找出一列中的序列与另一列相同的次数百分比

Question

我希望我能正确地表达这一点。我有一个包含两列的数据集，我试图在内存实验中进行比较。 Recall.CRESP 是一列，指定通过网格坐标选择的内存测试的正确答案。 Recall.RESP 显示参与者的响应。

这些列看起来像这样：

|Recall.CRESP                     | Recall.RESP                     |
|---------------------------------|---------------------------------|                 
|grid35grid51grid12grid43grid54   | grid35grid51grid12grid43grid54  |                
|grid22grid53grid35grid21grid44   | grid23grid53grid35grid21grid43  |
|grid12grid14grid15grid41grid23   | grid12grid24grid31grid41grid25  |
|grid15grid41grid33grid24grid55   | grid15grid41grid33grid14grid55  |          

我有以下代码行告诉我每行列彼此相同的次数百分比：

paste0((100*with(Data, mean(Recall.CRESP==Recall.RESP, na.rm = "TRUE"))), "%")

因此，例如，在我的数据集中，Recall.CRESP 列有 20% 的时间与 Recall.RESP 完全匹配，这表示受试者在 20% 的时间里在他们的记忆测试中得分 5 分（满分 5 分）。

但是，我希望能够通过两种方式对此进行扩展。第一个是在行相同时给我一个百分比，而不是在序列中存在部分匹配时给我一个百分比。例如 grid11gird42gird22grid51grid32 和 grid11gird15gird55grid42grid32 共享 2/5 的匹配，第一个和最后一个网格坐标相同。我不确定如何在 R 中为 2/5 的部分序列匹配（或 5 个中的任何其他结果）指定请求。另请记住，在此示例中，grid42 出现在两个序列中，但考虑到它在Recall.RESP 中的位置不正确，因此无法正确调用。在这些序列中，顺序很重要。

另一点是，到目前为止，我已经根据检查记忆项目前向召回的准确性来描述实验。然而，我也有单独的数据，参与者按倒序回忆。例如，来自Recall.CRESP 的grid11gird22gird33grid44grid55 和来自Recall.RESP 的grid51grid44grid33grid22grid11 正确匹配4/5 次。如何将代码转过来检查反向序列并计算 5 的百分比？

任何想法将不胜感激。

Answer 1

我会将字符串分成矩阵列，这样便于比较和操作：

# borrowing Oriol's nicely shared data
Recall.CRESP <- c('grid35grid51grid12grid43grid54',
                  'grid22grid53grid35grid21grid44',
                  'grid12grid14grid15grid41grid23',
                  'grid15grid41grid33grid24grid55')

Recall.RESP <- c('grid35grid51grid12grid43grid54',
                 'grid23grid53grid35grid21grid43',
                 'grid12grid24grid31grid41grid25',
                 'grid15grid41grid33grid14grid55')

# function to create matrices
matrixify = function(dat) {
    dat = do.call(rbind, strsplit(dat, split = "grid"))
    dat = dat[, -1]
    mode(dat) = "numeric"
    return(dat)
}

cresp_mat = matrixify(Recall.CRESP)
resp_mat = matrixify(Recall.RESP)

## an example of what we made: just the numbers in the right order
cresp_mat
#      [,1] [,2] [,3] [,4] [,5]
# [1,]   35   51   12   43   54
# [2,]   22   53   35   21   44
# [3,]   12   14   15   41   23
# [4,]   15   41   33   24   55

## Calculating results is now easy:
(forwards = rowMeans(cresp_mat == resp_mat))
# [1] 1.0 0.6 0.4 0.8

(reverse = rowMeans(cresp_mat == resp_mat[, 5:1]))
# [1] 0.2 0.2 0.0 0.2

当然，您可以将结果分配给原始数据的新列。

Answer 2

这是我的解决方案：

Recall.CRESP <- c('grid35grid51grid12grid43grid54',
                  'grid22grid53grid35grid21grid44',
                  'grid12grid14grid15grid41grid23',
                  'grid15grid41grid33grid24grid55')

Recall.RESP <- c('grid35grid51grid12grid43grid54',
                 'grid23grid53grid35grid21grid43',
                 'grid12grid24grid31grid41grid25',
                 'grid15grid41grid33grid14grid55')

df <- data.frame(Recall.CRESP, Recall.RESP, stringsAsFactors = F)
df$correctNormal <- NA
df$correctReverse <- NA

for (row in 1:nrow(df)) {
  crespVector <- unlist(strsplit(as.character(df[row, 1]), 'grid'))[-1]
  respVector <- unlist(strsplit(as.character(df[row, 2]), 'grid'))[-1]
  correctNormal <- 0
  correctReverse <- 0
  for (i in 1:length(crespVector)) {
    if (crespVector[i] == respVector[i]) correctNormal <- correctNormal + 1
    if (crespVector[i] == respVector[length(respVector) + 1 - i]) correctReverse <- correctReverse + 1
  }
  df$correctNormal[row] = correctNormal / 5
  df$correctReverse[row] = correctReverse / 5
}

df

##                     Recall.CRESP                    Recall.RESP correctNormal correctReverse
## 1 grid35grid51grid12grid43grid54 grid35grid51grid12grid43grid54           1.0            0.2
## 2 grid22grid53grid35grid21grid44 grid23grid53grid35grid21grid43           0.6            0.2
## 3 grid12grid14grid15grid41grid23 grid12grid24grid31grid41grid25           0.4            0.0
## 4 grid15grid41grid33grid24grid55 grid15grid41grid33grid14grid55           0.8            0.2