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Dataframe of factors transform speed?

10 messages · Benilton Carvalho, jim holtman, Charles C. Berry +1 more

#
Hello,

This is a speed question. I have a dataframe genoT:
[1]   1002 238304
'data.frame':   1002 obs. of  238304 variables:
 $ SNP_A.4261647: Factor w/ 3 levels "0","1","2": 3 3 3 3 3 3 3 3 3 3
...
 $ SNP_A.4261610: Factor w/ 3 levels "0","1","2": 1 1 3 3 1 1 1 2 2 2
...
 $ SNP_A.4261601: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1
...
 $ SNP_A.4261704: Factor w/ 3 levels "0","1","2": 3 3 3 3 3 3 3 3 3 3
...
 $ SNP_A.4261563: Factor w/ 3 levels "0","1","2": 3 1 2 1 2 3 2 3 3 1
...
 $ SNP_A.4261554: Factor w/ 3 levels "0","1","2": 1 1 NA 1 NA 2 1 1 2 1
...
 $ SNP_A.4261666: Factor w/ 3 levels "0","1","2": 1 1 2 1 1 1 1 1 1 2
...
 $ SNP_A.4261634: Factor w/ 3 levels "0","1","2": 3 3 2 3 3 3 3 3 3 2
...
 $ SNP_A.4261656: Factor w/ 3 levels "0","1","2": 1 1 2 1 1 1 1 1 1 2
...
 $ SNP_A.4261637: Factor w/ 3 levels "0","1","2": 1 3 2 3 2 1 2 1 1 3
...
 $ SNP_A.4261597: Factor w/ 3 levels "AA","AB","BB": 2 2 3 3 3 2 1 2 2 3
...
 $ SNP_A.4261659: Factor w/ 3 levels "AA","AB","BB": 3 3 3 3 3 3 3 3 3 3
...
 $ SNP_A.4261594: Factor w/ 3 levels "AA","AB","BB": 2 2 2 1 1 1 2 2 2 2
...
 $ SNP_A.4261698: Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1 1 ...
 $ SNP_A.4261538: Factor w/ 3 levels "AA","AB","BB": 2 3 2 2 3 2 2 1 1 2
...
 $ SNP_A.4261621: Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 1 1 1 1 1
...
 $ SNP_A.4261553: Factor w/ 3 levels "AA","AB","BB": 1 1 2 1 1 1 1 1 1 1
...
 $ SNP_A.4261528: Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1 1 ...
 $ SNP_A.4261579: Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 2 1 1 1 2
...
 $ SNP_A.4261513: Factor w/ 3 levels "AA","AB","BB": 2 1 2 2 2 NA 1 NA 2
1 ...
 $ SNP_A.4261532: Factor w/ 3 levels "AA","AB","BB": 1 2 2 1 1 1 3 1 1 1
...
 $ SNP_A.4261600: Factor w/ 2 levels "AB","BB": 2 2 2 2 2 2 2 2 2 2 ...
 $ SNP_A.4261706: Factor w/ 2 levels "AA","BB": 1 1 1 1 1 1 1 1 1 1 ...
 $ SNP_A.4261575: Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 1 1 2 2 1
...

Its columns are factors with different number of levels (from 1 to 3 -
that's what I got from read.table, i.e., it dropped missing levels). I
want to convert it to uniform factors with 3 levels. The 1st 10 rows
above show already converted columns and the rest are not yet converted.
Here's my attempt wich is a complete failure as speed:
+     for(j in 1:(10         )){ #-- this is to try 1st 10 cols and
measure the time, it otherwise is ncol(genoT) instead of 10

+        gt<-genoT[[j]]          #-- this is to avoid 2D indices
+        for(l in 1:length(gt at levels)){
+          levels(gt)[l] <- switch(gt at levels[l],AA="0",AB="1",BB="2")
#-- convert levels to "0","1", or "2"
+          genoT[[j]]<-factor(gt,levels=0:2)   #-- make a 3-level factor
and put it back
+        }
+     }
+ )
[1] 785.085   4.358 789.454   0.000   0.000

789s for 10 columns only!

