这个问题在这里已经有了答案:
Expand ranges defined by "from" and "to" columns
(9 个回答)
4年前关闭。
我有以下输入数据框:
df <- data.frame(x=c('a','b','c'),y=c(4,5,6),from=c(1,2,3),to=c(2,4,6))
df
x y from to
1 a 4 1 2
2 b 5 2 4
3 c 6 3 6
现在我想将每一行乘以 from 和 to 之间的值,即 ('a',4) 跨越两行,即
1,2
.预期结果如下所示:exp <- data.frame(x=c('a','a','b','b','b','c','c','c','c'),
y=c(4,4,5,5,5,6,6,6,6),
z=c(1,2,2,3,4,3,4,5,6))
exp
x y z
1 a 4 1
2 a 4 2
3 b 5 2
4 b 5 3
5 b 5 4
6 c 6 3
7 c 6 4
8 c 6 5
9 c 6 6
在没有循环的情况下完成此任务的最惯用方法是什么?
最佳答案
一种“非tidyverse”方式:
data.frame(
x = c('a', 'b', 'c'),
y = c(4, 5, 6),
from = c(1, 2, 3),
to = c(2, 4, 6),
stringsAsFactors = FALSE
) -> xdf
do.call(rbind.data.frame, lapply(1:nrow(xdf), function(i) {
data.frame(x = xdf$x[i], y=xdf$y[i], z=xdf$from[i]:xdf$to[i], stringsAsFactors=FALSE)
}))
一种“tidyverse”方式:
library(tidyverse)
data_frame(
x = c('a', 'b', 'c'),
y = c(4, 5, 6),
from = c(1, 2, 3),
to = c(2, 4, 6)
) -> xdf
rowwise(xdf) %>%
do(data_frame(x = .$x, y=.$y, z=.$from:.$to))
另一种尚未在下面进行基准测试的“tidyverse”方式:
xdf %>%
rowwise() %>%
do( merge( as_tibble(.), tibble(z=.$from:.$to), by=NULL) ) %>%
select( -from, -to ) # Omit this line if you want to keep all original columns.
既然你问了 abt 性能:
library(microbenchmark)
data.table::data.table(
x = c('a','b','c'),
y = c(4,5,6),
from = c(1,2,3),
to = c(2,4,6)
) -> xdt1
data.frame(
x = c('a', 'b', 'c'),
y = c(4, 5, 6),
from = c(1, 2, 3),
to = c(2, 4, 6),
stringsAsFactors = FALSE
) -> xdf1
data.table
ops 经常就地修改,所以在执行 op 之前保持一个公平的竞争环境并制作每个数据帧/表的副本。在大多数现代系统上,时间损失约为 100 纳秒。
microbenchmark(
data.table = {
xdt2 <- xdt1
xdt2[, diff:= (to - from) + 1]
xdt2 <- xdt2[rep(1:.N, diff)]
xdt2[,z := seq(from,to), by=.(x,y,from,to)]
xdt2[,c("x", "y", "z")]
},
base = {
xdf2 <- xdf1
do.call(rbind.data.frame, lapply(1:nrow(xdf2), function(i) {
data.frame(x = xdf2$x[i], y=xdf2$y[i], z=xdf2$from[i]:xdf2$to[i], stringsAsFactors=FALSE)
}))
},
tidyverse = {
xdf2 <- xdf1
dplyr::rowwise(xdf2) %>%
dplyr::do(dplyr::data_frame(x = .$x, y=.$y, z=.$from:.$to))
},
plyr = {
xdf2 <- xdf1
plyr::mdply(xdf2, function(x,y,from,to) data.frame(x,y,z=seq(from,to)))[c("x","y","z")]
},
times = 1000
)
## Unit: microseconds
## expr min lq mean median uq max neval
## data.table 920.361 1072.9265 1257.2321 1178.832 1280.2660 10628.552 1000
## base 677.069 761.3145 884.4136 825.472 915.8985 5366.515 1000
## tidyverse 15926.127 17231.5015 19201.4798 17994.919 20014.4140 166901.570 1000
## plyr 1938.838 2196.4205 2448.5314 2322.949 2501.5075 5735.255 1000
关于r - 如何根据值扩展数据框?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/46753597/