, replace = TRUE),
  sample(seq(from = 50, to = 60, by = 2), size = 31, replace = TRUE),
  sample(seq(from = 100, to = 130, by = 1), size = 31, replace = TRUE)
) %>% as.data.frame()

colnames(df_data) <- c("hour", "a", "a_lenght", "b", "b_length")


df_new <- df_data %>%
  mutate_if(vars(contains("length")), funs(ifelse(df_data$hour > 12, NA, .)))

但是我收到了一个Can't convert a list to function错误

Answer 1

总结已经做出的评论:

mutate_at使用列名称 - 当您可以提供按列名称的特征选择列的条件时,它最有用.这通常采用以下形式:mutate_at(vars(ends_with("length")),funs(...))

mutate_if使用列内容 - 当您可以提供基于列本身内容选择列的谓词函数时,它最有用.这通常采用以下形式:mutate_if(is.integer,funs(...))

但正如Galina所指出的那样,mutate_if是多功能的,可以使用grepl"欺骗"操作列名