欢迎您访问程序员文章站本站旨在为大家提供分享程序员计算机编程知识!
您现在的位置是: 首页  >  IT编程

Python实现简单过滤文本段的方法

程序员文章站 2024-04-03 12:32:52
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=39...

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);

更多关于Python相关内容感兴趣的读者可查看本站专题:《Python数据结构与算法教程》、《Python函数使用技巧总结》、《Python字符串操作技巧汇总》、《Python入门与进阶经典教程》及《Python文件与目录操作技巧汇总

希望本文所述对大家Python程序设计有所帮助。