传递一个字符向量作为参数传递给在plyr函数(Passing a character vector

我怀疑我做错了，但我想传递一个特征向量作为参数传递给一个函数ddply 。有一个关于消除报价等很多Q＆A，但没有它似乎为我工作（例如从字符向量R中删除引号和http://r.789695.n4.nabble.com/Pass-character-矢量到功能参数的-td3045226.html ）。

# reproducible data
df1<-data.frame(a=sample(1:50,10),b=sample(1:50,10),c=sample(1:50,10),d=(c("a","b","c","a","a","b","b","a","c","d")))
df2<-data.frame(a=sample(1:50,9),b=sample(1:50,9),c=sample(1:50,9),d=(c("e","f","g","e","e","f","f","e","g")))
df3<-data.frame(a=sample(1:50,8),b=sample(1:50,8),c=sample(1:50,8),d=(c("h","i","j","h","h","i","i","h")))

#make a list
list.1<-list(df1=df1,df2=df2,df3=df3)

# desired output
lapply(list.1, function(x)   ddply(x, .(d), function(x)  data.frame(am=mean(x$a), bm=mean(x$b), cm=mean(x$c))))

$df1
  d       am       bm       cm
1 a 31.00000 29.25000 18.50000
2 b 31.66667 24.33333 34.66667
3 c 18.50000  5.50000 24.50000
4 d 36.00000 39.00000 43.00000

$df2
  d       am       bm cm
1 e 18.25000 32.50000 18
2 f 27.66667 41.33333 24
3 g 25.00000  7.50000 42

$df3
  d       am       bm       cm
1 h 36.00000 25.00000 20.50000
2 i 25.33333 37.33333 24.33333
3 j 32.00000 32.00000 46.00000

但我的实际使用情况有很多新的栏目和不同类型的计算，我想在计算ddply功能。所以，我想要做的事，如：

# here's a simple version of a function that I want to send to ddply    
func <- "am=mean(x$a), bm=mean(x$b), cm=mean(x$c)"

# here's how I imagine it might work
lapply(list.1, function(x)   ddply(x, .(d), function(x)  data.frame(func)) )

# not the desired outcome... 
$df1
  d                                     func
1 a am=mean(x$a), bm=mean(x$b), cm=mean(x$c)
2 b am=mean(x$a), bm=mean(x$b), cm=mean(x$c)
3 c am=mean(x$a), bm=mean(x$b), cm=mean(x$c)
4 d am=mean(x$a), bm=mean(x$b), cm=mean(x$c)

$df2
  d                                     func
1 e am=mean(x$a), bm=mean(x$b), cm=mean(x$c)
2 f am=mean(x$a), bm=mean(x$b), cm=mean(x$c)
3 g am=mean(x$a), bm=mean(x$b), cm=mean(x$c)

$df3
  d                                     func
1 h am=mean(x$a), bm=mean(x$b), cm=mean(x$c)
2 i am=mean(x$a), bm=mean(x$b), cm=mean(x$c)
3 j am=mean(x$a), bm=mean(x$b), cm=mean(x$c)

我试过noquote ， deparse ， eval(as.symbol()) do.call(data.frame, ...)和一些方法在这里： https://github.com/hadley/devtools/wiki/评价上func无济于事。该解决方案可能会在这一点是显而易见的，但如果它不是，这里有一个较长的例子，更接近于我的使用情况下（即融化一切。！）：

# sample data
s <- 23 # number of samples
r <- 10 # number of runs per sample
el <- 17 # number of elements
mydata <- data.frame(ID = unlist(lapply(LETTERS[1:s], function(x) rep(x, r))),
                     run = rep(1:r, s))
# insert fake element data
mydata[letters[1:el]] <- lapply(1:el, function(i) rnorm(s*r, runif(1)*i^2))

# generate all combinations of 5 runs from  ten runs
su <- 5 # number of runs to sample from ten runs
idx <- combn(unique(mydata$run), su)

# RSE function
RSE <- function(x) {100*( (sd(x)/sqrt(length(x)))/mean(x) )}

# make a list of dfs for all samples for each combination of five runs
# to prepare to calculate RSEs
combys1 <- lapply(1:ncol(idx), function(i) mydata[mydata$run %in% idx[,i],] )

# make a list of dfs with RSE for each ID, for each combination of runs
combys2 <- lapply(1:length(combys1), function(i) ddply(combys1[[i]], "ID", summarise, RSEa=RSE(a), RSEb=RSE(b), RSEc=RSE(c), meana=mean(a), meanb=mean(b), meanc=mean(c)))

我要替换RSEa=RSE(a), RSEb=RSE(b), RSEc=RSE(c), meana=mean(a), meanb=mean(b), meanc=mean(c)的最后一行以上与对象doRSE从这里，以避免大量的打字：

