Yes, as @Tamas_Papp said, @printf is quite easy to use for this:
using Printf
ar = [[1, 85645, 197111, 197240], [1, 203753],[1, 24]]
loop = 0
for i =1:length(ar)
val = ar[i]' .-1
@printf("NetDegree : %-2d n%d ", length(val), loop)
@printf("\n")
for jj = 1:length(val)
vv = val[jj]
@printf("\t\t o%-6d I : %9.6f, %9.6f", val[jj], 5.000000, -5.000000)
@printf("\n")
end # end of jj
loop += 1
end # end of i
Output:
NetDegree : 4 n0
o0 I : 5.000000, -5.000000
o85644 I : 5.000000, -5.000000
o197110 I : 5.000000, -5.000000
o197239 I : 5.000000, -5.000000
NetDegree : 2 n1
o0 I : 5.000000, -5.000000
o203752 I : 5.000000, -5.000000
NetDegree : 2 n2
o0 I : 5.000000, -5.000000
o23 I : 5.000000, -5.000000