有很多方法可以做到这一点。这是一个。请参阅行内评论。
set.seed(357) # to make things reproducible, set random seed
N <- 10 # number of loops
xy <- vector("list", N) # create an empty list into which values are to be filled
# run the loop N times and on each loop...
for (i in 1:N) {
# generate a data.frame with 4 columns, and add a random number into each one
# random number depends on the mean specified
xy[[i]] <- data.frame(um10 = rnorm(1, mean = -10),
u0 = rnorm(1, mean = 0),
u10 = rnorm(1, mean = 10),
u100 = rnorm(1, mean = 100))
}
# result is a list of data.frames with 1 row and 4 columns
# you can bind them together into one data.frame using do.call
# rbind means they will be merged row-wise
xy <- do.call(rbind, xy)
um10 u0 u10 u100
1 -11.241117 -0.5832050 10.394747 101.50421
2 -9.233200 0.3174604 9.900024 100.22703
3 -10.469015 0.4765213 9.088352 99.65822
4 -9.453259 -0.3272080 10.041090 99.72397
5 -10.593497 0.1764618 10.505760 101.00852
6 -10.935463 0.3845648 9.981747 100.05564
7 -11.447720 0.8477938 9.726617 99.12918
8 -11.373889 -0.3550321 9.806823 99.52711
9 -7.950092 0.5711058 10.162878 101.38218
10 -9.408727 0.5885065 9.471274 100.69328
另一种方法是预先分配矩阵,添加值并将其强制为data.frame。
xy <- matrix(NA, nrow = N, ncol = 4)
for (i in 1:N) {
xy[i, ] <- rnorm(4, mean = c(-10, 0, 10, 100))
}
# notice that i name the column names post festum
colnames(xy) <- c("um10", "u0", "u10", "u100")
xy <- as.data.frame(xy)
可以使用'lapply'或'sapply'即'D1 < - data.frame(lapply(值,函数(x)的RNORM(10 ,mean = x))); colnames(d1)< - paste0(“V”,seq_along(values))' – akrun
嗨,感谢您的解决方案。如果能够获得for循环解决方案也是一件好事,因为这是我目前旨在改进的一项技能。谢谢! – George
您首先创建一个空矩阵或列表,然后为其分配值。即'output < - vector(“list”,4)';并在必要时使用你的代码和'cbind'或'rbind',即'do.call(cbind,output)' – akrun