有很多方法可以做到這一點。這是一個。請參閱行內評論。
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
另一種方法是預先分配矩陣,添加值並將其強制爲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,函數(x)的RNORM(10 ,mean = x))); colnames(d1)< - paste0(「V」,seq_along(values))' – akrun
嗨,感謝您的解決方案。如果能夠獲得for循環解決方案也是一件好事,因爲這是我目前旨在改進的一項技能。謝謝! – George
您首先創建一個空矩陣或列表,然後爲其分配值。即'output < - vector(「list」,4)';並在必要時使用你的代碼和'cbind'或'rbind',即'do.call(cbind,output)' – akrun