# prepare to calculate new colums with RSE and means
RSEs <- sapply(3:ncol(mydata), function(j) paste0("RSE",names(mydata[j]))) 
RSExs <- sapply(3:ncol(mydata), function(j) paste0("RSE(",names(mydata[j]),")")) 
doRSE <- paste0(sapply(1:length(RSEs), function(x) paste0(RSEs[x],"=",RSExs[x])), collapse=",", sep="")

我接受涉及基地，解决方案data.table和肮脏的把戏。好像这些都是接近我想要的，但我不能完全把它们翻译成我的问题：通特性参数和评价，使用字符的矢量多个变量力评价，使用对应于一个表达式作为字符的矢量函数参数

更新这里的渔获：我希望能够修改func在简单的例子（或doRSE在我的用例）创建了一堆，从现有列各种计算结果来探索数据的新列。我想要一个工作流程使得所形成的dataframes到有没有在原来的dataframes新列。对不起，这不是在原来的问题更加清晰。我看不出如何适应@Marius的回答要做到这一点，但是@ MNEL的是有帮助的（见下面的更新）

通过@ MNEL的优秀搞鬼工作，有一些小的修正，我可以得到我的用例所需的结果：

# @mnel's solution, adapted (no period before eval)
combys2 <- lapply(combys1, function(x) do.call(ddply,c(.data = quote(x), 
                           .variables = quote(.(ID)), .fun = quote(summarize),
                           eval(parse(text = sprintf('.(%s)', doRSE ))))))
head(combys2)

[[1]]
   ID       RSEa      RSEb     RSEc      RSEd     RSEe      RSEf     RSEg      RSEh      RSEi
1   A  168.30658  21.68632 5.657228  5.048057 4.162017 2.9581874 1.849009 0.6925148 0.4393491
2   B   26.55071  26.20427 4.782578  4.385409 2.342764 2.1813874 2.719625 1.1576681 0.6427935
3   C   73.83165  14.47216 8.154435  6.273202 3.046978 1.2179457 2.811405 1.1401837 0.8167067
4   D   31.96170  57.89260 9.438220  7.388410 3.755772 0.8601780 3.724875 0.8358204 0.9939387
5   E   63.22537  60.35532 5.839690 11.691304 3.828430 0.9217787 4.204300 0.8217187 0.7876634
6   F   56.37635  65.37907 4.149568  5.496308 2.227544 2.1548455 2.847291 1.1956212 0.2506518
7   G   69.32232  23.63214 4.255847  7.979225 4.917660 1.6185960 3.156521 0.3265555 0.8133279
8   H   29.82015  40.74184 7.372100  7.464792 2.749862 0.6054420 4.061368 0.9973909 1.3807720
9   I   50.58114  19.53732 2.989920  9.767678 4.000249 1.7451322 1.175397 0.9952093 0.9095086
10  J   92.96462  39.77475 6.140688 10.295668 3.407726 2.4663758 3.030444 0.5743419 0.9296482
11  K   90.72381  42.25092 2.483069  6.781054 3.142082 1.8080633 2.891740 1.1996176 0.8525290
12  L -385.24547  40.81267 4.506087  8.148382 2.976488 0.8304432 2.234134 0.2108664 0.4979777
13  M   22.77743  33.98332 2.913926  8.764639 2.307293 0.8366635 3.229944 1.0003125 0.3878567
14  N   66.75163  34.16087 6.611326 13.865377 1.285522 1.3863958 4.165575 0.7379386 0.4515194
15  O   37.37188 100.57479 5.738877  5.724862 2.839638 1.1366610 3.186332 0.7383855 0.3954544
16  P   17.08913  26.62210 6.060130  4.110893 2.688908 2.6970727 1.609043 1.3860834 0.8780010
17  Q   13.96392  74.92279 5.469304  8.467638 2.974131 1.2135436 3.284564 0.6232778 1.0759226
18  R   42.59899  30.75952 4.842832  8.764158 1.874020 1.5791048 3.427342 1.4479638 0.2964455
19  S   26.03307  15.56352 6.968717  7.783876 4.439733 2.0764179 4.683080 0.7459654 1.1268772
20  T   71.57945  33.81362 7.147049 11.201551 2.128315 2.2051611 2.419805 0.2688807 1.1559635
21  U   73.93002  11.77155 7.738910  7.207041 1.478491 1.4409844 4.042419 0.5883490 0.5585716
22  V   67.93166  39.54994 5.701551  8.636122 2.472963 1.6514199 2.627965 1.0359048 0.8747136
23  W   11.23057  12.51272 7.003448  7.424559 4.102693 0.6614847 2.246305 1.3422405 0.2665246
        RSEj      RSEk      RSEl      RSEm      RSEn      RSEo      RSEp      RSEq
1  0.6366733 0.3713819 2.1993487 0.3865293 0.5436581 0.9187585 0.4344699 0.8915868
2  0.3445095 0.2932025 1.8563179 0.5397595 1.0433388 0.3533622 0.1942316 0.1941072
3  0.2720344 0.5507595 2.0305726 0.4377259 0.8589854 0.5690906 0.1397337 0.4043247
4  0.6606667 0.6769112 3.4737352 0.5674656 1.2519256 0.8718298 0.1162969 0.8287504
5  0.4620774 0.5598069 1.9236112 0.7990046 0.9832732 0.6847352 0.4070675 0.9005185
6  0.7981610 0.4005493 0.9721068 0.2770989 1.7054674 0.3110139 0.4521183 0.8740444
7  0.3969116 0.4717575 4.1341106 0.7510628 0.9998299 0.5342292 0.4319642 1.1861705
8  0.2963956 0.2652221 0.4775827 0.2617120 0.8261874 0.5266087 0.1900943 0.2350553
9  0.2609359 0.5431035 2.6478440 0.1606919 0.7407281 0.6802262 0.1802069 0.7438792
10 0.4239787 0.8753544 3.4218030 0.5467869 0.7404017 0.5581173 0.3682014 0.6361436
11 0.4188502 0.8629862 4.4181479 0.1623873 0.8018811 0.5873609 0.3592134 0.5357984
12 0.5790265 0.5009210 3.7534287 0.1933726 0.5809601 0.5777868 0.3400925 0.4783890
13 0.3562582 0.2552756 2.1393219 0.1849345 0.5796194 0.6129469 0.3363311 0.4382125
14 0.7921502 0.6147990 2.9054634 0.5852325 1.4954072 0.9983203 0.2937837 0.7654504
15 0.5840424 0.2757707 1.5695675 0.3305385 0.8712636 0.5816490 0.1985457 0.7213289
16 0.3301280 0.3008273 2.9014987 0.4540833 0.5966479 0.9042004 0.1631630 0.7262141
17 0.5882511 0.2820978 3.0652666 0.4518936 1.3168151 0.4749311 0.2244693 0.6583083
18 0.4048816 0.3708787 3.2207478 0.2603412 1.3168318 0.3318745 0.3120436 0.6210711
19 0.4425123 0.3602076 3.7609863 0.5399527 0.8302572 0.3246904 0.1952143 0.2915325
20 0.5877835 0.6339015 1.6908570 0.3223056 0.5239339 0.6607198 0.2808094 0.3697380
21 0.4454056 0.7733354 4.3433420 0.4391075 0.5503594 0.5893406 0.2262403 0.2361512
22 0.9583940 0.6365843 3.0033951 0.6507968 0.8610046 0.6363198 0.2866719 0.5736855
23 0.4969730 0.3895182 2.0021608 0.3354475 1.4398250 0.7386870 0.2458906 0.3414804
...
...

