Reading non-uniform data into R - r

I struggling with reading non-uniform data into R.
I've achieved the following:
Used "readLines" to read the text file data in
Used "grep" to find the block of data that I want
Used the index from grep to create a variable (named "block") that contains only that block of data
All good so far - I now have the data I want. But - its a character variable with only one column that contains all the data.
This creates a sample of the variable I have made called "block" (first 3 rows):
line1 = c(" 114.24 -0.39 0.06 13.85 -0.06 1402.11 -1.48 0.0003 0.0000 35.468 1.02 -0.02 0.00 0 1 1 1 0 49.87 4 -290 0 0 -0.002 -0.010 0.155 999.00 11482.66 999.00 11482.66 16:52:24:119 255 13.89 50.00 0.00 -5.49 0.00")
line2 = c(" 114.28 -0.39 0.08 13.84 -0.06 1402.57 -1.48 0.0004 0.0000 35.479 1.29 -0.02 0.00 0 1 1 1 0 49.82 4 -272 0 0 -0.002 -0.011 0.124 999.00 11482.66 999.00 11482.66 16:52:24:150 255 13.89 50.00 0.00 -5.49 0.00")
line3 = c(" 114.31 -0.39 0.09 13.83 -0.06 1403.03 -1.47 0.0005 0.0000 35.492 1.42 -0.02 0.00 0 1 1 1 0 49.78 4 -263 0 0 -0.002 -0.011 0.046 999.00 11482.66 999.00 11482.66 16:52:24:197 255 13.89 50.00 0.00 -5.49 0.00")
block = c(line1,line2,line3)
My goal is to have this data as a data.frame with separate columns for each data point.
My attempts at using strsplit haven't helped (does the solution involve strsplit?)- what is the best approach here? Any suggestions/feedback welcome.
strsplit(block,"\s",fixed=F)

Either of the following should work for you:
## Creates a "data.table"
library(splitstackshape)
cSplit(data.table(x = block), "x", " ")
## Creates a "data.frame"
read.table(text = block, header = FALSE)
## Creates a character matrix
do.call(rbind, strsplit(block, "\\s+"))
## Like the above, but likely to be faster
library(stringi)
stri_split_regex(block, "\\s+", simplify = TRUE)
Note the "\\s+" for the last two options. The "+" is to match multiple spaces.

Actually - this looks like it might work.
Import raw data into R
But wanted to check if this was the best approach to this situation...?

Related

Finding Sync of Columns in R

Data
v1 <- c(52.9799999999814, 53.4200000000128, 52.0899999999965, 57.9700000000012,
60.679999999993, 0.300000000017462, 1.76999999998952, 61.1900000000023,
58.9599999999919, 1.73000000001048, 0.269999999989523, 6.92000000001281,
60.5299999999988, 60.859999999986, 59.5599999999977, 61.0600000000268,
60.6299999999756, 60.9700000000012, 60.1600000000035, 60.4599999999919,
60.0900000000256)
v2 <- c(52.679999999993, 53.140000000014, 52.8899999999849, 57.6700000000128,
60.5199999999895, 2.04000000000815, 61.890000000014, 59.5699999999779,
2.05999999999767, 6.98000000001048, 60.7399999999907, 60.7799999999988,
59.7300000000105, 60.9100000000035, 60.3299999999872, 60.5500000000175,
60.6600000000035, 60.3499999999767, 60.7300000000105, 60.6700000000128,
60.3799999999756)
tv3 <- data.frame(v1,v2)
tv3$v5 <- tv3$v2 - tv3$v1
tv3$v5
[1] -0.30 -0.28 0.80 -0.30 -0.16 1.74 60.12 -1.62 -56.90 5.25 60.47 53.86 -0.80
[14] 0.05 0.77 -0.51 0.03 -0.62 0.57 0.21 0.29
So you see, the difference should remain smaller, if it is larger, like in this case, at particular row, it gets 60.
So basically if we remove the 0.30 row in just V1 and shift it one cell up, The difference wouldn't hike upto 60.
So the 0.30 is noise value and that's what I have to figure and put it in V3
My desired results are the following.
v1 v2 V3
52.98 52.68 0.3
53.42 53.14 0.27
52.09 52.89
57.97 57.67
60.68 60.52
1.77 2.04
61.19 61.89
58.96 59.57
1.73 2.06
6.92 6.98
60.53 60.74
60.86 60.78
59.56 59.73
61.06 60.91
60.63 60.33
60.97 60.55
60.16 60.66
60.46 60.35
60.09 60.73
So notice here that all the sequence of columns seem to be in sync with just a difference of few points.
May be my case requires implementation of Needleman-Wunsch Algo

