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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：
? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 75 76 77## 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：
? 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20#!/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
? 1 2 3 4 5 6 7 8 9 10 11 12 13 14#!/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);
希望本文所述对大家Python程序设计有所帮助。