To me it seems like replacing 10 x 3 levels and then making a factor of
1002 element vector x 10 is a "negligible" amount of operations needed.

So, what's wrong with me? Any idea how to accelerate significantly the
transformation or (to go to the very beginning) to make read.table use a
fixed set of levels ("AA","AB", and "BB") and not to drop any (missing)
level?

R-devel_2006-08-26, Sun Solaris 10 OS - x86 64-bit

The machine is with 32G RAM and AMD Opteron 285 (2.? GHz) so it's not
it.

Thank you very much for the help,

Latchezar Dimitrov,
Analyst/Programmer IV,
Wake Forest University School of Medicine,
Winston-Salem, North Carolina, USA
#
it looks like that whatever method you used to genotype the 1002  
samples on the STY array gave you a transposed matrix of genotype  
calls. :-)

i'd use:

genoT = read.table(yourFile, stringsAsFactors = FALSE)

as a starting point... but I don't think that would be efficient (as  
you'd need to fix one column at a time - lapply).

i'd preprocess yourFile before trying to load it:

cat yourFile | sed -e 's/AA/1/g' | sed -e 's/AB/2/g' | sed -e 's/BB/3/ 
g' > outFile

and, now, in R:

genoT = read.table(outFile, header=TRUE)

b
On Jul 19, 2007, at 11:51 PM, Latchezar Dimitrov wrote:

            
#
Is this what you want?  It took 0.01 seconds to convert 20 rows of the
test data:
+     result[[as.character(i)]] <- sample(vals,1000, replace=TRUE,
prob=c(9000,1,1))
+ }
'data.frame':   1000 obs. of  20 variables:
 $ X1 : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X2 : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X3 : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X4 : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X5 : Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1 1 ...
 $ X6 : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X7 : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X8 : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X9 : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X10: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X11: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X12: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X13: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X14: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X15: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X16: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X17: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X18: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X19: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ X20: Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
+     x <- lapply(result.df, function(facts){
+         factor(match(as.character(facts), vals) - 1, levels=0:2)
+     })
+     result.df <- do.call('data.frame', x)
+ })
   user  system elapsed
   0.01    0.00    0.01
'data.frame':   1000 obs. of  20 variables:
 $ X1 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X2 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X3 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X4 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X5 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X6 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X7 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X8 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X9 : Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X10: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X11: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X12: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X13: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X14: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X15: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X16: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X17: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X18: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X19: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ X20: Factor w/ 3 levels "0","1","2": 1 1 1 1 1 1 1 1 1 1 ...

        
On 7/19/07, Latchezar Dimitrov <ldimitro at wfubmc.edu> wrote:

  
    
#
Hi,
It only looks like :-)

Otherwise it is correctly created dataframe of 1002 samples X (big
number) of columns (SNP genotypes). It worked perfectly until I decided
to put together to cohorts independently processed in R already. I got
stuck with my lack of foreseeing. Otherwise I would have put 3 dummy
lines w/ AA,AB, and AB on each one to make sure all 3 genotypes are
present and that's it! Lesson for the future :-)

Maybe I am not using columns and rows appropriately here but the
dataframe is correct (I have not used FORTRAN since FORTRAN IV ;-) - as
str says 1002 observ. of (big number) vars.
No it was not efficient at all. 'matter of fact nothing is more
efficient then loading already read data, alas :-(
... Too late ;-) As it must be clear now I have two dataframes I want to
put together with rbind(geno1,geno2). The issue again is
"uniformization" of factor variables w/ missing factors - they ended up
like levels AA,BB on one of the and levels AB,BB on the other which
means as.numeric of AA is 1 on the 1st and as.numeric of AB is 1 on the
second - complete mess. That's why I tried to make both uniform, i.e.
levels "AA","AB", and "BB" for every SNP and then rbind works.

In any case my 1st questions remains: "What's wrong with me?" :-)

Thanks,
Latchezar
#
set.seed(123)
genoT = lapply(1:240000, function(i) factor(sample(c("AA", "AB",  
"BB"), 1000, prob=sample(c(1, 1000, 1000), 3), rep=T)))
names(genoT) = paste("snp", 1:240000, sep="")
genoT = as.data.frame(genoT)
dim(genoT)
class(genoT)
system.time(out <- lapply(genoT, function(x) match(x, c("AA", "AB",  
"BB"))-1))
##
##
    user  system elapsed
119.288   0.004 119.339

(for all 240K)

best,
b

ps: note that "out" is a list.
On Jul 20, 2007, at 2:01 AM, Latchezar Dimitrov wrote:

            
#
On Thu, 19 Jul 2007, Latchezar Dimitrov wrote:

            
It looks like these are all numeric originally. Handling these as a
vector or matrix will speed things up a bit. You can then stitch
together a data.frame:

# simulate: 
#       genoT.names <- scan('data.file, what='a', nlines=1, <etc> ) 
# 	genoT <- scan('data.file',skip=1)
#
user  system elapsed
  20.978   2.036  49.714
Most of the _elapsed_ time is due to lags in copy-and-paste-ing in the 
commands.

HTH,

Chuck
Charles C. Berry                            (858) 534-2098
                                             Dept of Family/Preventive Medicine
E mailto:cberry at tajo.ucsd.edu	            UC San Diego
http://famprevmed.ucsd.edu/faculty/cberry/  La Jolla, San Diego 92093-0901
#
Hi,

Thanks for the help. My 1st question still unanswered though :-) Please
see bellow
Now this _is the problem. Everything before converting to data.frame
worked almost instantaneously however as.data.frame runs forever.
Obviously there is some scalability memory management issue. When I
tried my own method but creating a new result (instead of modifying the
old) dataframe it worked like a charm for the 1st 100 cols ~ .3s. I
figured 300,000 cols should be ~1000s. Nope! It ran for about 50,000(!)s
to finish about 42,000 cols only. 

BTW, what ver. of R is yours?

Now here's what I "discovered" further.

#-- create a 1-col frame:
    geno   <-
data.frame(c(geno.GASP[[1]],geno.JAG[[1]]),row.names=c(rownames(geno.GAS
P),rownames(geno.JAG)))

#-- main code I repeated it w/ j in 1:1000, 2001:3000, and 3001:4000,
i.e., adding a 1000 of cols to geno each time

system.time(
#   for(j in 1:(ncol(geno.GASP      ))){
    for(j in 3001:(4000              )){
      gt.GASP<-geno.GASP[[j]]
       for(l in 1:length(gt.GASP at levels)){
         levels(gt.GASP)[l] <-
switch(gt.GASP at levels[l],AA="0",AB="1",BB="2")
       }
       gt.JAG <-geno.JAG [[j]]
#      for(l in 1:length(gt.JAG @levels)){
#        levels(gt.JAG )[l] <- switch(gt.JAG
@levels[l],AA="0",AB="1",BB="2")
#      }
       geno[[j]]<-factor(c(as.numeric(factor(gt.GASP,levels=0:2))-1
###               factor(c(as.numeric(factor(gt.GASP,levels=0:2))-1
                          ,as.numeric(factor(gt.JAG, levels=0:2))-1
                          )
                        ,levels=0:2
                        )
    }
)

Times (each one is for a 1000 cols!):
[1] 26.673  0.032 26.705  0.000  0.000
[1] 77.186  0.037 77.225  0.000  0.000
[1] 128.165   0.042 128.209   0.000   0.000
[1] 180.940   0.047 180.989   0.000   0.000

See the big diff and the scaling I mentioned above?

Further more I removed geno[[j]] assignment leaving the operation
though, i.e., replaced it with ### line above. Times:

[1] 0.857 0.008 0.865 0.000 0.000

Huh!? What the heck! That's my second question :-) Any ideas?

I still believe my method is near optimal. Of course I have to somehow
get rid of the assignment bottleneck.

