Hi everyone,
I have a little bit of complicated task to finish with AWK. Here it is;
<data>
a1 a2 a3 a4 a5
b1 b2 b3 b4 b5
c1 c2 c3 c4 c5
d1 d2 d3 d4 d5
e1 e2 e3 e4 e5
</data>
lets say each data block contains 5 rows and 5 columns, what I need to do is this;
<data>
a1 a2 a3 a4 a5 a6
b1 b2 b3 b4 b5 b6
c1 c2 c3 c4 c5 c6
d1 d2 d3 d4 d5 d6
e1 e2 e3 e4 e5 e6
</data>
where the last fields are calculated as following;
a6 = c2*a2 + c3*a3 + c4*a4 + c5*a5
b6 = c2*b2 + c3*b3 + c4*b4 + c5*b5
c6 = c2*c2 + c3*c3 + c4*c4 + c5*c5
d6 = c2*d2 + c3*d3 + c4*d4 + c5*d5
e6 = c2*e2 + c3*e3 + c4*e4 + c5*e5
the algebra on the above calculation may not necessarily be simple as this.
Thanks for any help.
Please post a sample of the input and the desired output.
Mention the condition to be tested for.
Show how the real data is, how the "data" blocks are separated from one another.
Also, mention the "real" calculations.
Do not oversimplify. This often leads to frequent changes in requirements with subsequent confusion.
Here is two data blocks from my real data file
<data>
21 0.00 0.00 0.57 0.57
21 0.00 0.00 -0.19 0.19
6 -0.63 0.12 0.31 0.37
24 -0.44 0.15 0.25 0.30
-13 -0.23 0.37 0.13 0.14
</data>
<data>
1 0.00 0.00 0.10 0.10
-1 0.00 0.00 -0.66 0.66
6 -0.17 0.40 0.27 0.32
24 -0.48 -0.24 0.12 0.15
-13 0.17 -0.44 0.33 0.18
</data>
take the row where $1==6 to use its fields in further calculations
a6 = c2*a2 - c3*a3 - c4*a4 - c5*a5
b6 = c2*b2 - c3*b3 - c4*b4 - c5*b5
c6 = c2*c2 - c3*c3 - c4*c4 - c5*c5
d6 = c2*d2 - c3*d3 - c4*d4 - c5*d5
e6 = c2*e2 - c3*e3 - c4*e4 - c5*e5
After this operation these two data blocks will look like
<data>
21 0.00 0.00 0.57 0.57 a6
21 0.00 0.00 -0.19 0.19 b6
6 -0.63 0.12 0.31 0.37 c6
24 -0.44 0.15 0.25 0.30 d6
-13 -0.23 0.37 0.13 0.14 e6
</data>
<data>
1 0.00 0.00 0.10 0.10 a6
-1 0.00 0.00 -0.66 0.66 b6
6 -0.17 0.40 0.27 0.32 c6
24 -0.48 -0.24 0.12 0.15 d6
-13 0.17 -0.44 0.33 0.18 e6
</data>
where the numerical values of the last column (for the 2nd data block as an example)
a6 = (-0.17)*(0.00) -(0.40)*(0.00) -(0.27)*(0.10) -(0.32)*(0.10) = -0.0590
b6 = (-0.17)*(0.00) -(0.40)*(0.00) -(0.27)*(-0.66)-(0.32)*(0.66) = -0.0330
c6 = (-0.17)*(-0.17)-(0.40)*(0.40) -(0.27)*(0.27) -(0.32)*(0.32) = -0.3064
d6 = (-0.17)*(-0.48)-(0.40)*(-0.24)-(0.27)*(0.12) -(0.32)*(0.15) = 0.0972
e6 = (-0.17)*(0.17) -(0.40)*(-0.44)-(0.27)*(0.33) -(0.32)*(0.18) = 0.0004
Yoda
February 23, 2013, 11:06am
4
Try this code:
awk ' /<data>/ {
f = 1;
print $0
next;
} /<\/data>/ && s {
f = 0;
m = 0;
k = 12;
while(m < j)
{
for(i=1;i<=nf;i++)
{
printf "%s\t", a[i,++m];
if(i >= 2)
{
a6 -= (a[i,k] * a[i,m]);
++k;
}
}
printf "%.4f", a6;
printf "\n"
if(k > nf) k=12;
}
j = 0;
print $0;
} f == 1 {
for(i=1;i<=NF;i++)
{
a[i,++j] = $i;
}
if(a[1,11] == 6 )
s = 1;
if(a[1,11] != 6 )
s = 0;
nf = NF;
}' file
Hi bipi,
Yoda
February 23, 2013, 11:39am
6
Oops I forgot to reinitialize a6
Modified code:
awk ' /<data>/ {
f = 1;
print $0
next;
} /<\/data>/ && s {
f = 0;
m = 0;
k = 12;
while(m < j)
{
for(i=1;i<=nf;i++)
{
printf "%s\t", a[i,++m];
if(i >= 2)
{
a6 -= (a[i,k] * a[i,m]);
++k;
}
}
printf "%.4f", a6;
a6 = 0;
printf "\n"
if(k > nf) k=12;
}
j = 0;
print $0;
} f == 1 {
for(i=1;i<=NF;i++)
{
a[i,++j] = $i;
}
if(a[1,11] == 6 )
s = 1;
if(a[1,11] != 6 )
s = 0;
nf = NF;
}' file
Here is the O/P that I am getting:
$ ./hayreter
<data>
21 0.00 0.00 0.57 0.57 -0.3876
21 0.00 0.00 -0.19 0.19 -0.0114
6 -0.63 0.12 0.31 0.37 -0.6443
24 -0.44 0.15 0.25 0.30 -0.4837
-13 -0.23 0.37 0.13 0.14 -0.2814
</data>
<data>
1 0.00 0.00 0.10 0.10 -0.0590
-1 0.00 0.00 -0.66 0.66 -0.0330
6 -0.17 0.40 0.27 0.32 -0.3642
24 -0.48 -0.24 0.12 0.15 -0.0660
-13 0.17 -0.44 0.33 0.18 0.0582
this is getting better
for you to check your results, look at the table that I posted yesterday above
Yoda
February 23, 2013, 12:09pm
8
OK fixed it. But I don't know why it didn't work in the previous code!