Answer 1:

你可以使用语言中的一些丑陋的计算quote和plyr::.

阅读https://github.com/hadley/devtools/wiki/Computing-on-the-language可能会帮助你了解是否真的想这样做。

无论如何，这种方法可以是使用

使用.()来创建你的论点例如矢量和使用如何总结工作

 .(am=mean(a), bm=mean(b), cm=mean(c))

如果你真的想用一个字符串

 foo<- "am=mean(a), bm=mean(b), cm=mean(c)" eval(parse(text = sprintf('.(%s)', foo )))

使用quote宽松创建列表要传递给到do.call

例如

lapply(list.1, function(x) do.call(ddply,c(.data = quote(x), 
    .variables = quote(.(d)), .fun = quote(summarize),
      .(am=mean(a), bm=mean(b), cm=mean(c)))))

哦，孩子是丑陋的。

或者，你可以使用data.tables

library(data.table)


listDT <- lapply(list.1, data.table)


lapply(listDT, function(x) x[,lapply(.SD, mean), by = 'd'])

要么

mystuff <- sprintf('list(%s)', foo)
lapply(listDT, function(x) x[, eval(parse(text = mystuff)), by = 'd'])

但是，如果你在所有的data.tables所有相同的列，这将是更有效地创建一个大data.table（与列表中的每个元素的IDENTIFER），并在有效的。

Answer 2:

这里有一个ddply计算平均为未在所有列功能d在dataframes：

lapply(list.1,
       function(x) {
         ddply(
           x,
           .(d),
           function(df_part) {
             result_df <- data.frame(d=df_part$d[1])
             non_d_cols <- colnames(df_part)[! colnames(df_part) == "d"]
             for (col in non_d_cols) {
               col_mean <- mean(df_part[[col]])
               col_name <- paste0(col, "_mean")
               result_df[[col_name]] <- col_mean
             }
             return(result_df)
           })
       })

这似乎对我来说，做到这一点最简单的方法，它应该推广很好地你可能想在这些列做其他的计算。也许你可以通过在你需要计算平均值的列的特征向量参数，并使用该代替non_d_cols 。