Suppose i have the following Db:
db<-data.frame(para=c(round(rnorm(20,10,10),0)),sal1=c(rnorm(20,100,7)),sal2=c(rnorm(20,100,7)),sal3=c(rnorm(10,100,7)),sal4=rep(c("a","b"),5))
para sal1 sal2 sal3 sal4
1 13 104.73988 96.53538 107.03285 a
2 1 94.54826 88.79930 101.17248 b
3 26 102.36344 94.83702 91.11708 a
4 13 99.32913 95.90670 90.49470 b
Basically what i want:
paste(db$sal1,db$sal2,db$sal3, sep="-")
para sal1 sal2 sal3 sal4 newcol
1 8 105.11 101.38 100.01 a 105.11-101.38-100.01
2 2 109.55 88.98 104.12 b 109.55-88.98-104.12
3 25 100.12 103.84 102.43 a 100.12-103.84-102.43
4 15 105.22 90.95 100.67 b 105.22-90.95-100.67
5 21 97.57 97.78 103.89 a 97.57-97.78-103.89
6 -1 101.88 100.22 88.21 b 101.88-100.22-88.21
7 12 104.20 95.26 93.72 a 104.2-95.26-93.72
8 16 106.25 100.70 94.95 b 106.25-100.7-94.95
9 24 101.36 97.91 99.67 a 101.36-97.91-99.67
Actually the way i want:
colnam<-c("sal1","sal2","sal3")
colnameful<-paste0("db$",colnam)
paste using variable colnameful
Something like below i tried:
paste(colnameful, sep="-") # Useless
Please suggest, Thanks in advance.
Maybe something like:
colnam<-c("sal1","sal2","sal3")
db[[paste(colnam,collapse = "-")]] <- with(db,paste(sal1,sal2,sal3,sep = "-"))
...though you may find that you want to round the sal1, etc values before pasting them together.
If you only have the columns in a character vector you could try this instead of the with() piece:
do.call(paste,c(db[,colnam],sep = "-"))
EDIT:
Thanks to #joran and #Gregor (from the comments) who helped me understand that it is not good practice to use eval(parse(text = '')) . Also the function which I wrote was unnecessarily over complicated. So just found out a simple solution with only one loop (apply)
apply(db[colnam], 1, paste, collapse = "-")
Original Answer :
Based on the output showed , it seems you want to paste sal1 , sal2 and sal3 together but using colnam
colnam<-c("sal1","sal2","sal3")
You can use eval(parse( combination in sapply to get the values and then apply row-wise computation to paste the columns together
apply(sapply(paste0("db$",colnam), function(x) eval(parse(text = x))), 1,
function(x) paste(x, collapse = "-"))
Related
I have a list of column names that look like this...
colnames(dat)
1 subject
2 e.type
3 group
4 boxnum
5 edate
6 file.name
7 fr
8 active
9 inactive
10 reward
11 latency.to.first.active
12 latency.to.first.inactive
13 act0.600
14 act600.1200
15 act1200.1800
16 act1800.2400
17 act2400.3000
18 act3000.3600
19 inact0.600
20 inact600.1200
21 inact1200.1800
22 inact1800.2400
23 inact2400.3000
24 inact3000.3600
25 rew0.600
26 rew600.1200
27 rew1200.1800
28 rew1800.2400
29 rew2400.3000
30 rew3000.3600
I want to get the row sum for the columns that list act#, inact#, and reward#
This works...
for (row in 1:nrow(dat)) {
dat[row, "active"] = rowSums(dat[row,c(13:18)])
dat[row, "inactive"] = rowSums(dat[row,c(19:24)])
dat[row, "reward"] = rowSums(dat[row,c(25:30)])
}
But I don't want to hard coded it since the number of columns for the 3 sections may change. How can I do this without hard coding the column indexes?