For now the lesson is: "God bless lists"

Here is my final solution:
+     geno.GASP.L<-lapply(geno.GASP
+                        ,function(x){
+                           for(l in 1:length(x at levels)){levels(x)[l] <-
switch(x at levels[l],AA="0",AB="1",BB="2")}
+                           factor(x,levels=0:2)
+                         }
+                  )
+     geno.JAG.L <-lapply(geno.JAG
+                        ,function(x){
+ #                         for(l in 1:length(x at levels)){levels(x)[l] <-
switch(x at levels[l],AA="0",AB="1",BB="2")}
+                           factor(x,levels=0:2)
+                         }
+                  )
+ })
[1] 192.800   1.566 194.413   0.000   0.000   !!!!!!!!! :-)))))
+     class    (geno.GASP.L)<-"data.frame"
+     row.names(geno.GASP.L)<-row.names(geno.GASP)
+     class    (geno.JAG.L )<-"data.frame"
+     row.names(geno.JAG.L )<-row.names(geno.JAG )
+ })
[1] 12.156  0.001 12.155  0.000  0.000
+     geno<-rbind(geno.GASP.L,geno.JAG.L)
+ })
[1] 1542.340    9.072 2066.310    0.000    0.000

I logged my notes here as I was trying various things. Partly the reason
is my two questions:

"What was wrong with me?" and
"What the heck?!" remember above? :-)))

which  still remain unanswered :-(

I would have had a lot of fun if I had not to have this done by ...
Yesterday :-))

Thanks a lot for the help

Latchezar
#
One of the problems is that you are probably paging on your system
with an object that size (240000 x 1000).  This is about 1GB for a
single object:
+ genoT <- lapply(1:n, function(i) factor(sample(c("AA",
+ "AB", "BB"), 1000, prob=c(1000, 1, 1), rep=T)))
+ })
   user  system elapsed
  95.00    0.61  104.71
[1] 1045258752
I can create it on my 2GB machine as a list, but have problems
converting it to a dataframe because I don't have enough memory.

So unless you have at least 4GB on your system, it might take a long
time.  Look at your performance measurements on your system and see if
you have run out of physical memory and are paging.
On 7/21/07, Latchezar Dimitrov <ldimitro at wfubmc.edu> wrote:

  
    
#
Jim,

No, this is _not the problem. If you go to my 1st mail I have a monster
(at least was when I purchased it) with 32GB (sic :-) of RAM and 4 dual
core AMD64 285 (the fastest at that time and still pretty fast now :-) 

The machine stats paging when I run 2 copies of R working on two things
like that :-). If you look at my last e-mail I found a solution but
still have no clue why the heck x<-as.data.frame(y) where why is a list
of the same columns take real for ever and this the thing that killed me
before.

Thanks,
Latchezar
#
The problem is in the way that 'as.data.frame' works.  Use Rprof on a
small list and you will see where it is spending its time.

Now if you are really sure that all your data is consistent with being
a data frame,
you can create your own dataframe structure your self.  Not that I
would advocate it, but if you look at the output of 'dput' on a
dataframe, you can construct your own.

Here it took 20 seconds to create the test data with a list of 50,000
and only 2 seconds to create the data frame from that.
+ genoT <- lapply(1:n, function(i) factor(sample(c("AA",
+ "AB", "BB"), 1000, prob=c(1000, 1, 1), rep=T)))
+ })
   user  system elapsed
  20.85    0.12   22.83
+     row.names=c(NA, -length(genoT[[1]])), class='data.frame'))
   user  system elapsed
   2.00    0.08    2.11
'data.frame':   1000 obs. of  50000 variables:
 $ snp1    : Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp2    : Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp3    : Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp4    : Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp5    : Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp6    : Factor w/ 2 levels "AA","AB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp7    : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp8    : Factor w/ 2 levels "AA","BB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp9    : Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp10   : Factor w/ 3 levels "AA","AB","BB": 1 1 1 1 1 1 1 1 1 1 ...
 $ snp11   : Factor w/ 1 level "AA": 1 1 1 1 1 1 1 1 1 1 ...

        
On 7/21/07, Latchezar Dimitrov <ldimitro at wfubmc.edu> wrote: