基於 r 中的行值的條件 if 語句

Question

我是 R 的新手，我非常感謝您在這方面的幫助。

我有一個數據框，有 2 個級別是 11 個變量的“Y”和“N”指標。

我想要一個新列，當行值等於“Y”時連接列名

IE

Answer 1

在基數 R 中，我們可以使用which創建一個行/列索引矩陣，其中值是"Y" 。 使用tapply ，我們可以為每一行paste列名。

cols <- paste0('col', 1:9)
mat <- which(df[cols] == 'Y', arr.ind = TRUE)
df$new_col <- as.character(tapply(names(df)[mat[, 2]], mat[, 1], 
               paste, collapse = "_"))
df
#  col1 col2 col3 col4 col5 col6 col7 col8 col9 col10 col11                       new_col
#1    N    Y    N    Y    Y    Y    N    Y    Y     1   624 col2_col4_col5_col6_col8_col9
#2    N    Y    N    Y    Y    Y    N    Y    N     7   548      col2_col4_col5_col6_col8

---

使用tidyverse我們可以獲得長格式的數據， filter value "Y"行，並為每一行粘貼列值。

library(dplyr)

df %>%
  mutate(row = row_number()) %>%
  tidyr::pivot_longer(cols = -c(col10, col11, row)) %>%
  filter(value == 'Y') %>%
  group_by(row, col10, col11) %>%
  summarise(newcol = toString(name)) %>%
  ungroup() %>%
  select(-row)

數據

df <- structure(list(col1 = structure(c(1L, 1L), .Label = "N", class = "factor"), 
col2 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col3 = structure(c(1L, 1L), .Label = "N", class = "factor"), 
col4 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col5 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col6 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col7 = structure(c(1L, 1L), .Label = "N", class = "factor"), 
col8 = structure(c(1L, 1L), .Label = "Y", class = "factor"), 
col9 = structure(2:1, .Label = c("N", "Y"), class = "factor"), 
col10 = c(1, 7), col11 = c(623.53, 548.028)), row.names = c(NA, -2L),
class = "data.frame")

Answer 2

一個簡單的基本 R 方法是

df1$newcol <- apply(df1, 1, function(x){
  paste(names(df1)[x == "Y"], collapse = "_")
})

測試數據創建代碼。

set.seed(1234)
df1 <- t(replicate(2, sample(c("N", "Y"), 10, TRUE)))
df1 <- as.data.frame(df1)
df1 <- cbind(df1, matrix(1:4, 2))
names(df1) <- paste0("col", 1:ncol(df1))