Also, for example, I tried searching for the "act" named columns but it was also including the "active" column.
sub_dat <- dat[, 13:30]
result <- sapply(split.default(sub_dat, substr(names(sub_dat), 1, 3)), rowSums)
dat[, c('active', 'inactive', 'reward')] <- result
Easy-cheesy with witch select & matches from the tidyverse.
library(tidyverse)
data %>%
mutate(
sum_act = rowSums(select(., matches("act[0-9]"))),
sum_inact = rowSums(select(., matches("inact[0-9]"))),
sum_rew = rowSums(select(., matches("rew[0-9]")))
)
I made an example how it could be done:
t <- data.frame(c(1,2,3),c("a","b","c"))
colnames(t) <- c("num","char")
#with function append() you make a list of rows that fulfill your logical argument
whichRows <- append(which(t$char == "a"),which(t$char == "b"))
sum(t$num[whichRows])
or if I misunderstood you and you want to sum for every column separately then:
sum(t$num[which(t$char == "a")])
sum(t$num[which(t$char == "b")])
This quesiton is similar to my previous question (How to create a "householdID" for rows with shared "customerID" and "spouseID"?), although this version deals with a rats-nest mix of character and numeric strings instead of simply numeric IDs. I'm trying to create a "household ID" for all couples who appear in a larger dataframe. In short, each individual has a "customerID" and "spouseID". If a customerID is married, their spouse's ID appears in the "spouseID" column. If they are not married, the spouseID field is empty. Each member of a married couple will appear on its own row, resulting in the need for a common "householdID" that a couple shares.
What is the best way to and add a unique householdID that duplicates for couples? A small and over-simplified example of the original data is as follows. Note that the original IDs are far more complex, with varying lengths and patters of numbers and characters.
df <- data.frame(
prospectID=as.character(c("G1339jf", "6dhd54G1", "Cf14c", "Bvmkm1", "kda-1qati", "pwn9enr", "wj44v04t4t", "D15", "dkfs044nng", "v949s")),
spouseID=as.character( c( "", "wj44v04t4t", "", "pwn9enr", "", "Bvmkm1", "6dhd54G1", "", "v949s", "dkfs044nng")),
stringsAsFactors = FALSE)
> df
prospectID spouseID
1 G1339jf
2 6dhd54G1 wj44v04t4t
3 Cf14c
4 Bvmkm1 pwn9enr
5 kda-1qati
6 pwn9enr Bvmkm1
7 wj44v04t4t 6dhd54G1
8 D15
9 dkfs044nng v949s
10 v949s dkfs044nng
An example of my desired result is as follows:
> df
prospectID spouseID HouseholdID
1 G1339jf 1
2 6dhd54G1 wj44v04t4t 2
3 Cf14c 3
4 Bvmkm1 pwn9enr 4
5 kda-1qati 5
6 pwn9enr Bvmkm1 4
7 wj44v04t4t 6dhd54G1 2
8 D15 6
9 dkfs044nng v949s 7
10 v949s dkfs044nng 7
This is an edited solution due to comments made by OP.
Illustrative data:
df <- data.frame(
prospectID=as.character(c("A1jljljljl344asbvc", "A2&%$ll##fffh", "B1665453sskn:;", "B2gavQWEΩΩø⁄", "C1", "D1", "E1#+'&%", "E255646321", "F1", "G1")),
spouseID=as.character(c("A2&%$ll##fffh", "A1jljljljl344asöbvc", "B2gavQWEΩΩø⁄", "B1665453sskn:;_", "", "", "E255646321", "E1#+'&%", "", "")),
stringsAsFactors = FALSE)
First define a pattern to match:
patt <- paste(df$prospectID, df$spouseID, sep = "|")
Second, define a for loop; here, a little editing is necessary for the first and the last value. Maybe others can improve on this part:
for(i in 1:nrow(df)){
df$HousholdID[1] <- 1
df$HousholdID[i] <- ifelse(grepl(patt[i], df$prospectID[i+1]), 1, 0)
df$HousholdID[10] <- 1
}
The final step is to run cumsum:
df$HousholdID <- cumsum(df$HousholdID)
The result:
df
prospectID spouseID HousholdID
1 A1jljljljl344asbvc A2&%$ll##fffh 1
2 A2&%$ll##fffh A1jljljljl344asöbvc 1
3 B1665453sskn:; B2gavQWEΩΩø⁄ 2
4 B2gavQWEΩΩø⁄ B1665453sskn:;_ 2
5 C1 3
6 D1 4
7 E1#+'&% E255646321 5
8 E255646321 E1#+'&% 5
9 F1 6
10 G1 7
I have an R problem if you can help.
x <- data.frame("LocationCode" = c("ESC3","RIECAA6","SJHMAU","RIE104","SJH11","SJHAE","RIEAE1","WGH54","RIE205","GSBROB"), "HospitalNumber" = c("701190923R","2905451068","700547389X","AN11295201","1204541612","104010665","800565884R","620063158W","600029720K","1112391223"),"DisciplineName" = c("ESC Biochemistry", "RIE Haematology","SJH Biochemistry","RIE Biochemistry","SJH Biochemistry","WGH Biochemistry","ESC Biochemistry","WGH Biochemistry","SJH Biochemistry","RIE Haematology"))
From the dataframe above i do wish to add a new column (CRN) made up of all "HospitalNumber" rows with 9 digits plus 1 letter at the end (e.g 701190923R), create another column (TIT) with the rest of the rows which does not meet the 1st criteria
You can do this in base using the code
# Identify cases which match 9 digits then one letter
CRMMatch <- grepl("^\\d{9}[[:alpha:]]$", as.character(x$HospitalNumber))
#Create columns from Hospital number among the matches or those that do not match
x$CRN[CRMMatch] <- as.character(x$HospitalNumber)[CRMMatch]
x$TIT[!CRMMatch] <- as.character(x$HospitalNumber)[!CRMMatch]
# clean up by removing the variable created of matches
rm(CRMMatch)
A dplyr version could be
library(dplyr)
x <-
x %>%
mutate(CRN = if_else(grepl("^\\d{9}[[:alpha:]]$", as.character(HospitalNumber)),as.character(HospitalNumber), NA_character_),
TIT = if_else(!grepl("^\\d{9}[[:alpha:]]$", as.character(HospitalNumber)),as.character(HospitalNumber), NA_character_))
You can detect what you need with the instruction
library(stringr)
str_which(x$HospitalNumber,"[:digit:][:alpha:]")
and you get:
> str_which(x$HospitalNumber,"[:digit:][:alpha:]")
[1] 1 3 7 8 9
Then you know what positions you need and what you don't
Quite similar to Kerry Jackson's approach but using ifelse in base R. I have also converted your x$HospitalNumber from factor to character from the start, assuming that this is what you really want:
x[2] <- as.character( x[ , 2 ] )
x$CRN <- ifelse( grepl( "^\\d{9}[[:alpha:]]$", x$HospitalNumber) , x$HospitalNumber, "" )
x$TIT <- ifelse( x$CRN != "", "", x$HospitalNumber )
gives you
> x
LocationCode HospitalNumber DisciplineName CRN TIT
1 ESC3 701190923R ESC Biochemistry 701190923R
2 RIECAA6 2905451068 RIE Haematology 2905451068
3 SJHMAU 700547389X SJH Biochemistry 700547389X
4 RIE104 AN11295201 RIE Biochemistry AN11295201
5 SJH11 1204541612 SJH Biochemistry 1204541612
6 SJHAE 104010665 WGH Biochemistry 104010665
7 RIEAE1 800565884R ESC Biochemistry 800565884R
8 WGH54 620063158W WGH Biochemistry 620063158W
9 RIE205 600029720K SJH Biochemistry 600029720K
10 GSBROB 1112391223 RIE Haematology 1112391223
Supose I have this variables:
data <- data.frame(x=rnorm(10), y=rnorm(10))
form <- 'z = x*y'
How can I compute z (using data's variables) and add as a new variable to data?
I tried with parse() and eval() (base on an old question), but without success :/
Given what #Nico said is correct you might do:
d1 <- within(data, eval(parse(text=form)) )
d1
x y z
1 0.5939462 1.58683345 0.94249368
2 0.3329504 0.55848643 0.18594826
3 1.0630998 -1.27659221 -1.35714497
4 -0.3041839 -0.57326541 0.17437812
5 0.3700188 -1.22461261 -0.45312970
6 0.2670988 -0.47340064 -0.12644474
7 -0.5425200 -0.62036668 0.33656135
8 1.2078678 0.04211587 0.05087041
9 1.1604026 -0.91092165 -1.05703586
10 0.7002136 0.15802877 0.11065390
transform() is the easy way if using this interactively:
data <- data.frame(x=rnorm(10), y=rnorm(10))
data <- transform(data, z = x * y)
R> head(data)
x y z
1 -1.0206 0.29982 -0.30599
2 -1.6985 1.51784 -2.57805
3 0.8940 1.19893 1.07187
4 -0.3672 -0.04008 0.01472
5 0.5266 -0.29205 -0.15381
6 0.2545 -0.26889 -0.06842
You can't do this using form though, but within(), which is similar to transform(), does allow this, e.g.
