R:用 0 随机替换数据框的元素

Question

我正在使用 R 编程语言。

假设我有以下数据框:

var_1 = var_2 = var_3 = var_4 = var_5 =  c("1,2,3,4,5,6,7,8,9,10")

my_data = data.frame(var_1,var_2,var_3,var_4,var_5)

my_data = rbind(my_data, my_data[rep(1, 100), ])

rownames(my_data) = 1:nrow(my_data)

数据如下所示:

    head(my_data)

                 var_1                var_2                var_3                var_4                var_5
1 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10
2 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10
3 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10
4 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10
5 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10
6 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10

我的问题:我想用 0 随机替换此数据框中的元素 - 例如，最终结果应如下所示(为简洁起见，我只显示第一行) :

# desired result

                 var_1                var_2                var_3                var_4                var_5
1 1,0,3,0,5,6,0,0,9,10 1,2,0,4,5,0,0,8,9,0 1,0,3,0,0,0,0,8,9,0 1,2,3,4,0,6,7,0,0,10 1,2,0,4,5,0,7,8,0,10

我尝试使用以下代码行 (Replace random values in a column in a dataframe) 执行此操作:

my_data$var_1[sample(nrow(my_data),as.integer(0.5*nrow(my_data)) , replace = TRUE)] <- 0
my_data$var_2[sample(nrow(my_data),as.integer(0.5*nrow(my_data)), replace = TRUE)] <- 0
my_data$var_3[sample(nrow(my_data),as.integer(0.5*nrow(my_data)), replace = TRUE)] <- 0
my_data$var_4[sample(nrow(my_data),as.integer(0.5*nrow(my_data)), replace = TRUE)] <- 0
my_data$var_5[sample(nrow(my_data),as.integer(0.5*nrow(my_data)), replace = TRUE)] <- 0

但这是用 0 替换行中的所有元素(而不是仅替换行中的某些元素):

head(my_data)
                 var_1                var_2                var_3                var_4                var_5
1 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10                    0                    0                    0
2                    0                    0 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10                    0
3                    0 1,2,3,4,5,6,7,8,9,10                    0 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10
4                    0 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10                    0 1,2,3,4,5,6,7,8,9,10
5 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10
6 1,2,3,4,5,6,7,8,9,10                    0 1,2,3,4,5,6,7,8,9,10 1,2,3,4,5,6,7,8,9,10                    0

有人可以告诉我我做错了什么以及如何获得预期的结果吗？

谢谢!

Answer 1

这是一个版本，它允许您使用 Map 在每一列中分别指定成为 0 的概率向量 pnul。拆分字符串的 length 乘以 pnul 的元素以获得设置为零的 sample 的数量。您还可以将 pnul 设置为所有列中具有相同概率的标量。

pnul <- c(.0, .2, .5, .8, 1)

res <- Map(\(x, a) {
  S <- strsplit(x, ',')
  sapply(S, \(s) {
    s[sample(seq_along(s), length(s)*a)] <- '0'
    paste(s, collapse=',')
  })
}, my_data, pnul) |> as.data.frame()

head(res)
#                  var_1                var_2                var_3                var_4               var_5
# 1 1,2,3,4,5,6,7,8,9,10 0,0,3,4,5,6,7,8,9,10  1,2,0,4,0,0,7,8,0,0  0,0,0,0,0,0,0,8,9,0 0,0,0,0,0,0,0,0,0,0
# 2 1,2,3,4,5,6,7,8,9,10  1,0,3,4,5,6,7,8,9,0 1,0,3,0,5,0,0,0,9,10  0,0,0,0,0,0,7,8,0,0 0,0,0,0,0,0,0,0,0,0
# 3 1,2,3,4,5,6,7,8,9,10 1,0,0,4,5,6,7,8,9,10 1,0,0,0,0,6,7,0,9,10 0,0,0,0,5,0,0,0,0,10 0,0,0,0,0,0,0,0,0,0
# 4 1,2,3,4,5,6,7,8,9,10 1,2,3,0,5,6,7,0,9,10 0,0,3,0,5,0,7,0,9,10  0,0,0,4,0,0,7,0,0,0 0,0,0,0,0,0,0,0,0,0
# 5 1,2,3,4,5,6,7,8,9,10  1,0,3,4,5,6,7,8,9,0 0,2,0,4,5,0,7,0,0,10  1,0,0,0,0,0,0,8,0,0 0,0,0,0,0,0,0,0,0,0
# 6 1,2,3,4,5,6,7,8,9,10 0,2,3,4,5,6,0,8,9,10  1,2,3,0,5,0,7,0,0,0  0,0,0,4,5,0,0,0,0,0 0,0,0,0,0,0,0,0,0,0