Python实现简单过滤文本段的方法

  • Z时代
  • 2024-01-10
  • 分类：综合

本文实例讲述了Python实现简单过滤文本段的方法。分享给大家供大家参考，具体如下：

一、问题：

如下文本：

## Alignment 0: score=397.0 e_value=8.2e-18 N=9 scaffold1&scaffold106 minus
 0- 0:  10026549  10007782   2e-75
 0- 1:  10026550  10007781   8e-150
 0- 2:  10026552  10007780   1e-116
 0- 3:  10026555  10007778    0
 0- 4:  10026570  10007768    0
 0- 5:  10026579  10007758   4e-15
 0- 6:  10026581  10007738   2e-44
 0- 7:  10026587  10007734   9e-145
 0- 8:  10026591  10007732   2e-147
## Alignment 1: score=2304.0 e_value=1e-164 N=47 scaffold1&scaffold107 minus
 1- 0:  10026836  10007942   2e-84
 1- 1:  10026839  10007940    0
 1- 2:  10026840  10007938    0
 1- 3:  10026842  10007937   9e-82
 1- 4:  10026843  10007935   7e-79
 1- 5:  10026847  10007933   3e-119
 1- 6:  10026850  10007932   2e-87
 1- 7:  10026854  10007928   5e-22
 1- 8:  10026855  10007927   3e-101
 1- 9:  10026856  10007925   1e-106
 1- 10:  10026857  10007924    0
 1- 11:  10026858  10007922   9e-123
 1- 12:  10026859  10007921   1e-80
 1- 13:  10026860  10007920   8e-104
 1- 14:  10026862  10007918   4e-25
 1- 15:  10026863  10007917    0
 1- 16:  10026864  10007912   4e-40
 1- 17:  10026865  10007911    0
 1- 18:  10026866  10007910   7e-122
 1- 19:  10026867  10007908   2e-25
 1- 20:  10026868  10007907    0
 1- 21:  10026869  10007905    0
 1- 22:  10026870  10007904   3e-150
 1- 23:  10026871  10007903   5e-77
 1- 24:  10026874  10007901    0
 1- 25:  10026875  10007897    0
 1- 26:  10026876  10007896    0
 1- 27:  10026877  10007894    0
 1- 28:  10026880  10007893   3e-52
 1- 29:  10026881  10007892    0
 1- 30:  10026882  10007891    0
 1- 31:  10026883  10007890    0
 1- 32:  10026886  10007889   1e-50
 1- 33:  10026887  10007888   6e-157
 1- 34:  10026888  10007887    0
 1- 35:  10026889  10007884    0
 1- 36:  10026890  10007883   2e-18
 1- 37:  10026891  10007882   9e-64
 1- 38:  10026892  10007881    0
 1- 39:  10026895  10007880    0
 1- 40:  10026898  10007875    0
 1- 41:  10026900  10007874    0
 1- 42:  10026901  10007873    0
 1- 43:  10026902  10007871   2e-123
 1- 44:  10026903  10007870    0
 1- 45:  10026905  10007869    0
 1- 46:  10026909  10007868   1e-81
## Alignment 2: score=811.0 e_value=3.3e-43 N=17 scaffold1&scaffold111 minus
 2- 0:  10026595  10007449   6e-40
 2- 1:  10026599  10007448   4e-90
 2- 2:  10026600  10007447    0
 2- 3:  10026601  10007444   9e-55
 2- 4:  10026603  10007438   4e-78
 2- 5:  10026604  10007434   9e-122
 2- 6:  10026606  10007432   2e-162
 2- 7:  10026607  10007427    0
 2- 8:  10026608  10007426    0
 2- 9:  10026612  10007417    0
 2- 10:  10026613  10007415   8e-128
 2- 11:  10026614  10007414   3e-64
 2- 12:  10026615  10007409    0
 2- 13:  10026616  10007406    0
 2- 14:  10026617  10007403   1e-171
 2- 15:  10026618  10007402    0
 2- 16:  10026619  10007397   7e-18
........

要求：如果Alignment后面少于20行，把整个的去掉

二、实现方法：

python代码：

#!/usr/bin/python
sum = 0
sumdata = []
FD = open("/root/data.txt","r")
line = FD.readline()
while line:
 if line.find("Alignment") == 3:
 if sum >= 20:
 for i in sumdata:
 print i,
 sum=0
 sumdata=[line]
 else:
 sum = sum + 1
 sumdata.append(line)
 line=FD.readline()
 if len(line) == 0:
 if sum >= 20:
 for i in sumdata:
 print i,

附：

perl代码

#!/usr/bin/perl
open(FD,"/root/data.txt");
while (){
  if ($_ =~ /Alignment/){
    if($sum >= 20){
      print @sumdata;}
    $sum=0;
    @sumdata=($_);}
  else{
    $sum++;
    push(@sumdata,$_);}
}
print @sumdata if $sum >=20;
close(FD);

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希望本文所述对大家Python程序设计有所帮助。

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