更新时间:2023-12-01 09:47:52
您可以按照2列矩阵进行索引 - 第一列是行号,第二列是列号。
df [cbind(seq(cl),cl)]
pre>
#[1] 100 310 320 230 140这是一个向量化的操作,应该比循环遍历具有类似
sapply
的行更快,并抓取该行的适当值:#稍微更大的例子,1000行
set.seed(144)
df< - matrix(rnorm(3000),nrow = 1000)
cl< - sample(3,1000,replace = TRUE)
all.equal(df [cbind ),b(b)(c)),b(b)(b)(b) $ b microbenchmark(df [cbind(seq(cl),cl)],sapply(seq(nrow(df)),function(i)df [i,cl [i]])
#单位:微秒
#expr min lq平均值
#df [cbind(seq(cl),cl)] 23.828 26.335 34.26012 30.0350
#sapply(seq(nfd(df) (i)df [i,cl [i]])855.481 922.449 1178.47502 996.3815
#uq max neval
#38.0315 135.894 100
#1111.3960 3414.374 100
How do I construct a vector of values from
n
th column of some data frame, wheren
is a per-row value defined in some vector? Example:> df <- data.frame(a=c(100, 110, 120, 130, 140), b=c(200, 210, 220, 230, 240), c=c(300, 310, 320, 330, 340)) > df a b c 1 100 200 300 2 110 210 310 3 120 220 320 4 130 230 330 5 140 240 340 > cl <- c(1, 3, 3, 2, 1) > some.function(df, cl)
would result in:
[1] 100 310 320 230 140
You can index by a 2-column matrix -- the first column is the row number and the second is the column number.
df[cbind(seq(cl), cl)] # [1] 100 310 320 230 140
This is a vectorized operation that should be quicker than looping through the rows with something like
sapply
and grabbing the appropriate value from that row:# Slightly larger example, with 1000 rows set.seed(144) df <- matrix(rnorm(3000), nrow=1000) cl <- sample(3, 1000, replace=TRUE) all.equal(df[cbind(seq(cl), cl)], sapply(seq(nrow(df)), function(i) df[i, cl[i]])) # [1] TRUE library(microbenchmark) microbenchmark(df[cbind(seq(cl), cl)], sapply(seq(nrow(df)), function(i) df[i, cl[i]])) # Unit: microseconds # expr min lq mean median # df[cbind(seq(cl), cl)] 23.828 26.335 34.26012 30.0350 # sapply(seq(nrow(df)), function(i) df[i, cl[i]]) 855.481 922.449 1178.47502 996.3815 # uq max neval # 38.0315 135.894 100 # 1111.3960 3414.374 100