Speed up `strsplit` when possible output are known

I have a large data frame with a factor column that I need to divide into three factor columns by splitting up the factor names by a delimiter. Here is my current approach, which is very slow with a large data frame (sometimes several million rows):
data <- readRDS("data.rds")
data.df <- reshape2:::melt.array(data)
head(data.df)
## Time Location Class Replicate Population
##1 1 1 LIDE.1.S 1 0.03859605
##2 2 1 LIDE.1.S 1 0.03852957
##3 3 1 LIDE.1.S 1 0.03846853
##4 4 1 LIDE.1.S 1 0.03841260
##5 5 1 LIDE.1.S 1 0.03836147
##6 6 1 LIDE.1.S 1 0.03831485
Rprof("str.out")
cl <- which(names(data.df)=="Class")
Classes <- do.call(rbind, strsplit(as.character(data.df$Class), "\\."))
colnames(Classes) <- c("Species", "SizeClass", "Infected")
data.df <- cbind(data.df[,1:(cl-1)],Classes,data.df[(cl+1):(ncol(data.df))])
Rprof(NULL)
head(data.df)
## Time Location Species SizeClass Infected Replicate Population
##1 1 1 LIDE 1 S 1 0.03859605
##2 2 1 LIDE 1 S 1 0.03852957
##3 3 1 LIDE 1 S 1 0.03846853
##4 4 1 LIDE 1 S 1 0.03841260
##5 5 1 LIDE 1 S 1 0.03836147
##6 6 1 LIDE 1 S 1 0.03831485
summaryRprof("str.out")
$by.self
self.time self.pct total.time total.pct
"strsplit" 1.34 50.00 1.34 50.00
"<Anonymous>" 1.16 43.28 1.16 43.28
"do.call" 0.04 1.49 2.54 94.78
"unique.default" 0.04 1.49 0.04 1.49
"data.frame" 0.02 0.75 0.12 4.48
"is.factor" 0.02 0.75 0.02 0.75
"match" 0.02 0.75 0.02 0.75
"structure" 0.02 0.75 0.02 0.75
"unlist" 0.02 0.75 0.02 0.75
$by.total
total.time total.pct self.time self.pct
"do.call" 2.54 94.78 0.04 1.49
"strsplit" 1.34 50.00 1.34 50.00
"<Anonymous>" 1.16 43.28 1.16 43.28
"cbind" 0.14 5.22 0.00 0.00
"data.frame" 0.12 4.48 0.02 0.75
"as.data.frame.matrix" 0.08 2.99 0.00 0.00
"as.data.frame" 0.08 2.99 0.00 0.00
"as.factor" 0.08 2.99 0.00 0.00
"factor" 0.06 2.24 0.00 0.00
"unique.default" 0.04 1.49 0.04 1.49
"unique" 0.04 1.49 0.00 0.00
"is.factor" 0.02 0.75 0.02 0.75
"match" 0.02 0.75 0.02 0.75
"structure" 0.02 0.75 0.02 0.75
"unlist" 0.02 0.75 0.02 0.75
"[.data.frame" 0.02 0.75 0.00 0.00
"[" 0.02 0.75 0.00 0.00
$sample.interval
[1] 0.02
$sampling.time
[1] 2.68
Is there any way to speed up this operation? I note that there are a small (<5) number of each of the categories "Species", "SizeClass", and "Infected", and I know what these are in advance.
Notes:
stringr::str_split_fixed performs this task, but not any faster
The data frame is actually initially generated by calling reshape::melt on an array in which Class and its associated levels are a dimension. If there's a faster way to get from there to here, great.
data.rds at http://dl.getdropbox.com/u/3356641/data.rds
This should probably offer quite an increase:
library(data.table)
DT <- data.table(data.df)
DT[, c("Species", "SizeClass", "Infected")
:= as.list(strsplit(Class, "\\.")[[1]]), by=Class ]
The reasons for the increase:
data.table pre allocates memory for columns
every column assignment in data.frame reassigns the entirety of the data (data.table in contrast does not)
the by statement allows you to implement the strsplit task once per each unique value.
Here is a nice quick method for the whole process.
# Save the new col names as a character vector
newCols <- c("Species", "SizeClass", "Infected")
# split the string, then convert the new cols to columns
DT[, c(newCols) := as.list(strsplit(as.character(Class), "\\.")[[1]]), by=Class ]
DT[, c(newCols) := lapply(.SD, factor), .SDcols=newCols]
# remove the old column. This is instantaneous.
DT[, Class := NULL]
## Have a look:
DT[, lapply(.SD, class)]
# Time Location Replicate Population Species SizeClass Infected
# 1: integer integer integer numeric factor factor factor
DT
You could get a decent increase in speed by just extracting the parts of the string you need using gsub instead of splitting everything up and trying to put it back together:
data <- readRDS("~/Downloads/data.rds")
data.df <- reshape2:::melt.array(data)
# using `strsplit`
system.time({
cl <- which(names(data.df)=="Class")
Classes <- do.call(rbind, strsplit(as.character(data.df$Class), "\\."))
colnames(Classes) <- c("Species", "SizeClass", "Infected")
data.df <- cbind(data.df[,1:(cl-1)],Classes,data.df[(cl+1):(ncol(data.df))])
})
user system elapsed
3.349 0.062 3.411
#using `gsub`
system.time({
data.df$Class <- as.character(data.df$Class)
data.df$SizeClass <- gsub("(\\w+)\\.(\\d+)\\.(\\w+)", "\\2", data.df$Class,
perl = TRUE)
data.df$Infected <- gsub("(\\w+)\\.(\\d+)\\.(\\w+)", "\\3", data.df$Class,
perl = TRUE)
data.df$Class <- gsub("(\\w+)\\.(\\d+)\\.(\\w+)", "\\1", data.df$Class,
perl = TRUE)
})
user system elapsed
0.812 0.037 0.848
Looks like you have a factor, so work on the levels and then map back. Use fixed=TRUE in strsplit, adjusting to split=".".
Classes <- do.call(rbind, strsplit(levels(data.df$Class), ".", fixed=TRUE))
colnames(Classes) <- c("Species", "SizeClass", "Infected")
df0 <- as.data.frame(Classes[data.df$Class,], row.names=NA)
cbind(data.df, df0)

