r语言 - 如何查找任何列是否具有我正在寻找的特定值

Question

   id first  middle  last       Age
    1 Carol  Jenny   Smith      15
    2 Sarah  Carol   Roberts    20
    3 Josh   David   Richardson 22

我正在尝试在任何名称列(名字、中间名、姓氏(中找到特定名称。例如，如果我找到一个名字为 Carol 的人(不管它是名字/中间名/姓氏(，我想改变一列"Carol"并给出 1。所以我想要的是以下内容

   id first  middle  last       Age  Carol
    1 Carol   Jenny   Smith      15   1
    2 Sarah  Carol   Roberts    20    1
    3 Josh   David   Richardson 22    0

我一直在努力ifelse(c(first， middle， last( == "Carol" ， 1， 0 (或"卡罗尔"%in%首先...等但由于某种原因，我只能在一列而不是多列上工作。.谁能帮我？提前谢谢你！

Answer 1

我们可以使用rowSums

df$Carol <- as.integer(rowSums(df[2:4] == "Carol") > 0)
df
#  id first middle       last Age Carol
#1  1 Carol  Jenny      Smith  15     1
#2  2 Sarah  Carol    Roberts  20     1
#3  3  Josh  David Richardson  22     0

如果我们需要它作为一个函数

fun <- function(df, value) {
   as.integer(rowSums(df[2:4] == value) > 0)
}
fun(df, "Carol")
#[1] 1 1 0
fun(df, "Sarah")
#[1] 0 1 0

但这假设您要搜索的列位于位置 2:4 。

使列位置具有更大的灵活性

fun <- function(df, cols, value) {
   as.integer(rowSums(df[cols] == value) > 0)
 }
fun(df, c("first", "last","middle"), "Carol")
#[1] 1 1 0
fun(df, c("first", "last","middle"), "Sarah")
#[1] 0 1 0

Answer 2

这是一个tidyverse选项。我们首先将数据重塑为长格式，按id分组，并在至少一行中找到具有所需名称的id级别。然后我们重新塑造回宽格式。

library(tidyverse)
df %>% 
  gather(key, value, first:last) %>% 
  group_by(id) %>% 
  mutate(Carol = as.numeric(any(value=="Carol"))) %>% 
  spread(key, value)

     id   Age Carol first last       middle
1     1    15     1 Carol Smith      Jenny 
2     2    20     1 Sarah Roberts    Carol 
3     3    22     0 Josh  Richardson David

或者，作为一个函数：

find.target = function(data, target) {
  data %>% 
    gather(key, value, first:last) %>% 
    group_by(id) %>% 
    mutate(!!target := as.numeric(any(value==target))) %>% 
    spread(key, value) %>% 
    # Move new target column to end
    select(-target, target)
}
find.target(df, "Carol")
find.target(df, "Sarah")

您也可以一次执行多个操作。例如：

map(c("Sarah", "Carol", "David"), ~ find.target(df, .x)) %>% 
  reduce(left_join)

     id   Age first last       middle Sarah Carol David
1     1    15 Carol Smith      Jenny      0     1     0
2     2    20 Sarah Roberts    Carol      1     1     0
3     3    22 Josh  Richardson David      0     0     1

Answer 3

使用 tidyverse

library(tidyverse)
f1 <- function(data, wordToCompare, colsToCompare) {
          wordToCompare <- enquo(wordToCompare)
          data %>%
              select(colsToCompare) %>%
              mutate(!! wordToCompare :=  map(.,  ~ 
       .x == as_label(wordToCompare)) %>% 
           reduce(`|`) %>%
           as.integer)
              }
          
f1(df1, Carol, c("first", 'middle', 'last'))
# first middle       last Carol
#1 Carol  Jenny      Smith     1
#2 Sarah  Carol    Roberts     1
#3  Josh  David Richardson     0
f1(df1, Sarah, c("first", 'middle', 'last'))
#   first middle       last Sarah
#1 Carol  Jenny      Smith     0
#2 Sarah  Carol    Roberts     1
#3  Josh  David Richardson     0

<小时 />

或者这也可以通过pmap来完成

df1 %>%
  mutate(Carol = pmap_int(.[c('first', 'middle', 'last')],
          ~ +('Carol' %in% c(...))))
#   id first middle       last Age Carol
#1  1 Carol  Jenny      Smith  15     1
#2  2 Sarah  Carol    Roberts  20     1
#3  3  Josh  David Richardson  22     0

可以包装到函数中

f2 <- function(data, wordToCompare, colsToCompare) {
      wordToCompare <- enquo(wordToCompare)
      data %>%
           mutate(!! wordToCompare := pmap_int(.[colsToCompare],
          ~ +(as_label(wordToCompare) %in% c(...))))
  } 
f2(df1, Carol, c("first", 'middle', 'last'))
#  id first middle       last Age Carol
#1  1 Carol  Jenny      Smith  15     1
#2  2 Sarah  Carol    Roberts  20     1
#3  3  Josh  David Richardson  22     0

注意：两种整齐的方法都不需要任何重塑

<小时 />

使用 base R ，我们可以遍历"第一"、"中间"、"最后"列，并使用==进行比较，以获得逻辑vector的list，我们将其Reduce为具有|的单个逻辑vector，并将其强制为二进制+

df1$Carol <- +(Reduce(`|`, lapply(df1[2:4], `==`, 'Carol')))
df1
#  id first middle       last Age Carol
#1  1 Carol  Jenny      Smith  15     1
#2  2 Sarah  Carol    Roberts  20     1 
#3  3  Josh  David Richardson  22     0

<小时 />

注意：这篇文章有欺骗性。 例如在这里

数据

df1 <- structure(list(id = 1:3, first = c("Carol", "Sarah", "Josh"), 
middle = c("Jenny", "Carol", "David"), last = c("Smith", 
"Roberts", "Richardson"), Age = c(15L, 20L, 22L)),
  class = "data.frame", row.names = c(NA, 
 -3L))
          
   

Answer 4

使用apply族的解决方案

df$Carol = lapply(1:nrow(df), function(x) any(df[x,]=="Carol))

Answer 5

按照您的建议使用 mutate 和 if_else() 的另一个选项：

library(tidyverse)
data = read_table("   id first  middle  last       Age
    1 Carol  Jenny   Smith      15
    2 Sarah  Carol   Roberts    20
    3 Josh   David   Richardson 22")
data %>%
  mutate(carol = if_else(first == "Carol" | middle == "Carol" | last == "Carol",
                         "yes",
                         "no"))

结果：

# A tibble: 3 x 6
     id first middle last         Age carol
  <dbl> <chr> <chr>  <chr>      <dbl> <chr>
1     1 Carol Jenny  Smith         15 yes  
2     2 Sarah Carol  Roberts       20 yes  
3     3 Josh  David  Richardson    22 no