for 循环的函数写入 3 个单独的列 R 或 dplyr/reshape 解决方案？

Question

I'm a total beginner to for loops so I apologize if there's already a clear answer to this question but I wasn't able to find anything that I understood how to apply to this specific question. I also started to try a dplyr implementation at the end but couldn't figure that out either.

Here's my question: there's a function that derives 3 values from a vector. I'd like to write those 3 values to the same df as new columns. The function is timefit from the retimes library in R. If I run it on the whole df:

  a1 <-  timefit(data$RT)
  a1:
        mu: 480.3346 
     sigma: 77.8531 
       tau: 376.7426 

If I place the values into a df df <- data.frame(a1@par) :

      a1.par
mu    480.33462
sigma 77.85305
tau   376.74257

I'd like to run it separately for each subID based on another variable "location" (a factor with two levels). So that I end up with something like

subID location mu sigma tau
1      0        500 50   400
1      0        500 50   400
1      1        376 50   410
1      1        376 50   410
2      0        400 60   400
2      0        400 60   400
2      1        410 60   410  
2      1        410 60   410

I got started with

for (subID in data) {
  timefit(data$RT)
}

But I know that's not going to actually do what I need it to do. Values are extracted from the timefit model with @par into long format so I need to specify the function timefit to write to 3 separate column headers? Any suggestions?

Also, I thought about using ddply, but that last line is tripping me up, because the format is long but I need it to be wide. I've messed with reshape a bit, but I'm having trouble figuring it out

data <- data %>% 
  group_by(subID, location) %>%
  mutate(timefit_out = timefit(RT))

Thanks for your help!

Answer 1

You can use summarise instead of mutate here to generate a list-column containing a data.frame from each (subID, location) 's timefit . These data frames encode the mu , sigma , and tau from the result of timefit as columns. Then, use unnest to unnest this list-column to generate the result you want.

library(retimes)
library(dplyr)
library(tidyr)
result <- data %>% group_by(subID, location) %>%
                   summarise(timefit_out = list(data.frame(t(attr(timefit(RT),"par"))))) %>%
                   unnest()

Note that we extract the "par" attribute from the timefit class and then transpose it with t to form columns for mu , sigma , and tau .

Here, we assume that your input data is a data frame with columns subID , location , and the numeric column of reaction times RT that is input to timefit . A simulated example of such a dataset is given by:

data <- structure(list(subID = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L, 2L, 
2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L, 2L), 
location = c(0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 0L, 1L, 
1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 0L, 0L, 0L, 0L, 0L, 0L, 
0L, 0L, 0L, 0L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L), 
RT = c(0.341764254728332, 0.775535081513226, 0.281827432336286, 
0.23970171622932, 0.00226009078323841, 0.385179498931393, 
0.645917195128277, 0.812101020244882, 0.183301427634433, 
0.981765420176089, 0.656369511503726, 0.824469136772677, 
0.923240559641272, 0.598261737963185, 0.309975759591907, 
0.778991278028116, 0.757012664806098, 0.869985132943839, 
0.439378245733678, 0.8420404586941, 0.643788777757436, 0.381316626211628, 
0.123881611274555, 0.540528740268201, 0.661961955949664, 
0.0592848095111549, 0.904047027230263, 0.190083365887403, 
0.963809312786907, 0.0925120878964663, 0.117538752267137, 
0.451085010776296, 0.703220259631053, 0.378451474476606, 
0.305718191433698, 0.70383172808215, 0.699415655340999, 0.740436099236831, 
0.429179352009669, 0.205358384409919)), .Names = c("subID", 
"location", "RT"), row.names = c(NA, 40L), class = "data.frame")
##   subID location          RT
##1      1        0 0.341764255
##2      1        0 0.775535082
##3      1        0 0.281827432
##4      1        0 0.239701716
##5      1        0 0.002260091
##6      1        0 0.385179499
##7      1        0 0.645917195
##8      1        0 0.812101020
##9      1        0 0.183301428
##10     1        0 0.981765420
##11     1        1 0.656369512
##12     1        1 0.824469137
##13     1        1 0.923240560
##14     1        1 0.598261738
##15     1        1 0.309975760
##16     1        1 0.778991278
##17     1        1 0.757012665
##18     1        1 0.869985133
##19     1        1 0.439378246
##20     1        1 0.842040459
##21     2        0 0.643788778
##22     2        0 0.381316626
##23     2        0 0.123881611
##24     2        0 0.540528740
##25     2        0 0.661961956
##26     2        0 0.059284810
##27     2        0 0.904047027
##28     2        0 0.190083366
##29     2        0 0.963809313
##30     2        0 0.092512088
##31     2        1 0.117538752
##32     2        1 0.451085011
##33     2        1 0.703220260
##34     2        1 0.378451474
##35     2        1 0.305718191
##36     2        1 0.703831728
##37     2        1 0.699415655
##38     2        1 0.740436099
##39     2        1 0.429179352
##40     2        1 0.205358384

The values for RT in this example are generated using runif so they are between 0 and 1 . Your values are much different, but that should not matter here.

Using this data, we get:

print(result)
##Source: local data frame [4 x 5]
##Groups: subID [2]
##
##  subID location        mu     sigma         tau
##  <int>    <int>     <dbl>     <dbl>       <dbl>
##1     1        0 0.5275058 0.2553621 0.007086207
##2     1        1 0.2609386 0.1583494 0.085449559
##3     2        0 0.5205647 0.1994942 0.027329115
##4     2        1 0.4632886 0.2881343 0.008026460

Answer 2

What you are probably looking for, if you are looking for a dplyr solution is do . It allows returns of data.frames, though may require a bit of manipulation. Specifically, it is designed to work over groups, rather than (necessarily) rows. So, you will have to set groups if you want it to return with some of the original information (and depending on the structure of your function).

For this, I am generating a simple data set:

myData <-
  data.frame(
    RT = 1:4
  )

You will also need to construct a function that returns the values you want as a data.frame. For your use, you will probably calculate the result of timefit in the function, then extract each of the values as a column to return:

myFunc <- function(x){
  data.frame(a= x + 1, b = x + 2, c = x + 3)
}

Then, group by the columns you want to separate by (and return), and call do :

myData %>%
  group_by(RT) %>%
  do((myFunc(.$RT)))

Which, in this case, returns this:

     RT     a     b     c
1     1     2     3     4
2     2     3     4     5
3     3     4     5     6
4     4     5     6     7