Passing an expression to a nested grouping in data.table

I have a data.table object similar to this one
library(data.table)
c <- data.table(CO = c(10000,10000,10000,20000,20000,20000,20000),
SH = c(1427,1333,1333,1000,1000,300,350),
PRC = c(6.5,6.125,6.2,0.75,0.5,3,3.5),
DAT = c(0.5,-0.5,0,-0.1,NA_real_,0.2,0.5),
MM = c("A","A","A","A","A","B","B"))
and I am trying to perform calculations using nested grouping, passing an expression as an argument. Here is a simplified version of what I have:
setkey(c,MM)
mycalc <- quote({nobscc <- length(DAT[complete.cases(DAT)]);
list(MKTCAP = tail(SH,n=1)*tail(PRC,n=1),
SQSUM = ifelse(nobscc>=2, sum(DAT^2,na.rm=TRUE), NA_real_),
COVCOMP = ifelse(nobscc >= 2, head(DAT,n=1), NA_real_),
NOBS = nobscc)})
myresults <- c[,.SD[,{setkey=CO; eval(mycalc)},by=CO],by=MM]
which produces
MM CO MKTCAP SQSUM COVCOMP NOBS
[1,] A 10000 8264.6 0.50 0.5 3
[2,] A 20000 500.0 NA NA 1
[3,] B 20000 1225.0 0.29 0.2 2
In the example above I have two elements of the list which use the ifelse construct (in the actual code there are 3), all doing the same test : if the number of observations is greater than 2, then a certain calculation (which is different for each element of the list, and each could be written as a function) is to be performed, otherwise I want the value of the these elements to be NA. Another thing these elements have in common is that they use one and the same column of my data.table: the one called DAT.
So my question is: is there any way I can do the ifelse test only once, and if it is FALSE, pass the value NA to the respective elements of the list, and if TRUE, evaluate a different expression for each of the elements of the list?
NOTE: My goal is to reduce the system.time (system and elapsed). If this modification will not reduce time and calculations, bearing in mind I have 72 million observations, that's an acceptable answer. I also welcome suggestions to change other parts of the code.
EDIT: Results of summaryRprof()
$by.total
total.time total.pct self.time self.pct
"system.time" 18.94 99.79 0.00 0.00
".Call" 18.92 99.68 0.10 0.53
"[" 18.92 99.68 0.04 0.21
"[.data.table" 18.92 99.68 0.02 0.11
"eval" 18.80 99.05 0.24 1.26
"ifelse" 18.30 96.42 0.46 2.42
"lm" 17.70 93.26 0.58 3.06
"sapply" 8.06 42.47 0.36 1.90
"model.frame" 7.74 40.78 0.16 0.84
"model.frame.default" 7.58 39.94 0.98 5.16
"lapply" 6.62 34.88 0.70 3.69
"FUN" 4.24 22.34 1.10 5.80
"model.matrix" 4.04 21.29 0.02 0.11
"model.matrix.default" 4.02 21.18 0.26 1.37
"match" 3.66 19.28 0.86 4.53
".getXlevels" 3.12 16.44 0.12 0.63
"na.omit" 2.40 12.64 0.24 1.26
"%in%" 2.30 12.12 0.34 1.79
"simplify2array" 2.24 11.80 0.12 0.63
"na.omit.data.frame" 2.16 11.38 0.14 0.74
"[.data.frame" 2.12 11.17 1.18 6.22
"deparse" 1.80 9.48 0.66 3.48
"unique" 1.80 9.48 0.54 2.85
"[[" 1.52 8.01 0.12 0.63
"[[.data.frame" 1.40 7.38 0.54 2.85
".deparseOpts" 1.34 7.06 0.96 5.06
"paste" 1.32 6.95 0.16 0.84
"lm.fit" 1.20 6.32 0.64 3.37
"mode" 1.14 6.01 0.14 0.74
"unlist" 1.12 5.90 0.56 2.95
Instead of forming and operating on data subsets like this:
setkey(c,MM)
myresults <- c[, .SD[,{setkey=CO; eval(mycalc)},by=CO], by=MM]
You could try doing this:
setkeyv(c, c("MM", "CO"))
myresults <- c[, eval(mycalc), by=key(c)]
This should speed up your code, since it avoids all of the nested subsetting of .SD objects, each of which requires its own call to [.data.table.
On your original question, I doubt the ifelse evaluations are taking much time, but if you want to avoid them, you could take them out of mycalc and use := to overwrite the desired values with NA:
mycalc <- quote(list(MKTCAP = tail(SH,n=1)*tail(PRC,n=1),
SQSUM = sum(DAT^2,na.rm=TRUE),
COVCOMP = head(DAT,n=1),
NOBS = length(DAT[complete.cases(DAT)])))
setkeyv(c, c("MM", "CO"))
myresults <- c[, eval(mycalc), by=key(c)]
myresults[NOBS<2, c("SQSUM", "COVCOMP"):=NA_real_]
## Or, alternatively
# myresults[NOBS<2, SQSUM:=NA_real_]
# myresults[NOBS<2, COVCOMP:=NA_real_]

Extracting rows based on the value

I have tab delim text file which contains the following columns:
Probe A_sig A_Pval
ILMN_122 12.31 0.04
ILMN_456 56.12 0
ILMN_198 981.2 0.06
ILMN_980 876.0 0.001
ILMN_542 123.9 0.16
ILMN_567 134.1 0
ILMN_452 213.4 0.98
ILMN_142 543.8 0.04
ILMN_765 187.4 0.05
Now I want to take out those rows which has the Pval <.05. The output should look like
Probe A_sig A_Pval
ILMN_122 12.31 0.04
ILMN_980 876.0 0.001
ILMN_142 543.8 0.04
Can anyone please help me?
I'll answer this but it's a basic question that is probably repeated elsewhere on this list.
Load data.
DAT <- read.table(text="Probe A_sig A_Pval
ILMN_122 12.31 0.04
ILMN_456 56.12 0
ILMN_198 981.2 0.06
ILMN_980 876.0 0.001
ILMN_542 123.9 0.16
ILMN_567 134.1 0
ILMN_452 213.4 0.98
ILMN_142 543.8 0.04
ILMN_765 187.4 0.05", h=T)
You can use indexing as in:
DAT[DAT$A_Pval <.05, ]
However this returns the zero vales as well. That isn't what you're output looks like. If you don't want the zeros use logical operator & as well as in:
DAT[DAT$A_Pval <.05 & DAT$A_Pval!=0, ]
I suggest you take a look at some manuals and this (LINK) reference card to help get you started.
my_dataframe[my_dataframe$A_Pval < 0.05,]
The trailing comma is important.

How to efficiently use Rprof in R?

I would like to know if it is possible to get a profile from R-Code in a way that is similar to matlab's Profiler. That is, to get to know which line numbers are the one's that are especially slow.
What I acchieved so far is somehow not satisfactory. I used Rprof to make me a profile file. Using summaryRprof I get something like the following:
$by.self
self.time self.pct total.time total.pct
[.data.frame 0.72 10.1 1.84 25.8
inherits 0.50 7.0 1.10 15.4
data.frame 0.48 6.7 4.86 68.3
unique.default 0.44 6.2 0.48 6.7
deparse 0.36 5.1 1.18 16.6
rbind 0.30 4.2 2.22 31.2
match 0.28 3.9 1.38 19.4
[<-.factor 0.28 3.9 0.56 7.9
levels 0.26 3.7 0.34 4.8
NextMethod 0.22 3.1 0.82 11.5
...
and
$by.total
total.time total.pct self.time self.pct
data.frame 4.86 68.3 0.48 6.7
rbind 2.22 31.2 0.30 4.2
do.call 2.22 31.2 0.00 0.0
[ 1.98 27.8 0.16 2.2
[.data.frame 1.84 25.8 0.72 10.1
match 1.38 19.4 0.28 3.9
%in% 1.26 17.7 0.14 2.0
is.factor 1.20 16.9 0.10 1.4
deparse 1.18 16.6 0.36 5.1
...
To be honest, from this output I don't get where my bottlenecks are because (a) I use data.frame pretty often and (b) I never use e.g., deparse. Furthermore, what is [?
So I tried Hadley Wickham's profr, but it was not any more useful considering the following graph:
Is there a more convenient way to see which line numbers and particular function calls are slow?
Or, is there some literature that I should consult?
Any hints appreciated.
EDIT 1:
Based on Hadley's comment I will paste the code of my script below and the base graph version of the plot. But note, that my question is not related to this specific script. It is just a random script that I recently wrote. I am looking for a general way of how to find bottlenecks and speed up R-code.
The data (x) looks like this:
type word response N Classification classN
Abstract ANGER bitter 1 3a 3a
Abstract ANGER control 1 1a 1a
Abstract ANGER father 1 3a 3a
Abstract ANGER flushed 1 3a 3a
Abstract ANGER fury 1 1c 1c
Abstract ANGER hat 1 3a 3a
Abstract ANGER help 1 3a 3a
Abstract ANGER mad 13 3a 3a
Abstract ANGER management 2 1a 1a
... until row 1700
The script (with short explanations) is this:
Rprof("profile1.out")
# A new dataset is produced with each line of x contained x$N times
y <- vector('list',length(x[,1]))
for (i in 1:length(x[,1])) {
y[[i]] <- data.frame(rep(x[i,1],x[i,"N"]),rep(x[i,2],x[i,"N"]),rep(x[i,3],x[i,"N"]),rep(x[i,4],x[i,"N"]),rep(x[i,5],x[i,"N"]),rep(x[i,6],x[i,"N"]))
}
all <- do.call('rbind',y)
colnames(all) <- colnames(x)
# create a dataframe out of a word x class table
table_all <- table(all$word,all$classN)
dataf.all <- as.data.frame(table_all[,1:length(table_all[1,])])
dataf.all$words <- as.factor(rownames(dataf.all))
dataf.all$type <- "no"
# get type of the word.
words <- levels(dataf.all$words)
for (i in 1:length(words)) {
dataf.all$type[i] <- as.character(all[pmatch(words[i],all$word),"type"])
}
dataf.all$type <- as.factor(dataf.all$type)
dataf.all$typeN <- as.numeric(dataf.all$type)
# aggregate response categories
dataf.all$c1 <- apply(dataf.all[,c("1a","1b","1c","1d","1e","1f")],1,sum)
dataf.all$c2 <- apply(dataf.all[,c("2a","2b","2c")],1,sum)
dataf.all$c3 <- apply(dataf.all[,c("3a","3b")],1,sum)
Rprof(NULL)
library(profr)
ggplot.profr(parse_rprof("profile1.out"))
Final data looks like this:
1a 1b 1c 1d 1e 1f 2a 2b 2c 3a 3b pa words type typeN c1 c2 c3 pa
3 0 8 0 0 0 0 0 0 24 0 0 ANGER Abstract 1 11 0 24 0
6 0 4 0 1 0 0 11 0 13 0 0 ANXIETY Abstract 1 11 11 13 0
2 11 1 0 0 0 0 4 0 17 0 0 ATTITUDE Abstract 1 14 4 17 0
9 18 0 0 0 0 0 0 0 0 8 0 BARREL Concrete 2 27 0 8 0
0 1 18 0 0 0 0 4 0 12 0 0 BELIEF Abstract 1 19 4 12 0
The base graph plot:
Running the script today also changed the ggplot2 graph a little (basically only the labels), see here.
Alert readers of yesterdays breaking news (R 3.0.0 is finally out) may have noticed something interesting that is directly relevant to this question:
Profiling via Rprof() now optionally records information at the statement level, not just the function level.
And indeed, this new feature answers my question and I will show how.
Let's say, we want to compare whether vectorizing and pre-allocating are really better than good old for-loops and incremental building of data in calculating a summary statistic such as the mean. The, relatively stupid, code is the following:
# create big data frame:
n <- 1000
x <- data.frame(group = sample(letters[1:4], n, replace=TRUE), condition = sample(LETTERS[1:10], n, replace = TRUE), data = rnorm(n))
# reasonable operations:
marginal.means.1 <- aggregate(data ~ group + condition, data = x, FUN=mean)
# unreasonable operations:
marginal.means.2 <- marginal.means.1[NULL,]
row.counter <- 1
for (condition in levels(x$condition)) {
for (group in levels(x$group)) {
tmp.value <- 0
tmp.length <- 0
for (c in 1:nrow(x)) {
if ((x[c,"group"] == group) & (x[c,"condition"] == condition)) {
tmp.value <- tmp.value + x[c,"data"]
tmp.length <- tmp.length + 1
}
}
marginal.means.2[row.counter,"group"] <- group
marginal.means.2[row.counter,"condition"] <- condition
marginal.means.2[row.counter,"data"] <- tmp.value / tmp.length
row.counter <- row.counter + 1
}
}
# does it produce the same results?
all.equal(marginal.means.1, marginal.means.2)
To use this code with Rprof, we need to parse it. That is, it needs to be saved in a file and then called from there. Hence, I uploaded it to pastebin, but it works exactly the same with local files.
Now, we
simply create a profile file and indicate that we want to save the line number,
source the code with the incredible combination eval(parse(..., keep.source = TRUE)) (seemingly the infamous fortune(106) does not apply here, as I haven't found another way)
stop the profiling and indicate that we want the output based on the line numbers.
The code is:
Rprof("profile1.out", line.profiling=TRUE)
eval(parse(file = "http://pastebin.com/download.php?i=KjdkSVZq", keep.source=TRUE))
Rprof(NULL)
summaryRprof("profile1.out", lines = "show")
Which gives:
$by.self
self.time self.pct total.time total.pct
download.php?i=KjdkSVZq#17 8.04 64.11 8.04 64.11
<no location> 4.38 34.93 4.38 34.93
download.php?i=KjdkSVZq#16 0.06 0.48 0.06 0.48
download.php?i=KjdkSVZq#18 0.02 0.16 0.02 0.16
download.php?i=KjdkSVZq#23 0.02 0.16 0.02 0.16
download.php?i=KjdkSVZq#6 0.02 0.16 0.02 0.16
$by.total
total.time total.pct self.time self.pct
download.php?i=KjdkSVZq#17 8.04 64.11 8.04 64.11
<no location> 4.38 34.93 4.38 34.93
download.php?i=KjdkSVZq#16 0.06 0.48 0.06 0.48
download.php?i=KjdkSVZq#18 0.02 0.16 0.02 0.16
download.php?i=KjdkSVZq#23 0.02 0.16 0.02 0.16
download.php?i=KjdkSVZq#6 0.02 0.16 0.02 0.16
$by.line
self.time self.pct total.time total.pct
<no location> 4.38 34.93 4.38 34.93
download.php?i=KjdkSVZq#6 0.02 0.16 0.02 0.16
download.php?i=KjdkSVZq#16 0.06 0.48 0.06 0.48
download.php?i=KjdkSVZq#17 8.04 64.11 8.04 64.11
download.php?i=KjdkSVZq#18 0.02 0.16 0.02 0.16
download.php?i=KjdkSVZq#23 0.02 0.16 0.02 0.16
$sample.interval
[1] 0.02
$sampling.time
[1] 12.54
Checking the source code tells us that the problematic line (#17) is indeed the stupid if-statement in the for-loop. Compared with basically no time for calculating the same using vectorized code (line #6).
I haven't tried it with any graphical output, but I am already very impressed by what I got so far.
Update: This function has been re-written to deal with line numbers. It's on github here.
I wrote this function to parse the file from Rprof and output a table of somewhat clearer results than summaryRprof. It displays the full stack of functions (and line numbers if line.profiling=TRUE), and their relative contribution to run time:
proftable <- function(file, lines=10) {
# require(plyr)
interval <- as.numeric(strsplit(readLines(file, 1), "=")[[1L]][2L])/1e+06
profdata <- read.table(file, header=FALSE, sep=" ", comment.char = "",
colClasses="character", skip=1, fill=TRUE,
na.strings="")
filelines <- grep("#File", profdata[,1])
files <- aaply(as.matrix(profdata[filelines,]), 1, function(x) {
paste(na.omit(x), collapse = " ") })
profdata <- profdata[-filelines,]
total.time <- interval*nrow(profdata)
profdata <- as.matrix(profdata[,ncol(profdata):1])
profdata <- aaply(profdata, 1, function(x) {
c(x[(sum(is.na(x))+1):length(x)],
x[seq(from=1,by=1,length=sum(is.na(x)))])
})
stringtable <- table(apply(profdata, 1, paste, collapse=" "))
uniquerows <- strsplit(names(stringtable), " ")
uniquerows <- llply(uniquerows, function(x) replace(x, which(x=="NA"), NA))
dimnames(stringtable) <- NULL
stacktable <- ldply(uniquerows, function(x) x)
stringtable <- stringtable/sum(stringtable)*100
stacktable <- data.frame(PctTime=stringtable[], stacktable)
stacktable <- stacktable[order(stringtable, decreasing=TRUE),]
rownames(stacktable) <- NULL
stacktable <- head(stacktable, lines)
na.cols <- which(sapply(stacktable, function(x) all(is.na(x))))
stacktable <- stacktable[-na.cols]
parent.cols <- which(sapply(stacktable, function(x) length(unique(x)))==1)
parent.call <- paste0(paste(stacktable[1,parent.cols], collapse = " > ")," >")
stacktable <- stacktable[,-parent.cols]
calls <- aaply(as.matrix(stacktable[2:ncol(stacktable)]), 1, function(x) {
paste(na.omit(x), collapse= " > ")
})
stacktable <- data.frame(PctTime=stacktable$PctTime, Call=calls)
frac <- sum(stacktable$PctTime)
attr(stacktable, "total.time") <- total.time
attr(stacktable, "parent.call") <- parent.call
attr(stacktable, "files") <- files
attr(stacktable, "total.pct.time") <- frac
cat("\n")
print(stacktable, row.names=FALSE, right=FALSE, digits=3)
cat("\n")
cat(paste(files, collapse="\n"))
cat("\n")
cat(paste("\nParent Call:", parent.call))
cat(paste("\n\nTotal Time:", total.time, "seconds\n"))
cat(paste0("Percent of run time represented: ", format(frac, digits=3)), "%")
invisible(stacktable)
}
Running this on the Henrik's example file, I get this:
> Rprof("profile1.out", line.profiling=TRUE)
> source("http://pastebin.com/download.php?i=KjdkSVZq")
> Rprof(NULL)
> proftable("profile1.out", lines=10)
PctTime Call
20.47 1#17 > [ > 1#17 > [.data.frame
9.73 1#17 > [ > 1#17 > [.data.frame > [ > [.factor
8.72 1#17 > [ > 1#17 > [.data.frame > [ > [.factor > NextMethod
8.39 == > Ops.factor
5.37 ==
5.03 == > Ops.factor > noNA.levels > levels
4.70 == > Ops.factor > NextMethod
4.03 1#17 > [ > 1#17 > [.data.frame > [ > [.factor > levels
4.03 1#17 > [ > 1#17 > [.data.frame > dim
3.36 1#17 > [ > 1#17 > [.data.frame > length
#File 1: http://pastebin.com/download.php?i=KjdkSVZq
Parent Call: source > withVisible > eval > eval >
Total Time: 5.96 seconds
Percent of run time represented: 73.8 %
Note that the "Parent Call" applies to all the stacks represented on the table. This makes is useful when your IDE or whatever calls your code wraps it in a bunch of functions.
I currently have R uninstalled here, but in SPlus you can interrupt the execution with the Escape key, and then do traceback(), which will show you the call stack. That should enable you to use this handy method.
Here are some reasons why tools built on the same concepts as gprof are not very good at locating performance problems.
A different solution comes from a different question: how to effectively use library(profr) in R:
For example:
install.packages("profr")
devtools::install_github("alexwhitworth/imputation")
x <- matrix(rnorm(1000), 100)
x[x>1] <- NA
library(imputation)
library(profr)
a <- profr(kNN_impute(x, k=5, q=2), interval= 0.005)
It doesn't seem (to me at least), like the plots are at all helpful here (eg plot(a)). But the data structure itself does seem to suggest a solution:
R> head(a, 10)
level g_id t_id f start end n leaf time source
9 1 1 1 kNN_impute 0.005 0.190 1 FALSE 0.185 imputation
10 2 1 1 var_tests 0.005 0.010 1 FALSE 0.005 <NA>
11 2 2 1 apply 0.010 0.190 1 FALSE 0.180 base
12 3 1 1 var.test 0.005 0.010 1 FALSE 0.005 stats
13 3 2 1 FUN 0.010 0.110 1 FALSE 0.100 <NA>
14 3 2 2 FUN 0.115 0.190 1 FALSE 0.075 <NA>
15 4 1 1 var.test.default 0.005 0.010 1 FALSE 0.005 <NA>
16 4 2 1 sapply 0.010 0.040 1 FALSE 0.030 base
17 4 3 1 dist_q.matrix 0.040 0.045 1 FALSE 0.005 imputation
18 4 4 1 sapply 0.045 0.075 1 FALSE 0.030 base
Single iteration solution:
That is the data structure suggests the use of tapply to summarize the data. This can be done quite simply for a single run of profr::profr
t <- tapply(a$time, paste(a$source, a$f, sep= "::"), sum)
t[order(t)] # time / function
R> round(t[order(t)] / sum(t), 4) # percentage of total time / function
base::! base::%in% base::| base::anyDuplicated
0.0015 0.0015 0.0015 0.0015
base::c base::deparse base::get base::match
0.0015 0.0015 0.0015 0.0015
base::mget base::min base::t methods::el
0.0015 0.0015 0.0015 0.0015
methods::getGeneric NA::.findMethodInTable NA::.getGeneric NA::.getGenericFromCache
0.0015 0.0015 0.0015 0.0015
NA::.getGenericFromCacheTable NA::.identC NA::.newSignature NA::.quickCoerceSelect
0.0015 0.0015 0.0015 0.0015
NA::.sigLabel NA::var.test.default NA::var_tests stats::var.test
0.0015 0.0015 0.0015 0.0015
base::paste methods::as<- NA::.findInheritedMethods NA::.getClassFromCache
