作者rom1025 (περ'ιπεπια)
看板Perl
标题Re: [问题] 如何下 statics-regression 的参数
时间Tue Dec 20 09:25:35 2011
※ 引述《gilberthsu (雪循)》之铭言：
: 因为有 700x20000 以上的资料需要作 regression
: 资料格式是 .csv 档
: expression,Ctcf_0,H2az_0, ...
: 1.639966339,0,13.08689976, ...
: 0.278085646,16.25440025,12.86340046, ...
: -1.070784985,0,23.0781002, ...
: 0.790170416,0,12.53590012, ...
: -1.837946653,0,14.08780003, ...
: ...
: 因此没办法用作者的范例方法，一个一个 key 进去
: 後来看了 package 内发现可以用 array 的方式丢参数
: my $reg = Statistics::Regression->new($regname, \@var_names)
: 不过怎麽测试都测不出来要怎麽放参数
: 以下是我试着写的 code
: #!/usr/bin/perl -w
: use Statistics::Regression;
: my $reg = Statistics::Regression->new($expr_data, \@genename);
: open training, "training.csv";
: $genename = <training>;
: chomp $genename;
: @genename = split /,/, $genename; # 1st row is gene's name
: $expr = shift @genename; #1st data is sample expression
: unshift @genename, "const"; #put const column in array
: while (<training>){
: chomp;
: @train = split /,/, $_;
: $expr_data = shift @train;
: unshift @train, 1.0;
: my $reg->include( $expr_data, \@train );
: }
: $reg->print();
: close(training);
: 感谢大家指点
#!/usr/bin/perl -w
use Statistics::Regression;
open training, "training.csv";
$genename = <training>;
chomp $genename;
@genename = split /,/, $genename; # 1st row is gene's name
$expr = shift @genename; #1st data is sample expression
unshift @genename, "const"; #put const column in array
my $reg = Statistics::Regression->new($expr,[@genename]);#取得$expr跟@genename
while (<training>){
chomp;
@train = split /,/, $_;
$expr_data = shift @train;
unshift @train, 1.0;
$reg->include( $expr_data, [@train] ); #去除my，@train改成AOA型式
}
$reg->print();
close(training);
--
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1F：推 gilberthsu:感谢！原来只需要用 AOA ... 12/20 12:19