Question

我有一些文本文件。第四列表示蛋白质残基，第五列表示蛋白质残基的类型.C表示线圈，T表示转弯，S表示弯曲。我想计算每个文件中C + T + S的平均值：公式为total no:of C+T+S/total no:of residues.

FILE1.TXT

  311  446 E L  C           0   0   78      0, 0.0  -239,-0.1     0, 0.0     0, 0.0   0.000 360.0 360.0 360.0-115.3   -6.6   19.5  -28.9
  312  447 E K  C     +     0   0  109      1,-0.2  -240,-0.0  -239,-0.0     0, 0.0   0.747 360.0  90.4  57.7  22.2   -8.5   22.8  -28.5
  313  448 E K  S    S+     0   0  194      2,-0.0    -1,-0.2     0, 0.0     0, 0.0  -0.107  72.2 141.1-120.6   7.9  -10.0   21.6  -31.8
  314  449 E Q  C     -     0   0   28      1,-0.1     2,-0.6    15,-0.0  -242,-0.1  -0.216  56.0-123.3 -57.3 138.8  -12.1   20.3  -28.9
  315  450 E V  E     +K  327   0B  82     12,-0.8    12,-2.9     2,-0.0     2,-0.4  -0.786  33.3 172.1 -96.1 120.7  -15.8   20.2  -29.2
  316  451 E E  E     +K  326   0B  83     -2,-0.6    10,-0.2    10,-0.2     2,-0.2  -0.987   1.7 163.8-128.9 140.8  -18.0   22.0  -26.7
  317  452 E T  E     -K  325   0B  68      8,-2.3     8,-2.5    -2,-0.4     2,-0.4  -0.656  37.5 -96.6-133.1-171.0  -21.7   22.6  -26.7
  318  453 E R  E     -K  324   0B 158      6,-0.2     6,-0.2    -2,-0.2     4,-0.0  -0.949  35.1-119.5-118.2 134.9  -24.4   23.6  -24.2
  319  454 E T  C >   -     0   0   29      4,-2.8     3,-2.7    -2,-0.4     4,-0.1  -0.217  43.2 -90.7 -65.2 165.3  -26.6   21.2  -22.4
  320  455 E A  T 3  S+     0   0  122      1,-0.3    -1,-0.1     2,-0.1    -2,-0.0   0.771 132.2  45.8 -47.6 -27.4  -30.3   21.4  -23.0
  321  456 E D  T 3  S-     0   0  122      2,-0.0    -1,-0.3     0, 0.0    -2,-0.1  -0.032 128.4 -91.1-110.0  30.5  -30.3   23.7  -20.0
  322  457 E G  S <  S+     0   0   47     -3,-2.7     2,-0.3     1,-0.2    -2,-0.1   0.723  74.9 143.9  73.3  25.7  -27.3   25.9  -20.9

FILE2.TXT

  140  361 B G  C           0   0   53      0, 0.0     2,-0.2     0, 0.0   -53,-0.1   0.000 360.0 360.0 360.0  97.3   25.2  -16.6   -6.6
  141  362 B A  C     -     0   0   70    -55,-0.1   -56,-0.0     1,-0.0     0, 0.0  -0.649 360.0 -33.9-148.3  84.1   28.0  -18.6   -4.8
  142  363 B R  C     -     0   0  224     -2,-0.2   -43,-0.1     1,-0.1   -55,-0.1   1.000  68.7-149.8  66.4  76.9   31.1  -16.5   -4.0
  143  364 B A  C     +     0   0   25    -57,-0.1     2,-0.3   -43,-0.1   -43,-0.2  -0.357  27.9 158.9 -70.8 155.7   31.6  -13.9   -6.7
  144  365 B H  B     -K   99   0B  51    -45,-2.7   -45,-2.6    -2,-0.0     2,-0.1  -0.947  40.1 -66.6-163.2-179.4   35.1  -12.8   -7.5
  145  366 B S  C           0   0   40     -2,-0.3   -47,-0.2   -47,-0.2   -44,-0.1  -0.467 360.0 360.0 -80.5 153.0   37.6  -11.3  -10.0
  146  367 B S  C           0   0   74     -2,-0.1    -1,-0.1   -46,-0.1   -48,-0.1   0.787 360.0 360.0 -70.5 360.0   38.5  -13.0  -13.3

file3.txt

  256 1417 C T  E     +f  129   0C  82     -2,-0.4     2,-0.3  -128,-0.2  -126,-0.2  -0.911  15.9 171.9-119.8 151.1   22.7   -9.3   39.1
  257 1418 C V  E     -f  130   0C  74   -128,-1.7  -126,-2.3    -2,-0.4     2,-0.3  -0.968  27.7-119.0-148.2 152.8   22.9  -13.0   38.4
  258 1419 C K  E     -f  131   0C 147     -2,-0.3     2,-0.5  -128,-0.2  -126,-0.2  -0.752  10.2-157.2-104.3 149.0   20.7  -15.8   39.6
  259 1420 C E  E     -f  132   0C  18   -128,-2.8  -126,-1.8    -2,-0.3  -125,-0.5  -0.987   9.4-178.4-118.2 119.5   21.4  -18.8   41.8

期望的输出

  file1   8/12= 0.67
  file2   6/7 = 0.86
  file3   0/4 = 0.00

Answer 1

使用awk

awk '$5~/(C|T|S)/{sum[FILENAME]++}{line[FILENAME]=FNR}
     END{for (i in line)
           printf "%s %s/%s = %.2f\n",i,sum[i]+0,line[i],sum[i]/line[i]
        }'  file*.txt

file1.txt 8/12 = 0.67
file2.txt 6/7 = 0.86
file3.txt 0/4 = 0.00

用awk做简单的计算

1 个答案: