基于 r 中行值的条件 if 语句

Question

我是 R 的新手，非常感谢您在这方面的帮助。

我有一个数据框，2 个级别是 'Y' 和 'N' 指标，针对 11 个变量。

我想要一个新列，当行值等于 'Y'

时连接列名

即

Answer 1

在 base R 中，我们可以使用 which 创建一个 row/column 索引矩阵，其中值为 "Y"。使用 tapply，我们可以 paste 每行的列名。

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 我们可以得到长格式的数据，filter 行，其中 value 是 "Y"，并为每一行粘贴列值。

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)

数据

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

一个简单的基础 R 方法是

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

测试数据创建代码。

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))

基于 r 中行值的条件 if 语句

Conditional if statement based on row values in r

r

dplyr

data-wrangling