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# 1  
Old 09-21-2017
Matrix to 3 col sorted

Hello experts, I have matrices sorted by position, there are 400k rows, 3000 columns.

Code:
ID      CHR     POS     M1      M2      M3      M4      M5
ID1     1       1       4.6     2.6     2.1     3.5     4.2
ID2     1       100     3.6     2.9     3.2     2.6     2.5
ID3     1       1000    4.1     2.9     2.4     2.8     2.5
ID4     1       2000    4.2     2.4     2.8     3.7     2.8
ID5     1       3000    4.6     2.9     3.7     3.5     2.7
ID6     1       4000    5.2     3       3.2     4.2     3.7
ID7     1       5000    5.8     2.2     2.8     3.1     2.7

I`m looking for a sorted long-format output by position, grouped by markers (Ms)

Code:
M1 ID1__1__1 4.6
M1 ID2__1__100 3.6
M1 ID3__1__1000 4.1
M1 ID4__1__2000 4.2
M1 ID5__1__3000 4.6
M1 ID6__1__4000 5.2
M1 ID7__1__5000 5.8
M2 ID1__1__1 2.6
M2 ID2__1__100 2.9
M2 ID3__1__1000 2.9
M2 ID4__1__2000 2.4
M2 ID5__1__3000 2.9
M2 ID6__1__4000 3
M2 ID7__1__5000 2.2
M3 ID1__1__1 2.1
M3 ID2__1__100 3.2
M3 ID3__1__1000 2.4
M3 ID4__1__2000 2.8
M3 ID5__1__3000 3.7
M3 ID6__1__4000 3.2
M3 ID7__1__5000 2.8
M4 ID1__1__1 3.5
M4 ID2__1__100 2.6
M4 ID3__1__1000 2.8
M4 ID4__1__2000 3.7
M4 ID5__1__3000 3.5
M4 ID6__1__4000 4.2
M4 ID7__1__5000 3.1

I have two working methods

1. This step requires a final sorting which for 400000x3000 datapoints will be very row or run out of memory.


Code:
awk -F"\t" 'NR==1{split($0,lines);}NR>1{for(i=4;i<length(lines);i++) print lines[i],$1"__"$2"__"$3,$i}' tmp | sort -k1,1


2. This step is very very slow as it requires to take the entire matrix into memory before making splits.

Code:
awk -F"\t" '{a[NR]=$0}END{split(a[1],lines); for(i=4;i<=length(lines);i++) { for(j=2;j<=length(a);j++) { split(a[j],haps); print lines[i],haps[1]"__"haps[2]"__"haps[3],haps[i]}}}'  tmp


Is there a more efficient way to get the output I need? Thanks in advance.
# 2  
Old 09-21-2017
Code:
rm -rf output
mkdir output
awk '
NR==1 { for (i=1; i<=NF; i++) a[i]=$i ;}
NR > 1 {
   for (i=4; i<=NF; i++) print "touch output/\"" a[i], $1 "__" $2 "__" $3, $i "\"";
}
' infile | sh

ls -1 output > outfile

This User Gave Thanks to rdrtx1 For This Post:
# 3  
Old 09-21-2017
Interesting idea! Let the filesystem do a part of the sorting, and let ls sort the other parts.
More fine-tuned so ls sorts less:
Code:
outdir=./output
mkdir -p $outdir
awk '
NR==1 {
  for (i=4; i<=NF; i++) { line[i]=$i; printf "mkdir -p \"%s\"\n",$i }
  next
}
{
  for (i=4; i<=NF; i++) printf ">\"%s/%s__%s__%s %s\"\n",line[i],$1,$2,$3,$i
}
' infile > tmpfile
(
cd $outdir
sh tmpfile
for i in *
do
  ls -1 "$i" | sed "s#^#${i} #"
done
) | tee outfile


Last edited by MadeInGermany; 09-21-2017 at 02:37 PM..
This User Gave Thanks to MadeInGermany For This Post:
# 4  
Old 09-21-2017
Me too, I found this idea at least creative, but 1,2E9 directory entries, albeit in 3000 directories, might be somewhat, hmm, demanding on the file system...
On the other hand, I thought of awk distributing the columns into 3000 files, one column each, and then concatening them, but opening and appending 3000 files 400k times also would not be too fast. Right now, I'm lost...
# 5  
Old 09-22-2017
Whatever strategy you choose, according to your expectation, you will get an output file that is near 5 times bigger than your input file.

input file :
rows = 400k
columns = 3000
Total amount of datas = 400k x 3000 = 1,2 billions

output file :
rows = (400k -1) x ( 3000 -3 )
"-1" because of the header row
"-3" because the 3 first contains the datas you want
columns = 5 (i take into account the datas, not the formatting and "__" stuff)
Total amount of datas = 5 x (400k -1) x ( 3000 -3 ) = 5,99...billions

Indeed, since you will be repeating the datas of the first 3 column for each and every subsequent column. (so (3 + 2 ) x (3000 - 3) ) rather than having it once for all of them ( "+ 2" is because you even want to add the header and the values of the subsequent column).

So just because of your prerequisits and expectations, for sure you will have to write more data, and thus you will need the corresponding number of I/O writing, independently of the strategy you choose.

If you have a huge amount of data to write, then there is and uncompressible amount of time to do it.
Of course I/O operations are quicker to do in RAM or SSD than on a standard hard drive, but still ...

It would be cheaper to save the transposed matrix and request on it
So you will have it twice : in line and in columns ... of course it will cost you 1,2 billions of datas more.
But 1,2 billions x 2 = 2,4 billions is still more than twice smaller than 5,9 billions of datas !

Anyway, processing such an amount of data using files is inappropriate : this is what Database have been designed for...
My 2 cents ...

Smilie

PS// The reality might even be worse since i didn't even take into account the size of the datas, just the number of them ...

Last edited by ctsgnb; 09-22-2017 at 12:18 PM..
This User Gave Thanks to ctsgnb For This Post:
# 6  
Old 09-24-2017
Hi senhia83,
maybe a little bit faster than your second solution:
Code:
awk 'NR==1 {C=NF;Y=$0;Z=$4;next}{X[NR]=$0;print Z,$1"__"$2"__"$3,$4}END{for(i=5;i<C;++i){$0=Y;Z=$i;j=1;while(++j<=NR){$0=X[j];print Z,$1"__"$2"__"$3,$i}}}'

or:
Code:
awk 'NR==1 {C=NF;Y=$0;Z=$4;next}
{X[NR]=$0;print Z,$1"__"$2"__"$3,$4}
END{
 for(i=5;i<C;++i){
   $0=Y;Z=$i;j=1;
   while(++j<=NR){
     $0=X[j];print Z,$1"__"$2"__"$3,$i
   }
 }
}' file

Regards.
This User Gave Thanks to disedorgue For This Post:
# 7  
Old 09-27-2017
Thanks ! the filesystem ended up crashing with the files approach ..but a very innovative idea nonetheless !
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