awk ' /<data>/ {
f = 1;
print $0
next;
} /<\/data>/ && s {
f = 0;
m = 0;
k = 12;
while(m < j)
{
for(i=1;i<=nf;i++)
{
printf "%s\t", a[i,++m];
if(i >= 2)
{
b = (a[i,k] * a[i,m]);
++k;
}
}
a6 = b[2] - b[3] - b[4] - b[5];
printf "%.4f", a6;
a6 = 0;
printf "\n"
if(k > nf) k=12;
}
j = 0;
print $0;
} f == 1 {
for(i=1;i<=NF;i++)
{
a[i,++j] = $i;
}
if(a[1,11] == 6 )
s = 1;
if(a[1,11] != 6 )
s = 0;
nf = NF;
}' file
Current O/P:
$ ./hayreter
<data>
21 0.00 0.00 0.57 0.57 -0.3876
21 0.00 0.00 -0.19 0.19 -0.0114
6 -0.63 0.12 0.31 0.37 0.1495
24 -0.44 0.15 0.25 0.30 0.0707
-13 -0.23 0.37 0.13 0.14 0.0084
</data>
<data>
1 0.00 0.00 0.10 0.10 -0.0590
-1 0.00 0.00 -0.66 0.66 -0.0330
6 -0.17 0.40 0.27 0.32 -0.3064
24 -0.48 -0.24 0.12 0.15 0.0972
-13 0.17 -0.44 0.33 0.18 0.0004
</data>
1 Like
Try this (with assumptions about your data):
awk '/^[ \t]*<data>/{data=1;countrow=0;rec[countrow++]=$0;next}
/^[ \t]*<\/data>/{
if(output) {
print rec[0]
n=split(rec[foundat],pattline)
for(i=1;i<countrow;i++) {
split(rec,otherline)
c=0
for(j=2;j<=n;j++) {
if(j==2) {c=pattline[j]*otherline[j];continue}
c-=(pattline[j]*otherline[j])
}
print rec,c
}
print
}
else
for(i in rec) delete rec
output=foundat=data=0;next
}
data{
if(!output && $1=="6") { output=1; foundat=countrow }
rec[countrow++]=$0
}' file
1 Like
Both scripts works quite fine, I thank to you both so much.
Yoda
February 23, 2013, 11:31pm
11
Here is a brief explanation:
awk ' /<data>/ { # Search for pattern: <data>
f = 1; # If found set f = 1
print $0 # Print <data> tag
next; # Stop processing current record
} /<\/data>/ && s { # Search for pattern: </data>
f = 0; # If found set f = 0
m = 0; # m = 0 (counter for fetching all array elements)
k = 12; # k = 12 (index of 2nd element in row c)
while(m < j) # While m < j
{
for(i=1;i<=nf;i++) # For each records in 2D array
{
printf "%s\t", a[i,++m]; # Print records
if(i >= 2) # If array index i > 2 (starting from second element)
{
b = (a[i,k] * a[i,m]); # Multiply current record with row c record.
++k;
}
}
a6 = b[2] - b[3] - b[4] - b[5]; # Subtract elements in array b and assign to a6
printf "%.4f", a6; # Print a6
a6 = 0; # Reinitialize a6 = 0
printf "\n"
if(k > nf) k=12; # if k index goes beyond number of recs, reset to 12
}
j = 0;
print $0; # Print </data> tag
} f == 1 { # If f == 1
for(i=1;i<=NF;i++) # Creating a 2D array with elements in <data> tag.
{
a[i,++j] = $i;
}
if(a[1,11] == 6 ) # If row c first element == 6
s = 1; # Set s = 1
if(a[1,11] != 6 ) # If row c first element != 6
s = 0; # Set s = 0
nf = NF; # Set nf = number of records in line NF
}' file
Thanks again for this brief (but quite detailed) explanation.
Yoda
February 23, 2013, 11:50pm
13
Here are the indices (i,j) for each data elements when stored in array:
1 (1,1) 0.00 (2,2) 0.00 (3,3) 0.10 (4,4) 0.10 (5,5)
-1 (1,6) 0.00 (2,7) 0.00 (3,8) -0.66 (4,9) 0.66 (5,10)
6 (1,11) -0.17 (2,12) 0.40 (3,13) 0.27 (4,14) 0.32 (5,15)
24 (1,16) -0.48 (2,17) -0.24 (3,18) 0.12 (4,19) 0.15 (5,20)
-13 (1,21) 0.17 (2,22) -0.44 (3,23) 0.33 (4,24) 0.18 (5,25)
This is why in the code I am using j index 11 - 15 to identify row c elements. Hence you have to change the j index as per your requirement. I hope you understood.
don't worry, I perfectly understood what you mean
