Python实现简单过滤文本段的方法

Python  /  管理员 发布于 7年前   189

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/pythonsum = 0sumdata = []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/perlopen(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。