Question

数据集调用：有一个数据帧df，如下所示：

# df
 No. Sub_No. Req_No.
 130   1      231
 130   2      ""
 130   3      ""
 121   1      ""
 140   1      398
 140   2       ""

期望的输出

# df
Type  No. Sub_No. Req_No.
 A    130   1      231
 A    130   2      ""
 A    130   3      ""
      121   1      ""
 A    140   1      398
 A    140   2      ""

您只能定义类型：＆＃34; A＆＃34;当它有＆＃34; Req_No＆＃34;
实际数据集与另一个类型和超过100万行混合
我只有接近：

df["Type"] <- ifelse(grepl("[[:digit:]]",df$Req_No.),"A","-")

显然它可以实现所需的数据集。

Answer 1

假设""是NA和'否'作为分组变量

library(dplyr)
df %>% 
  group_by(No.) %>% 
  mutate(Type = if(all(is.na(Req_No.))) NA else "A")

如果是空白（""）而不是NA

df %>%
   group_by(No.) %>%
   mutate(Type = if(all(Req_No.=="")) "" else "A")
# A tibble: 6 x 4
# Groups: No. [3]
#    No. Sub_No. Req_No. Type 
#  <int>   <int> <chr>   <chr>
#1   130       1 231     A    
#2   130       2 ""      A    
#3   130       3 ""      A    
#4   121       1 ""      ""   
#5   140       1 398     A    
#6   140       2 ""      A

数据

df <- structure(list(No. = c(130L, 130L, 130L, 121L, 140L, 140L), Sub_No. = c(1L, 
2L, 3L, 1L, 1L, 2L), Req_No. = c(231L, NA, NA, NA, 398L, NA)), .Names = c("No.", 
 "Sub_No.", "Req_No."), class = "data.frame", row.names = c(NA, 
-6L))

#df with blanks
df <- structure(list(No. = c(130L, 130L, 130L, 121L, 140L, 140L), Sub_No. = c(1L, 
 2L, 3L, 1L, 1L, 2L), Req_No. = c("231", "", "", "", "398", ""
 )), .Names = c("No.", "Sub_No.", "Req_No."), row.names = c(NA, 
 -6L), class = "data.frame")

如何从ifelse的结果中将字符串填充到另一行

1 个答案:

数据