更改data.table子集的随机选择的值

Question

基本上是这个问题的扩展,因为我注意到,如果你第二次进行子集化,就不可能改变列的值.

random.length  <-  sample(x = 15:30, size = 1)
dt <- data.table(city=sample(c("Cape Town", "New York", "Pittsburgh", "Tel Aviv", "Amsterdam"), size=random.length, replace = TRUE), score = sample(x=1:10, size = random.length, replace=TRUE)) 
set.seed(1)
dt[sample(.N,3), score :=9999]
set.seed(1)
dt[sample(.N,3),]

这可以按预期工作,并将三个随机选择的城市的分数更改为9999.虽然如果您在第一步中进行分组,然后进行采样并尝试分配新的分数值,那么这是不可能的.

set.seed(1)
dt[city == "New York",][sample(.N,1), score := 55555]
set.seed(1)
dt[city == "New York",][sample(.N,1)]

我想要实现的是,我可以更改某个列的值,该列是某个子集的一部分,并从该子集中随机选择.

Answer 1

dt[city == "New York"]返回一个全新的对象,您可以通过引用进行更新.但是,这并没有影响dt.即

dt[expr, col := val] != dt[expr][, col := val]

第一个表达式的更新dt,其中expr的计算结果为TRUE.第二个更新返回的子集dt[expr].除非您将结果分配回变量,否则无法取回结果.

Answer 2

which除了以上所有建议之外,您还可以对索引进行采样(可以使用函数计算):

dt[sample(which(city == "New York"), 1), score:=555L]
dt
#           city score
#  1:   Tel Aviv     8
#  2:  Amsterdam     3
#  3:  Cape Town    10
#  4:   New York     1
#  5:  Cape Town    10
#  6: Pittsburgh     2
#  7: Pittsburgh     8
#  8:  Amsterdam    10
#  9:  Amsterdam     8
# 10:  Amsterdam     4
# 11:   Tel Aviv     7
# 12:  Amsterdam     2
# 13: Pittsburgh     1
# 14:  Amsterdam     3
# 15: Pittsburgh     2
# 16:   New York     7
# 17:   Tel Aviv    10
# 18:   New York    10
# 19:  Cape Town     1
# 20:  Amsterdam     7
# 21:  Amsterdam     3
# 22:   New York   555
# 23:  Cape Town     6
# 24:   New York     1
# 25:   Tel Aviv    10
#           city score