0.0030 0.0030 0.0030 0.0030
NA::doTryCatch NA::tryCatchList NA::tryCatchOne base::crossprod
0.0030 0.0030 0.0030 0.0045
base::try base::tryCatch methods::getClassDef methods::possibleExtends
0.0045 0.0045 0.0045 0.0045
methods::loadMethod methods::is imputation::dist_q.matrix methods::validObject
0.0075 0.0090 0.0120 0.0136
NA::.findNextFromTable methods::addNextMethod NA::.nextMethod base::lapply
0.0166 0.0346 0.0361 0.0392
base::sapply imputation::impute_fn_knn methods::new imputation::kNN_impute
0.0392 0.0392 0.0437 0.0557
methods::callNextMethod kernlab::as.kernelMatrix base::apply kernlab::kernelMatrix
0.0572 0.0633 0.0663 0.0753
methods::initialize NA::FUN base::standardGeneric
0.0798 0.0994 0.1325
From this, I can see that the biggest time users are kernlab::kernelMatrix and the overhead from R for S4 classes and generics.
Preferred:
I note that, given the stochastic nature of the sampling process, I prefer to use averages to get a more robust picture of the time profile:
prof_list <- replicate(100, profr(kNN_impute(x, k=5, q=2),
interval= 0.005), simplify = FALSE)
fun_timing <- vector("list", length= 100)
for (i in 1:100) {
fun_timing[[i]] <- tapply(prof_list[[i]]$time, paste(prof_list[[i]]$source, prof_list[[i]]$f, sep= "::"), sum)
}
# Here is where the stochastic nature of the profiler complicates things.
# Because of randomness, each replication may have slightly different
# functions called during profiling
sapply(fun_timing, function(x) {length(names(x))})
# we can also see some clearly odd replications (at least in my attempt)
> sapply(fun_timing, sum)
[1] 2.820 5.605 2.325 2.895 3.195 2.695 2.495 2.315 2.005 2.475 4.110 2.705 2.180 2.760
[15] 3130.240 3.435 7.675 7.155 5.205 3.760 7.335 7.545 8.155 8.175 6.965 5.820 8.760 7.345
[29] 9.815 7.965 6.370 4.900 5.720 4.530 6.220 3.345 4.055 3.170 3.725 7.780 7.090 7.670
[43] 5.400 7.635 7.125 6.905 6.545 6.855 7.185 7.610 2.965 3.865 3.875 3.480 7.770 7.055
[57] 8.870 8.940 10.130 9.730 5.205 5.645 3.045 2.535 2.675 2.695 2.730 2.555 2.675 2.270
[71] 9.515 4.700 7.270 2.950 6.630 8.370 9.070 7.950 3.250 4.405 3.475 6.420 2948.265 3.470
[85] 3.320 3.640 2.855 3.315 2.560 2.355 2.300 2.685 2.855 2.540 2.480 2.570 3.345 2.145
[99] 2.620 3.650
Removing the unusual replications and converting to data.frames:
fun_timing <- fun_timing[-c(15,83)]
fun_timing2 <- lapply(fun_timing, function(x) {
ret <- data.frame(fun= names(x), time= x)
dimnames(ret)[[1]] <- 1:nrow(ret)
return(ret)
})
Merge replications (almost certainly could be faster) and examine results:
# function for merging DF's in a list
merge_recursive <- function(list, ...) {
n <- length(list)
df <- data.frame(list[[1]])
for (i in 2:n) {
df <- merge(df, list[[i]], ... = ...)
}
return(df)
}
# merge
fun_time <- merge_recursive(fun_timing2, by= "fun", all= FALSE)
# do some munging
fun_time2 <- data.frame(fun=fun_time[,1], avg_time=apply(fun_time[,-1], 1, mean, na.rm=T))
fun_time2$avg_pct <- fun_time2$avg_time / sum(fun_time2$avg_time)
fun_time2 <- fun_time2[order(fun_time2$avg_time, decreasing=TRUE),]
# examine results
R> head(fun_time2, 15)
fun avg_time avg_pct
4 base::standardGeneric 0.6760714 0.14745123
20 NA::FUN 0.4666327 0.10177262
12 methods::initialize 0.4488776 0.09790023
9 kernlab::kernelMatrix 0.3522449 0.07682464
8 kernlab::as.kernelMatrix 0.3215816 0.07013698
11 methods::callNextMethod 0.2986224 0.06512958
1 base::apply 0.2893367 0.06310437
7 imputation::kNN_impute 0.2433163 0.05306731
14 methods::new 0.2309184 0.05036331
10 methods::addNextMethod 0.2012245 0.04388708
3 base::sapply 0.1875000 0.04089377
2 base::lapply 0.1865306 0.04068234
6 imputation::impute_fn_knn 0.1827551 0.03985890
19 NA::.nextMethod 0.1790816 0.03905772
18 NA::.findNextFromTable 0.1003571 0.02188790
Results
From the results, a similar but more robust picture emerges as with a single case. Namely, there is a lot of overhead from R and also that library(kernlab) is slowing me down. Of note, since kernlab is implemented in S4, the overhead in R is related since S4 classes are substantially slower than S3 classes.
I'd also note that my personal opinion is that a cleaned up version of this might be a useful pull request as a summary method for profr. Although I'd be interested to see others' suggestions!

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