Question

有一个data.table

library(data.table)
car <- data.table(no = 1:100, turn = sample(1:5,100,replace = TRUE), 
              dis = sample(1:10,100,replace = TRUE))

我想改变＆＃34; dis＆＃34;在{em> n 出现turn == 3时，为-1，即第三次＆＃34;转向＆＃34;是3。

我可以选择turn == 3的第三行：

car[turn == 3, .SD[3]]

然而，我没有设法更新＆＃34; dis＆＃34;在这一行：

car[turn == 3, .SD[3]][, dis := -1]

相关的问与答：Conditionally replacing column values with data.table。

Answer 1

一些替代方案。使用rowid或cumsum创建组内行的计数器。在i。

中将计数器添加到您的条件中

我使用稍小的玩具数据集，只是为了更容易跟踪更改：

d <- data.table(x = 1:3, y = 1:12)

d[rowid(x) == 3 & x == 3, y := -1]

# @mt1022
d[cumsum(x == 3) == 3 & (x == 3), y := -1]

# @docendo discimus
d[(ix <- x == 3) & cumsum(ix) == 3, y := -1]

虽然OP没有提到速度是一个问题，但我仍然很想在更大的矢量上计算不同的方法。不出所料，@ Frank的方法是最快的，尤其是当搜索的唯一值的数量增加时：

frank << docendo < henrik < mt022

microbenchmark(henrik = d[rowid(x) == 3 & x == 3, y := -1],
               mt1022 = d[cumsum(x == 3) == 3 & (x == 3), y := -1],
               docendo = d[(ix <- x == 3) & cumsum(ix) == 3, y := -1],
               frank = d[d[x == 3, which = TRUE][3], y := -1], unit = "relative")

d <- data.table(x = sample(1:3, 1e6, replace = TRUE), y = 1:1e6)
# Unit: relative
#    expr      min       lq     mean   median       uq      max neval cld
#  henrik 4.417303 4.369407 4.133514 4.319839 4.329658 1.260394   100  b 
#  mt1022 5.461961 5.285562 5.174559 5.186404 5.239738 1.608712   100   c
# docendo 3.572646 3.624369 3.788678 3.589705 3.576637 1.733272   100  b 
#   frank 1.000000 1.000000 1.000000 1.000000 1.000000 1.000000   100 a 

d <- data.table(x = sample(1:30, 1e6, replace = TRUE), y = 1:1e6)
# Unit: relative
#    expr      min       lq     mean   median       uq      max neval cld
#  henrik 22.64881 19.54375 18.81963 18.91335 19.78559 5.507692   100  bc
#  mt1022 24.58258 21.17535 19.84417 20.96256 22.76020 3.625263   100   c
# docendo 19.40044 16.75912 16.23321 16.47953 18.06264 4.234100   100  b 
#   frank  1.00000  1.00000  1.00000  1.00000  1.00000 1.000000   100 a

d <- data.table(x = sample(1:300, 1e6, replace = TRUE), y = 1:1e6)
# Unit: relative
#    expr      min       lq     mean   median       uq       max neval cld
#  henrik 31.81237 32.51122 28.79490 30.35766 28.63560  8.236282   100  b 
#  mt1022 34.71984 35.45341 33.20405 33.57394 31.50914 21.556367   100   c
# docendo 27.99046 28.15855 26.56954 26.60644 25.20044  7.847163   100  b 
#   frank  1.00000  1.00000  1.00000  1.00000  1.00000  1.000000   100 a

# Unit: milliseconds
#    expr       min        lq      mean    median       uq        max neval cld
#  henrik 60.655582 76.455531 83.061266 77.632036 78.57818 203.224042   100   c
#  mt1022 66.701182 84.133034 87.967300 84.937201 85.72464 201.167914   100   c
# docendo 52.938545 67.214360 71.558130 68.003891 68.51897 184.178346   100  b 
#   frank  1.977821  2.494039  2.629852  2.663577  2.76089   3.613905   100 a

Answer 2

这是另一种方式：

car[car[turn == 3, which = TRUE][3], dis := -1 ]

评论。对于此类操作，您可能需要启用verbose，以便查看是否进行了任何更改。例如，如果我们寻找第111个...... ....

car[car[turn == 3, which = TRUE][111], dis := -1, verbose = TRUE ]
# Detected that j uses these columns: dis 
# Assigning to 0 row subset of 100 rows

显示编辑了0行。

另请参阅Using .I to return row numbers with data.table package有关使用which = TRUE。

的信息

如何在另一个变量的第n次出现时更改变量？

2 个答案: