Question

我有一个包含200多个变量的数据框（下面是一个示例）：

  | x | P      | Var1_mean | Var2_mean | Var3_mean | Var1_sd | Var2_sd | Var3_sd
------------------------------------------------------------------------------
1 | A | P1     | 100       | 50.47     | 298.2     | 2.33    | 0.04    | 8.77
2 | A | P2     | 98        | 18        | 350.33    | 2.32    | 0.04    | 10.3
3 | B | P1     | 100       | 30.93     | 152.73    | 2.33    | 0.04    | 4.49
4 | B | P2     | 100       | 25.33     | 237.67    | 2.33    | 0.04    | 6.99
5 | C | P1     | 99.9      | 25.07     | 184.93    | 2.32    | 0.04    | 5.44
6 | C | P2     | 100       | 18.33     | 132.33    | 2.32    | 0.04    | 3.89

每个变量对参考周期P1和测量周期P2都有N个观测值（A，B，C等）。

我希望每个观察结果为每个变量计算2个时期之间的差异，并将其除以参考期的stdev。

使用上面的例子：

df <- data.frame(x=c("A","A","B","B","C","C"),
                 P=c("P1","P2","P1","P2","P1","P2"),        
             Var1_mean=c(100.0,98,100.0,100.0,99.9,100.0),
             Var2_mean = c(50.47,18,30.93,25.33,25.07,18.33),
             Var3_mean = c(298.2,350.33,152.73,237.67,184.93,132.33),
             Var1_sd = c(2.33,2.32,2.33,2.33,2.32,2.32),
             Var2_sd = c(0.04,0.04,0.04,0.04,0.04,0.04),
             Var3_sd = c(8.77,10.3,4.49,6.99,5.44,3.89))

Z.A.Var1 <- (df$Var1_mean[df$x=="A" & df$P=="P1"] - df$Var1_mean[df$x=="A" & df$P=="P2"])
            / df$Var1_sd[df$x=="A" & df$P=="P1"]

Z.A.Var2 <- (df$Var2_mean[df$x=="A" & df$P=="P1"] - df$Var2_mean[df$x=="A" & df$P=="P2"])
            / df$Var2_sd[df$x=="A" & df$P=="P1"]

等等。

我可以使用＆＃34; for＆＃34;计算循环，扫描观察和变量，但运行起来很麻烦且很慢。

是否有人建议如何以更智能的方式执行此操作，例如使用类似的dplyr操作系统？

Answer 1

尝试使用parseInt

data.table

以上假设您的200多个变量位于library(data.table) dat <- data.table(df) dat[, .(calc_V1 = ((Var1_mean[P == "P1"] - Var1_mean[P == "P2"])/Var1_sd[P == "P1"]), calc_V2 = ((Var2_mean[P == "P1"] - Var2_mean[P == "P2"])/Var2_sd[P == "P1"]), calc_V3 = ((Var3_mean[P == "P1"] - Var3_mean[P == "P2"])/Var3_sd[P == "P1"])), by = .(x)] x calc_V1 calc_V2 calc_V3 1: A 0.85836910 811.75 -5.944128 2: B 0.00000000 140.00 -18.917595 3: C -0.04310345 168.50 9.669118列中。相反，如果您的真实数据集非常宽（200多个变量中的每一个都有两列），您需要采用不同的方法：

Answer 2

我总是尽量避免 for 循环。我会在两个表中发送P1和P2变量，在另一个表中发送SD。然后使用矩阵运算。像这样：

desired.stuff <- (P1.stuff - P2.stuff) / sd.stuff

此解决方案适用于任意数量的变量，只要P1和P2具有相同数量的变量，您就可以了。

这是使用data.table语法的解决方案：

library(data.table)
df <- data.table(df) #make a data.table
P1.stuff <- df[P=="P1", c(1:3), with=F] #select the P1 variables
P2.stuff <- df[P=="P2", c(3:5), with=F] #select the P2 variables
sd.stuff <- df[P=="P1", c(6:8), with=F)] #select de P1 SDs
desired.stuff <- P1.stuff - P2.stuff / sd.stuff

此外，由于您似乎正在进行Cohen的D计算，您可能需要查看effsize包： https://cran.r-project.org/web/packages/effsize/effsize.pdf

Answer 3

可以使用tidyverse个包来重塑和聚合数据：

library(dplyr)
library(tidyr)

df.new <- gather(df, variable, value, -x, -P) %>% # convert data to 'long' format
    separate(variable, c('variable', 'measure')) %>% # separate out variable number and measurement (mean and sd) columns
    spread(measure, value) %>% # make each row have a mean and sd column
    group_by(variable, x) %>% # group by variable and observation and...
    summarize(result = diff(mean) / sd[P == 'P1']) # compute the requested summary, storing in 'result'

  variable      x        result
     <chr> <fctr>         <dbl>
1     Var1      A   -0.85836910
2     Var1      B    0.00000000
3     Var1      C    0.04310345
4     Var2      A -811.75000000
5     Var2      B -140.00000000
6     Var2      C -168.50000000
7     Var3      A    5.94412771
8     Var3      B   18.91759465
9     Var3      C   -9.66911765

使用dplyr或apply对多个变量的多个观察应用计算

3 个答案: