我制作了这个最小的可重复示例来举例说明我的问题。我已经设法解决了这个问题,但我相信有更优雅的编码方式。
问题是关于基于多个标准的二元分类。为了收到甜甜圈(编码为 1),需要至少 3(或更多)的分数:至少一个“a”标准项目,至少两个“b”标准项目和至少三个“c”标准项。如果不满足这些要求,则不会奖励任何甜甜圈(编码为 0)。
这是我的解决方案。你会如何更简洁/优雅地编码它?
require(dplyr)
df <- data.frame("a1" = c(3,2,2,5),
"a2" = c(2,1,3,1),
"b1" = c(2,1,5,4),
"b2" = c(1,2,1,4),
"b3" = c(3,2,3,4),
"c1" = c(3,3,1,3),
"c2" = c(4,2,3,4),
"c3" = c(3,3,4,1),
"c4" = c(1,2,3,4),
stringsAsFactors = FALSE)
df_names <- names(df[, 1:9])
a_items <- names(df[, 1:2])
b_items <- names(df[, 3:5])
c_items <- names(df[, 6:9])
df_response <- df %>%
select(df_names) %>%
mutate_all(
funs(case_when(
. >=3 ~ 1,
is.na(.) ~ 0,
TRUE ~ 0))) %>%
mutate(a_crit = case_when( rowSums(.[ ,a_items]) >=1 ~ 1, # one a item needed
TRUE ~ 0)) %>%
mutate(b_crit = case_when( rowSums(.[ ,b_items]) >=2 ~ 1, # two b items needed
TRUE ~ 0)) %>%
mutate(c_crit = case_when( rowSums(.[ ,c_items]) >=3 ~ 1, # three c items needed
TRUE ~ 0)) %>%
mutate(overal_crit = case_when( a_crit == 1 & b_crit == 1 & c_crit == 1 ~ 1,
TRUE ~ 0))
df_response$overal_crit
Run Code Online (Sandbox Code Playgroud)
我会进行简单的mutate通话
library(dplyr)
df %>%
mutate(a_crit = as.integer(rowSums(.[a_items] >= 3) >= 1),
b_crit = as.integer(rowSums(.[b_items] >= 3) >= 2),
c_crit = as.integer(rowSums(.[c_items] >= 3) >= 3),
overal_crit = as.integer((a_crit + b_crit + c_crit) == 3))
# a1 a2 b1 b2 b3 c1 c2 c3 c4 a_crit b_crit c_crit overal_crit
#1 3 2 2 1 3 3 4 3 1 1 0 1 0
#2 2 1 1 2 2 3 2 3 2 0 0 0 0
#3 2 3 5 1 3 1 3 4 3 1 1 1 1
#4 5 1 4 4 4 3 4 1 4 1 1 1 1
Run Code Online (Sandbox Code Playgroud)