英文:
Create binary table where the number of rows are values from a variable
问题
我有一个包含两个计数变量（var1和var2）的数据集，它们代表两个时间点的存活个体数。
我需要创建一个包含新的二进制变量的表格，其中行数对应于var1的值，并在这个变量内，根据var2的值分配"1"（例如，如果var1中有10，var2中有3，那么我需要10行，其中3行为"1"，其余为"0"）。
是否有任何可以帮助的函数？
谢谢。
英文:
I have a dataset with a two variables with counts (var1 and var2), that represent alive individuals at two times.
I need to create a table with a new binary variable, where the number of rows correspond to the values of var1, and within this variable, I need to assign "1" depending on values of var2 (e.g. if there is a 10 in var1 and a 3 in var2, then I would need 10 rows, from which 3 rows have "1", and the rest have "0").
Any function that would help?
Thanks
答案1
得分: 1
我可能误解了你的需求，但以下是我做的：
exemple <- data.frame(id = 1:3, var1 = c(10, 5, 1), var2 = c(3, 3, 0))
apply(exemple, 1, \(x) data.frame(id = rep(x[["id"]], times = x[["var1"]]),
var_binaire = rep(c(1, 0), times = c(x[["var2"]], x[["var1"]] - x[["var2"]])))) |>
do.call(what = "rbind")
它似乎可以达到效果，它使用了rep函数重复生成1和0。apply被用于在行方向上执行操作。因此apply在此处创建了一个列表，每个元素都是一个具有var1行和var_binaire = 1 var2次的数据框。然后，do.call("rbind")将它们合并成一个数据框。以下是结果：
id var_binaire
1 1 1
2 1 1
3 1 1
4 1 0
5 1 0
6 1 0
7 1 0
8 1 0
9 1 0
10 1 0
11 2 1
12 2 1
13 2 1
14 2 0
15 2 0
16 3 0
我曾以为可以在不使用apply的情况下完成，但我无法想出方法。也许会有更好的解决方案出现。希望对你有所帮助。
英文:
I may have misunderstood what you wanted but here is what I did:
exemple <- data.frame(id = 1:3, var1 = c(10, 5, 1), var2 = c(3, 3, 0))
apply(exemple, 1, \(x) data.frame(id = rep(x[["id"]], times = x[["var1"]]),
var_binaire = rep(c(1, 0), times = c(x[["var2"]], x[["var1"]] - x[["var2"]])))) |>
do.call(what = "rbind")
It seems to do the trick, it creates a data.frame by repeating 1's and 0's with function rep. Apply is used with margin = 1 to perform rowwise operations. Hence apply create here a list with each element a data.frame with var1 lines and var_binaire = 1 var2 times. Then, do.call("rbind") combines all in a data.frame. Here is the result:
exemple
id var1 var2
1 1 10 3
2 2 5 3
3 3 1 0
output
id var_binaire
1 1 1
2 1 1
3 1 1
4 1 0
5 1 0
6 1 0
7 1 0
8 1 0
9 1 0
10 1 0
11 2 1
12 2 1
13 2 1
14 2 0
15 2 0
16 3 0
I thought it would be possible without apply, but I couldn't figure out how. Maybe a better solution will come up. I hope this helped.
答案2
得分: 1
与@Guillaume Mulier非常相似的一种方法，但使用向量化：
x <- data.frame(id = 1:3, var1 = c(10, 5, 1), var2 = c(3, 3, 0))
with(x, data.frame(id=rep(id, var1),
mort=rep(rep(1:0, nrow(x)),
c(matrix(c(var2, var1 - var2), 2, byrow=TRUE)))))
# id mort
#1 1 1
#2 1 1
#3 1 1
#4 1 0
#5 1 0
#6 1 0
#7 1 0
#8 1 0
#9 1 0
#10 1 0
#11 2 1
#12 2 1
#13 2 1
#14 2 0
#15 2 0
#16 3 0
请注意，这段代码是R语言的一部分，主要用于数据框（data.frame）操作。
英文:
A way quite similar to @Guillaume Mulier but vectorizing;
x <- data.frame(id = 1:3, var1 = c(10, 5, 1), var2 = c(3, 3, 0))
with(x, data.frame(id=rep(id, var1),
mort=rep(rep(1:0, nrow(x)),
c(matrix(c(var2, var1 - var2), 2, byrow=TRUE)))))
# id mort
#1 1 1
#2 1 1
#3 1 1
#4 1 0
#5 1 0
#6 1 0
#7 1 0
#8 1 0
#9 1 0
#10 1 0
#11 2 1
#12 2 1
#13 2 1
#14 2 0
#15 2 0
#16 3 0
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