The average number of files in the shell

Question

I want to calculate the average of 15 files: - ifile1.txt, ifile2.txt, ....., ifile15.txt. The number of columns and rows of each file is the same. Part of the data looks like

ifile1.txt      ifile2.txt       ifile3.txt
3  5  2  2 .    1  2  1  3 .    4  3  4  1 .
1  4  2  1 .    1  3  0  2 .    5  3  1  5 .
4  6  5  2 .    2  5  5  1 .    3  4  3  1 .
5  5  7  1 .    0  0  1  1 .    4  3  4  0 .
.  .  .  . .    .  .  .  . .    .  .  .  . .  

I would like to find above a new file that will show the average of these 15 fils.

ofile.txt
2.66   3.33  2.33 2      . (i.e. average of 3 1 4, average of 5 2 3 and so on)
2.33   3.33  1    2.66   .
3      5     4.33 1.33   .
3      2.33  4    0.66   .
.      .     .    .      .

I tried to do the following, but getting an error

awk'{for (i=1; i<=NF; i++)} rows=FNR;cols=NF} END 
{for (i=1; i<=rows; i++){for (j=1; j<=cols; j++) 
s+=$i;print $0,s/NF;s=0}}' ifile* > ofile.txt

Answer 1

As written:

awk'{for (i=1; i<=NF; i++)} rows=FNR;cols=NF} END
…

you get the command "not found" as an error, because you must leave a space between awkand the script inside the quotation marks. When you fix this, you begin to run into problems because there are two }and only one in the first line of the script {.

, 2D-, , . . 2D- END.

awk 'FNR == 1 { nfiles++; ncols = NF }
     { for (i = 1; i < NF; i++) sum[FNR,i] += $i
       if (FNR > maxnr) maxnr = FNR
     }
     END {
         for (line = 1; line <= maxnr; line++)
         {
             for (col = 1; col < ncols; col++)
                  printf "  %f", sum[line,col]/nfiles;
             printf "\n"
         }
     }' ifile*.txt

:

ifile1.txt

3 5 2 2
1 4 2 1
4 6 5 2
5 5 7 1

ifile2.txt

1 2 1 3
1 3 0 2
2 5 5 1
0 0 1 1

ifile3.txt

4 3 4 1
5 3 1 5
3 4 3 1
4 3 4 0

script :

  2.666667  3.333333  2.333333
  2.333333  3.333333  1.000000
  3.000000  5.000000  4.333333
  3.000000  2.666667  4.000000

2, %.2f %f.

Answer 2

$ { head -n1 ifile1.txt; paste ifile*.txt;} | awk 'NR==1{d=NF; next;} {for (i=1;i<=d;i++) {s=0; for (j=i;j<=NF;j+=d) s+=$j; printf "%.2f%s",s/(NF/d),j==NF+d?"\n":"\t";}}'
2.67    3.33    2.33    2.00
2.33    3.33    1.00    2.67
3.00    5.00    4.33    1.33
3.00    2.67    4.00    0.67

script . - script . , .

• { head -n1 ifile1.txt; paste ifile*.txt;}

  ifile1.txt. paste , ..:

  $ paste ifile*.txt
  3  5  2  2      1  2  1  3      4  3  4  1
  1  4  2  1      1  3  0  2      5  3  1  5
  4  6  5  2      2  5  5  1      3  4  3  1
  5  5  7  1      0  0  1  1      4  3  4  0
  

• |

  awk. awk :

• NR==1{d=NF; next;}

  d. next.

• for (i=1;i<=d;i++) {s=0; for (j=i;j<=NF;j+=d) s+=$j; printf "%.2f%s",s/(NF/d),j==NF+d?"\n":"\t";}

  .

script:

{
    head -n1 ifile1.txt
    paste ifile*.txt
} | 
awk '
    NR==1 {d=NF; next;}

    {
        for (i=1;i<=d;i++)
        {
            s=0; for (j=i;j<=NF;j+=d)
                s+=$j;
            printf "%.2f%s",s/(NF/d),j==NF+d?"\n":"\t";
        }
    }

Answer 3

, . $0 i END, .

awk '{rows=FNR; cols=NF; for (i = 1; i <= NF; i++) { total[FNR, i] += $i }}
     FILENAME != lastfn { count++; lastfn = FILENAME }
     END { for (i = 1; i <= rows; i++) { 
                for (j =  1; j <= cols; j++) {
                    printf("%s ", total[i, j]/count)
                }
                printf("\n")
            }
        }' ifile* > ofile.txt