通过每个数据帧中的行数对数据帧列表进行重新排序

Question

Say I have three dataframes in my.list , each with different numbers of rows. I would like to reorder this list so that the first element of the list is the dataframe with the highest number of rows (in the below example, d2 ).

d1 <- data.frame(y1 = c(1, 2, 3),
                 y2 = c(4, 5, 6))
d2 <- data.frame(y1 = c(3, 2, 1, 3, 2),
                 y2 = c(6, 5, 4, 2, 5))
d3 <- data.frame(y1 = c(2, 1),
                 y2 = c(3, 2))

my.list <- list(d1, d2, d3)

The expected output:

str(mylist[[1]]) ## 'data.frame':   5 obs. of  2 variables:
                     $ y1: num  3 2 1 3 2
                     $ y2: num  6 5 4 2 5

The reason for this: I'm repeatedly plotting data from the first element in several lists of dataframes, and would like to make sure I'm plotting the dataframes with the most data points when I call plot(my.list[[1]]) .

Probably a cleaner solution would be to, within the plot call, search for the element/dataframe with the highest number of rows and plot that, but I'm not sure how easy that would be.

One potentially complicating factor is that there will occasionally be a list of dataframes where there is more than one dataframe sharing the highest number of rows. In that case, it wouldn't matter which one is called--they'd both do fine--but I'm not sure whether that creates an issue here.

Answer 1

假设您的列表称为“ lst”

lst= lst[order(sapply(lst,nrow),decreasing = T)]

Answer 2

Use vapply to get the number of rows in each data frame, then use rev(order(...)) to sort them from most rows to least.

nrow_each <- vapply(my.list, nrow, numeric(1))
my.list[rev(order(nrow_each))]

Or do it in one, difficult-to-read line

my.list[rev(order(vapply(my.list, nrow, numeric(1))))]