【发布时间】：2021-10-26 15:24:46
【问题描述】：
我有一个数据框
stim1 stim2 Chosen Rejected
1: 2 1 2 1
2: 3 2 2 3
3: 3 1 1 3
4: 2 3 3 2
5: 1 3 1 3
我的目标是在每次试验中添加一列，指定刺激是最近（在之前的试验中）被选择还是被拒绝。
想要的结果
stim1 stim2 Chosen Rejected Previous_stim1 Previous_stim2
1: 2 1 2 1 NaN NaN
2: 3 2 2 3 NaN Chosen
3: 3 1 1 3 Rejected Rejected
4: 2 3 3 2 Chosen Rejected
5: 1 3 1 3 Chosen Chosen
任何帮助将不胜感激！
更新
TarJae 提出了一个非常有用的建议，有助于对我正确共享的数据框进行分类。我没有提到它确实是更大数据框的一部分，并且由于某种原因，这种方法很快就会停止正确分类
stim1 stim2 Chosen Rejected Previous_stim1 Previous_stim2
1: 2 1 2 1 <NA> <NA>
2: 3 2 2 3 <NA> Chosen
3: 3 1 1 3 Rejected Rejected
4: 2 3 3 2 Chosen Rejected
5: 1 3 1 3 Chosen Chosen
6: 2 1 1 2 Chosen Chosen
7: 2 3 2 3 Chosen Chosen
8: 3 1 1 3 Chosen Chosen
9: 2 1 2 1 Chosen Chosen
例如，在第 6 行 stim1==2。最近，2 被拒绝（第 4 行），但方法将其归类为已选择。
有什么想法会发生什么吗？
再次感谢大家的帮助。
更新 2
非常感谢您的帮助。但是假设我还有一个“结果”专栏。
stim1 stim2 Chosen Rejected outcome Previous_stim 1 Previous_stim 2
1: 15 13 15 13 1 <NA> <NA>
2: 13 14 14 13 1 Rejected <NA>
3: 14 15 14 15 1 Chosen Chosen
4: 14 13 14 13 0 Chosen Rejected
5: 13 15 13 15 0 Rejected Rejected
6: 14 15 14 15 1 Chosen Rejected
7: 15 13 15 13 1 Rejected Chosen
8: 14 15 14 15 0 Chosen Chosen
I want to encode whether it was
1) most recently chosen and outcome=1 (can be coded as 1)
2) most recently chosen and outcome=0 (can be coded as 2)
3) most recently rejected and outcome=1 (can be coded as 3)
4) most recently rejected and outcome=0 (can be coded as 4)
有没有一种简单的方法来修改代码以实现这一点？
期望的输出
stim1 stim2 Chosen Rejected outcome Previous_stim 1 Previous_stim 2 Left_type right_type
1 2 3 2 3 1 <NA> <NA> NaN NaN
2 1 3 3 1 1 <NA> Rejected NaN 3
3 2 1 1 2 1 Chosen Rejected 1 3
4 1 2 1 2 0 Chosen Rejected 1 3
5 3 1 3 1 1 Chosen Chosen 1 2
最后跟进
最后，我想添加一列，检查在之前的试验中选择的刺激（我将其称为相关刺激的最近被拒绝的试验）是否与我当前的替代刺激相同
例如如果有
stim1 stim2 Chosen Rejected Previous_stim1 Previous_stim2
1: 2 1 2 1 NaN NaN
2: 3 2 2 3 NaN Chosen
3: 3 1 1 3 Rejected Rejected
4: 2 3 3 2 Chosen Rejected
5: 1 3 1 3 Chosen Chosen
这就是我将如何更新我的表格
in trial 3, previous_stim1 (i.e 3) was previously rejectedin favor of 2 (from trial 2) and not in favor of 1 (which is the current alternative) and so Current_alternative_left=0.
Similarly, previous_stim2 (i.e 1)was previously
rejected but that was rejected in favor of 2 (from trial 1)
and so current_alternative_right=0
On the other hand, in trial 4 stim1=2
was previously chosen relative to the same
刺激，因为它目前正在对抗 (3) 所以 current_alternative_right=1
期望的输出
stim1 stim2 Chosen Rejected outcome Previous_stim 1 Previous_stim 2 Left_type right_type
1 2 3 2 3 1 <NA> <NA> NaN NaN
2 1 3 3 1 1 <NA> Rejected NaN 3
3 2 1 1 2 1 Chosen Rejected 1 3
4 1 2 1 2 0 Chosen Rejected 1 3
5 3 1 3 1 1 Chosen Chosen 1 2
Current_alternative_left Current_alternative_right
NaN NaN
NaN 0
0 0
1 0
1 0
我是 data.table 的新手，但我尝试复制 ThomasisCoding 函数以将其也返回
h <- function(stim, cr) {
stim_chosen <- rep(NA,length(stim))
for (k in seq_along(stim)[-1]) {
ind <- which(cr[1:(k - 1), , drop = FALSE] == stim[k], arr.ind = TRUE)
if (length(ind)) {
stim_chosen[k] <- stim[tail(ind,1)[,"row"]]
}
}
stim_chosen
}
setDT(df)[ ,
paste0("Chosen_Last", 1:2) := lapply(
.(stim1, stim2),
h,
cr = cbind(Chosen,Rejected)
)
]
虽然这并没有给我正确的答案。有人知道我哪里出错了吗？
【问题讨论】：
-
为什么是
Previous_stim1row2NaN? -
因为以前没有选择或拒绝它的实例
-
对不起，只是为了了解逻辑。
stim1第 1 行 =2。这不是stim1row2=3 的前一个实例吗？ -
当我查看以前的实例时，我只检查选择和拒绝的列。所以第 2 行 ==3 和 3 中的 stim1 从未出现过，所以它的 NaN。但是 stim2==2 是在之前的试验中选择的，所以 Previous_stim2=chosen
-
那么，
Previous_stim1[5]将是Rejected？
标签： r dataframe dplyr data-wrangling