我不知道为什么这不起作用。
po <- unite(Need_info[1:5,],"lala",c(5,6,13,14,15,16),na.rm = TRUE,remove = TRUE)
返回
问题可能是列是因素。尝试使用:
library(dplyr)
library(tidyr)
Need_info %>%
mutate_if(is.factor, as.character) %>%
unite("lala",c(5,6,13,14,15,16),na.rm = TRUE,remove = TRUE)
使用可重现的示例:
df <- data.frame(a = c(letters[1:5], NA), b = c(NA, letters[11:15]))
df %>% unite("lala", c(1, 2), na.rm =TRUE, remove = TRUE)
# lala
#1 1_NA
#2 2_1
#3 3_2
#4 4_3
#5 5_4
#6 NA_5
转换为字符后:
df %>%
mutate_all(as.character) %>%
unite("lala", c(1, 2), na.rm = TRUE, remove = TRUE)
# lala
#1 a
#2 b_k
#3 c_l
#4 d_m
#5 e_n
#6 o