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我创建了一个计算 dPrime 的函数。该函数将数据框作为其参数。这可以正常工作，但是必须将列称为“stimDiff”和“stimSame”，因为函数使用这些特定名称计算 dPrime。我想将此函数应用于具有多个主题的数据框，并且能够计算每个主题的 dPrime，结果是一个新的数据框，其中包含每个主题的 dPrime 分数。测试数据框如下所示：

stimDiff0 <- c(rep("diff", 20), rep("same", 5))
stimSame0 <- c(rep("diff", 10), rep("same", 15))

stimDiff1 <- c(rep("diff", 10), rep("same", 15))
stimSame1 <- c(rep("diff", 10), rep("same", 15))

stimDiff2 <- c(rep("diff", 19), rep("same", 6))
stimSame2 <- c(rep("diff", 11), rep("same", 14))

stimDiff3 <- c(rep("diff", 21), rep("same", 4))
stimSame3 <- c(rep("diff",  9), rep("same", 16))

stimDiff4 <- c(rep("diff", 18), rep("same", 7))
stimSame4 <- c(rep("diff", 12), rep("same", 13))

stimDiff5 <- c(rep("diff", 22), rep("same", 3))
stimSame5 <- c(rep("diff", 14), rep("same", 11))

stimDiff <- c(stimDiff0, stimDiff1, stimDiff2,
              stimDiff3, stimDiff4, stimDiff5)
stimSame <- c(stimSame0, stimSame1, stimSame2,
              stimSame3, stimSame4, stimSame5)
subject <- rep(0:5, each = 25)

x <- data.frame(subject = subject, stimDiff = stimDiff, stimSame = stimSame)

我正在尝试使用以下代码按主题数据框获取 dPrim：

tapply(c(x$stimDiff, x$stimSame), x$subject, data = x, FUN = dPrime)

我收到以下错误：

Error en tapply(list(x$stimDiff, x$stimSame), x$subject, data = x, FUN = dPrime) : 
arguments must have same length

我知道有些软件包可以计算 dPrime。我这样做是为了学习如何编写函数。我更愿意使用基础 R 找到解决方案。

这是函数 dPrime 的代码：

dPrime <- function(x) {

# Calculate number of same, diff and total responses
# for the stimuli that were actually different
stimDiffRdiff <- nrow(x[x$stimDiff == 'diff', ])
stimDiffRsame <- nrow(x[x$stimDiff == 'same', ])
stimDiffTotal <- length(x$stimDiff)

# Calculate number of same, diff and total responses
# for the stimuli that were actually the same
stimSameRdiff <- nrow(x[x$stimSame == 'diff', ])
stimSameRsame <- nrow(x[x$stimSame == 'same', ])
stimSameTotal <- length(x$stimSame)

# Hit rate = the number of correct responses 'diff'
# when the stimuli were actually diff, divided by 
# the total number of responses
hitRate <- stimDiffRdiff / stimDiffTotal

# Miss rate = the number of incorrect responses
# 'same' when the stimuli were actually diff
# divided by the total number of responses
missRate <- stimDiffRsame / stimDiffTotal

# False alarm = the number responses 'diff'
# when the stimuli were actually the same
# divided by the total number of responses
falseAlarm <- stimSameRdiff / stimSameTotal

# Correct rejection = the number of responses
# same when the stimuli were actually the same
# divided by the number of total responses
corrReject <- stimSameRsame / stimSameTotal

# Calculate z-critical values for hit rate
# and false alarm rate
zHitRate <- qnorm(hitRate)
zFalseAlarm <- qnorm(falseAlarm)

# Calculate d prime
dPrime <- zHitRate - zFalseAlarm

print(dPrime)
}
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3 回答 3

2

以@jvcasill 的原始功能和其他用户的响应为基础：

dPrime <- function (data, subj = 1, stimDiff = 2, stimSame = 3) {
    # dPrime() returns a vector of the length of the number of subjects
    #+ in data[, subj] that contains the sensitivity index "d'" for each.
    # `data`:     data frame
    # `subj`:     index of "subject" column in `data`; default is 1
    # `stimDiff`: index of "stimDiff" column in `data`; default is 2
    # `stimSame`: index of "stimSame" column in `data`; default is 3
    if (is.data.frame(data)) {
        # Divide `data` by subject with split(), as have done others who've
        #+ responded to this question
        data.by.subj   <- split(data, data[, subj])
        # Calculate number of subjects and create vector of same length
        #+ to return
        n.subj         <- length(data.by.subj)
        dPrime.by.subj <- vector(mode = "double", length = n.subj)
        # Loop through "data.by.subj" subject by subject and calculate d'
        for (subj in seq_len(n.subj)) {
            # For clarity, create temporary data set with data of
            #+ current "subj"
            data.tmp      <- data.by.subj[[subj]]
            stimDiffRdiff <- nrow(data.tmp[data.tmp[, stimDiff] == "diff", ])
            stimDiffRsame <- nrow(data.tmp[data.tmp[, stimDiff] == "same", ])
            stimDiffTotal <- length(data.tmp[, stimDiff])
            stimSameRdiff <- nrow(data.tmp[data.tmp[, stimSame] == "diff", ])
            stimSameRsame <- nrow(data.tmp[data.tmp[, stimSame] == "same", ])
            stimSameTotal <- length(data.tmp[, stimSame])
            hitRate       <- stimDiffRdiff / stimDiffTotal
            missRate      <- stimDiffRsame / stimDiffTotal
            falseAlarm    <- stimSameRdiff / stimSameTotal
            # The following appears unused in the original function
            # corrReject  <- stimSameRsame / stimSameTotal
            zHitRate      <- qnorm(hitRate)
            zFalseAlarm   <- qnorm(falseAlarm)
            dPrime        <- zHitRate - zFalseAlarm
            dPrime.by.subj[subj] <- dPrime
        }
        # For clarity, give each d' value in vector to be returned,
        #+ "dPrime.by.subj", name of corresponding subject
        names(dPrime.by.subj) <- names(data.by.subj)
        return(dPrime.by.subj)
    } else stop("'data' is not a data frame")
}

请注意，我不确定它返回的值（对于@jvcasill 提供的示例数据集）是否与使用@Splendour 方法获得的值相同，是否有意义。

于 2014-12-11T01:55:31.293 回答
1

尝试 data.table（使用长度函数而不是 dPrime）：

library(data.table)
xt = data.table(x)
xt[,list(len=length(c(stimSame,stimDiff))),by=subject]
   subject len
1:       0  50
2:       1  50
3:       2  50
4:       3  50
5:       4  50
6:       5  50

以 R 为底：

sapply(split(x, x$subject), dPrime)
[1] 1.094968
[1] 0
[1] 0.8572718
[1] 1.352917
[1] 0.6329951
[1] 1.024018
        0         1         2         3         4         5 
1.0949683 0.0000000 0.8572718 1.3529167 0.6329951 1.0240176

重复输出是因为 dPrime 函数中的“print(dPrime)”语句。您应该用 return(dPrime) 替换它。更好的是，由于 dPrime 也是一个函数，您应该将 'dPrime <- zHitRate - zFalseAlarm' 语句中的 dPrime 替换为其他名称，例如 'ret'：

ret = dPrime <- zHitRate - zFalseAlarm
return(ret)
于 2014-12-10T17:05:44.370 回答
1

这是基本 R 中的一个（有点不雅的）解决方案：

将数据框拆分为列表，每个主题一个：

by.subject <- split(x, x$subject)

计算每个块的 dPrime，返回一个命名的数字向量：

dPrime.values <- unlist(lapply(by.subject, dPrime), use.names=T)

构造一个新的数据框：

df <- data.frame(dPrime=dPrime.values)
df$subject <- as.numeric(rownames(df))
于 2014-12-10T21:45:06.557 回答