Sort by first row - awk


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# 1  
Sort by first row - awk

how can i sort the table based on first row? thanks in advance

input
Code:
name    d       b       c       a
l       l1      l2      l3    l4
l1      1       2       3       4
l2      2       2       2       1
l3      1       1       2       2

ouput
Code:
name    a       b       c       d
l1     l4       l2      l3     l1
l1      4       2       3       1
l2      1       2       2       2
l3      2       1       2       1

# 2  
This is not as easy as it seems in the first place. Given you spell "Name" with uppercase "N" and that "N" sorts below all lower case letters, you could try like
Code:
awk -f transp.awk file | LC_ALL=C sort | awk -f transp.awk
Name    a       b       c       d
l       l4      l2      l3      l1
l1      4       2       3       1
l2      1       2       2       2
l3      2       1       2       1

with transp.awk
Code:
                {MX=NF
                 for (i=1; i<=NF; i++) C[NR, i]=$i
                }
END             {for (j=1; j<=NR; j++)  {for (i=1; i<=MX; i++) printf "%s\t", C[i,j]
                                         printf "\n"
                                        }
                }

You may want to add some refinements, like setting MX to the max(NF), or doing some error checking... some awk versions offer their internal sort; that might help as well...
This User Gave Thanks to RudiC For This Post:
# 3  
Umm. sorry. It seems the script is transposing the table not sorting.
# 4  
Yes it does. Did you consider reading the entire post and using the first line of my proposal, i.e. transpose - sort - transpose?
# 5  
ok. thanks a lot.
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