Question

我在创建一个根据P列分组并计算所有＆＃34; YES＆＃34;对于已保存在列表中的列，例如＆＃34; list_col＆＃34;。我很确定我会使用apply系列中的函数，但不确定如何按特定列进行分组（在这种情况下为Col P）

P <- as.character(c("A", "A", "A", "A", "A", "B", "B", "B", "B", "B"))
a <- as.character(c(NA,"Yes","Yes",NA,NA,NA,"Yes","Yes","Yes",NA))
b <- as.character(c(NA,"Yes",NA,NA,"Yes",NA,"Yes","Yes",NA,NA))
c <- as.character(c(NA,NA,NA,NA,"Yes",NA,"Yes",NA,NA,NA))

df_sample <- as.data.frame(cbind(P, a, b, c))
df_sample

list_col <- c("a","b","c")

理想情况下，我会根据上面的示例数据框和下面的答案寻找以下答案，并根据＆＃34; list_col＆＃34;

更改列

P   a   b   c
A   2   2   1
B   3   2   1

任何帮助将不胜感激

Answer 1

以下是dplyr，

的方法

library(dplyr)

df_sample %>% 
    group_by(P) %>% 
    select(list_col) %>% 
    summarise_all(funs(sum(. == 'Yes', na.rm = TRUE)))

#Adding missing grouping variables: `P`
# A tibble: 2 x 4
#       P     a     b     c
#  <fctr> <int> <int> <int>
#1      A     2     2     1
#2      B     3     2     1

Answer 2

在data.table：

library(data.table)
list_col <- c("a","b","c")
setDT(df_sample)[, (lapply(.SD, function(x) sum(x=="Yes", na.rm = TRUE))), by = P, .SDcols = list_col]

#  P a b c
#1: A 2 2 1
#2: B 3 2 1

或者，基本R解决方案仍使用lapply：

res <-lapply(split(df_sample[,list_col], df_sample$P), function(x) colSums(x=="Yes", na.rm = TRUE))
do.call(rbind, res) 
# a b c
#A 2 2 1
#B 3 2 1

为了它的价值，我的机器上有一个微基准测试：

microbenchmark::microbenchmark(splitlapply = do.call(rbind, lapply(split(df_sample[,list_col], df_sample$P), function(x) colSums(x=="Yes", na.rm = TRUE))),
+                                dt = sampleDT[, (lapply(.SD, function(x) sum(x=="Yes", na.rm = TRUE))), by = P, .SDcols = list_col])
Unit: microseconds
        expr     min        lq      mean    median        uq      max neval
 splitlapply 455.841  505.0715  546.6699  529.3225  561.2315  889.436   100
          dt 861.722 1052.9920 1114.2752 1111.7040 1166.7695 1707.761   100

Answer 3

使用melt

中的reshape

library(reshape)    
df=na.omit(melt(df_sample,id.vars='P'))
table(df$P,df$variable)

  a b c
A 2 2 1
B 3 2 1

Answer 4

df_sample <- as.data.frame(cbind(P, a, b, c), stringsAsFactors = FALSE)
for (i in list_col){
  df_r <- df_sample[, i] == 'Yes' & !is.na(df_sample[, i])
  df_sample[df_r, i] <- df_sample$P[df_r]
}
sapply(df_sample[, list_col], table)

使用列表中的指定列分组依据和计数

4 个答案: