如何对多列序列应用单个条件以在中创建单个列

Question

我有一个类似于以下内容的数据集：

Age      Monday Tuesday Wednesday 
6-9        a     b        a
6-9        b     b        c
6-9              c        a
9-10       c     c        b
9-10       c     a        b

使用R，我想要一个二进制变量来表示整个行是否包含"a“(1表示完整的a，0表示不包含)，如下所示：

Age      Monday Tuesday Wednesday  Entire a
6-9        a              a          1
6-9        b     b        c          0
6-9              c        a          0
9-10       c     c        b          0
9-10       a     a        a          1

注意:我的数据还包含行中缺少的值。我感兴趣的栏目是“要素”。我使用了以下代码，但是这些代码并不起作用：

L <- dataframe %>%
    select(Age,Monday:Wednesday) %>%
    mutate (Entire a = ifelse(c(Monday:Wednesday)=="a",1,0,na.rm=TRUE))

Answer 1

我会使用dplyr解决方案：

library(dplyr)

my.data <- data.frame(
  age = c("6-9", "6-9", "6-9", "9-10", "9-10", "9-10"),
  Monday = c("a", "b", NA, "c", "a", "a"),
  Tuesday = c("a", "b", "a", "c", "a", NA),
  Wednesday = c("a", "c", "a", "c", "a", NA)
)

my.data %>%
  mutate(
    `Entire a` = apply(.[, 2:4], 1, function(x) all(x == "a", na.rm = T) %>% as.numeric)
  )

# age Monday Tuesday Wednesday Entire a
# 1  6-9      a       a         a        1
# 2  6-9      b       b         c        0
# 3  6-9   <NA>       a         a        1
# 4 9-10      c       c         c        0
# 5 9-10      a       a         a        1
# 6 9-10      a    <NA>      <NA>        1

all()函数中的na.rm参数将控制是否忽略缺少的值。

Answer 2

我们可以使用==创建一个逻辑矩阵，并将rowSums转换为binary

colnm <- names(dataframe)[-1]
dataframe$Entire_a <- +(rowSums(replace(dataframe[colnm], 
       dataframe[colnm] == '', 'a') == 'a')  == length(colnm))
dataframe$Entire_a
#[1] 1 0 0 0 1

或者，另一种选择是使用paste，然后使用grep

+(grepl("^a+$", do.call(paste, c(dataframe[colnm], sep=""))))
#[1] 1 0 0 0 1

如果缺少的值为NA且不为空('')，则使用

+(rowSums(replace(dataframe[colnm], is.na(dataframe[colnm]), 'a') == 'a')  == 3)

数据

dataframe <- structure(list(Age = c("6-9", "6-9", "6-9", "9-10", "9-10"), 
    Monday = c("a", "b", "", "c", "a"), Tuesday = c("", "b", 
    "c", "c", "a"), Wednesday = c("a", "c", "a", "b", "a")), 
    row.names = c(NA, 
-5L), class = "data.frame")

Answer 3

我们可以使用purrr中的pmap_int来执行这个逐行操作。

如果空值('')尚未设置，请将其设置为NA。

library(dplyr)
library(purrr)

dataframe %>%
    na_if('') %>%
    mutate(Entire_a = pmap_int(select(., Monday:Wednesday), 
                         ~+all(c(...) == 'a', na.rm = TRUE)))

#   Age Monday Tuesday Wednesday Entire_a
#1  6-9      a    <NA>         a        1
#2  6-9      b       b         c        0
#3  6-9   <NA>       c         a        0
#4 9-10      c       c         b        0
#5 9-10      a       a         a        1