我有如下数据框:
S A B C D E
1 N N N N N
2 N Y Y N N
3 Y N Y N N
4 Y N Y Y Y
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我在哪里需要创建一个新列F,其中包含来自A,B,C,D和E多个列的出现次数最多的字符?
输出应如下所示:
S A B C D E F
1 N N N N N N
2 N Y Y N N N
3 Y N Y N N N
4 Y N Y Y Y Y
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我们可以创建一个Mode函数并将其应用于行
df1$F <- apply(df1[-1], 1, Mode)
df1
# S A B C D E F
#1 1 N N N N N N
#2 2 N Y Y N N N
#3 3 Y N Y N N N
#4 4 Y N Y Y Y Y
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或者另一个选择是
df1$F <- c('N', 'Y')[max.col(table(c(row(df1[-1])), unlist(df1[-1])), 'first')]
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哪里
Mode <- function(x) {
ux <- unique(x)
ux[which.max(tabulate(match(x, ux)))]
}
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或使用 tidyverse
library(tidyverse)
df1 %>%
mutate(F = pmap_chr(.[-1], ~ Mode(c(...))))
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或者另一个选择是
gather(df1, key, F, - S) %>%
group_by(S, F) %>%
summarise(n = n()) %>%
slice(which.max(n)) %>%
ungroup %>%
dplyr::select(F) %>%
bind_cols(df1, .)
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或者我们转置数据集,Mode按每个列应用,然后将输出作为新列绑定到原始数据集
t(df1[-1]) %>%
as.data.frame %>%
summarise_all(Mode) %>%
unlist %>%
bind_cols(df1, F = .)
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或一个选项 data.table
library(data.table)
setDT(df1)[, F := names(which.max(table(unlist(.SD)))), S][]
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注意:这些是常规方法,而不仅仅是检查单个案例
如果我们需要一种高效的方法,没有任何方法ifelse,我们也可以通过
df1$F <- c("Y", "N")[(rowSums(df1[-1] == "N") > 2) + 1]
df1$F
#[1] "N" "N" "N" "Y"
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或搭配 Reduce
c("Y", "N")[(Reduce(`+`, lapply(df1[-1], `==`, "N")) > 2) + 1]
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或另一种方法是
c("Y", "N")[(str_count(do.call(paste0, df1[-1]), "N") > 2) + 1]
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df1 <- structure(list(S = 1:4, A = c("N", "N", "Y", "Y"), B = c("N",
"Y", "N", "N"), C = c("N", "Y", "Y", "Y"), D = c("N", "N", "N",
"Y"), E = c("N", "N", "N", "Y")), class = "data.frame", row.names = c(NA,
-4L))
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