R> within(data, eval(parse(text = form)))
x y z
1 -0.8833 -0.05256 0.046428
2 1.6673 1.61101 2.686115
3 1.1261 0.16025 0.180453
4 0.9726 -1.32975 -1.293266
5 -1.6220 -0.51079 0.828473
6 -1.1981 2.62663 -3.147073
7 -0.3596 -0.01506 0.005416
8 -0.9700 0.21865 -0.212079
9 1.0626 1.30377 1.385399
10 -0.8020 -1.04639 0.839212
though it involves some amount of jiggery-pokery with the language which to my mind is not elegant. Effectively, you are doing something like this:
R> eval(eval(parse(text = form), data), data, parent.frame())
[1] 0.046428 2.686115 0.180453 -1.293266 0.828473 -3.147073 0.005416
[8] -0.212079 1.385399 0.839212
(and assigning the result to the named component in data.)
Does form have to come like this, as a character string representing some expression to be evaluated?
I suspect I'm Doing It Wrong, but I'd like to pass a character vector as an argument to a function in ddply. There's a lot of Q&A on removing quotes, etc. but none of it seems to work for me (eg. Remove quotes from a character vector in R and http://r.789695.n4.nabble.com/Pass-character-vector-to-function-argument-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
$df3
d am bm cm
1 h 36.00000 25.00000 20.50000
2 i 25.33333 37.33333 24.33333
3 j 32.00000 32.00000 46.00000
But my actual use-case has many new columns and different types of calculations that I want to calculate in the ddply function. So I want to do something like:
# here's a simple version of a function that I want to send to ddply
func <- "am=mean(x$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)
I've tried noquote, deparse, eval(as.symbol()), do.call(data.frame, ...) and some of the methods here: https://github.com/hadley/devtools/wiki/Evaluation on func to no avail. The solution might be obvious at this point (ie. melt everything!), but in case it's not, here's a longer example that's closer to my use case:
# 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)))
I want to replace RSEa=RSE(a), RSEb=RSE(b), RSEc=RSE(c), meana=mean(a), meanb=mean(b), meanc=mean(c) in the last line above with the object doRSE from here, to avoid lots of typing:
# 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="")
I'm open to solutions involving base, data.table and dirty tricks. Seems like these are close to what I want, but I can't quite translate them to my problem:
Pass character argument and evaluate,
Force evaluation of multiple variables using vector of character,
Using a vector of characters that correspond to an expression as an argument to a function
UPDATE Here's the catch: I want to be able to modify the func in the simple example (or doRSE in my use-case) to create a bunch of new columns that result from various calculations on the existing columns to explore the data. I want a workflow that allows the resulting dataframes to have new columns that were not in the original dataframes. Sorry that wasn't more clear in the original question. I can't see how to adapt #Marius' answer to do this, but #mnel's is helpful (see update below)
Working through #mnel's excellent dirty tricks, with some minor fixes I can get the desired result on my use-case:
# #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
...
...
You can do some ugly computing on the language using quote and plyr::.
Reading https://github.com/hadley/devtools/wiki/Computing-on-the-language will probably help understand whether you really want to do this.
Anyway, an approach could be to use
use .() to create your vector of arguments eg and use how summarize works
.(am=mean(a), bm=mean(b), cm=mean(c))
and if you really wanted to use a character string
foo<- "am=mean(a), bm=mean(b), cm=mean(c)"
eval(parse(text = sprintf('.(%s)', foo )))
Use quote liberally to create your list to be passed to to do.call
for example
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)))))
Oh boy is that ugly.
Or, you could use data.tables
library(data.table)
listDT <- lapply(list.1, data.table)
lapply(listDT, function(x) x[,lapply(.SD, mean), by = 'd'])
or
mystuff <- sprintf('list(%s)', foo)
lapply(listDT, function(x) x[, eval(parse(text = mystuff)), by = 'd'])
However, if you had all the same columns in all your data.tables, it would be more efficient to create one large data.table (with an identifer for each element of the list) and work on that.
Here's a ddply function that calculates the mean for all the columns that aren't d in your 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)
})
})
That seems to me like the simplest way to do it, and it should generalize well to other calculations you might want to do on those columns. Maybe you could pass in a character vector argument of the columns you want to calculate the mean for, and use that in place of non_d_cols.