創建文件迅速從單個列

我有一個看起來像這樣的數據：創建文件迅速從單個列

-1 1:-0.394668 2:-0.794872 3:-1 4:-0.871341 5:0.9365 6:0.75597 
1 1:-0.463641 2:-0.897436 3:-1 4:-0.871341 5:0.44378 6:0.121824 
1 1:-0.469432 2:-0.897436 3:-1 4:-0.871341 5:0.32668 6:0.302529 
-1 1:-0.241547 2:-0.538462 3:-1 4:-0.871341 5:0.9994 6:0.987166 
1 1:-0.757233 2:-0.948718 3:-1 4:-0.871341 5:-0.33904 6:0.915401 
1 1:-0.167147 2:-0.589744 3:-1 4:-0.871341 5:0.95078 6:0.991566

第一列是一流的，而接下來的6列功能。我想爲個別功能創建6個文件。例如

my_input_feat1.txt將包含

-1 1:-0.394668 
    1 1:-0.463641 
    ... 
    1 1:-0.757233 
    1 1:-0.167147

my_input_feat2.txt將包含

-1 2:-0.794872 
... 
1 2:-0.589744

等。我有一個Perl代碼，但它的速度非常慢。有沒有更快的方法？通常輸入文件將包含100K行。

use strict; 
use Data::Dumper; 
use Carp; 
my $input = $ARGV[0] || "myinput.txt"; 




my $INFILE_file_name = $input;  # input file name 

open (INFILE, '<', $INFILE_file_name) 
    or croak "$0 : failed to open input file $INFILE_file_name : $!\n"; 

    my $out1 = $input."_feat_1.txt"; 
    my $out2 = $input."_feat_2.txt"; 
    my $out3 = $input."_feat_3.txt"; 
    my $out4 = $input."_feat_4.txt"; 
    my $out5 = $input."_feat_5.txt"; 
    my $out6 = $input."_feat_6.txt"; 

    unlink($out1); 
    unlink($out2); 
    unlink($out3); 
    unlink($out4); 
    unlink($out5); 
    unlink($out6); 

    print "$out1\n"; 

while (<INFILE>) { 
    chomp; 
    my @els = split(/\s+/,$_); 
    my $lbl = $els[0]; 

    my $OUTFILE1_file_name = $out1;  # output file name 
    open (OUTFILE1, '>>', $OUTFILE1_file_name) 
     or croak "$0 : failed to open output file $OUTFILE1_file_name : $!\n"; 
    print OUTFILE1 "$lbl $els[1]\n"; 
    close (OUTFILE1);   # close output file 

    my $OUTFILE2_file_name = $out2;  # output file name 
    open (OUTFILE2, '>>', $OUTFILE2_file_name) 
     or croak "$0 : failed to open output file $OUTFILE2_file_name : $!\n"; 
    print OUTFILE2 "$lbl $els[2]\n"; 
    close (OUTFILE2);   # close output file 

    # Etc.. until OUTFILE 6 

} 

close (INFILE);

來源

2011-01-11 neversaint

#!/usr/bin/sh 

for i in `seq 1 $1`; do 
    cut -f1,$i $2 > ${2}_$i; 
done

或

#!/usr/bin/perl 

use warnings; use strict; 

my $input_file = $ARGV[0]; 
my %handles; 

while (<>) { 
    my ($class, @features) = split /\s+/; 

    for my $i (1 .. @features) { 
     open $handles{$i}, '>', $input_file . "_$i" or die $! 
     unless exists $handles{$i}; 

     print {$handles{$i}} join(' ', $class, $features[$i - 1]), "\n";  
    } 
} 

while (my (undef, $handle) = each %handles) { 
    close $handle or die $!; 
}

來源

2011-01-11 08:29:07

shell腳本是否正常？

awk '{print $1" "$2}' data.txt > feat1_file.txt 
awk '{print $1" "$3}' data.txt > feat2_file.txt 
awk '{print $1" "$4}' data.txt > feat3_file.txt 
awk '{print $1" "$5}' data.txt > feat4_file.txt 
awk '{print $1" "$6}' data.txt > feat5_file.txt 
awk '{print $1" "$7}' data.txt > feat6_file.txt

來源

2011-01-11 07:52:58 eumiro

謝謝。但我想有一個輸入名稱的變量。這樣我可以根據不同的輸入名稱動態創建這樣的文件。請注意，功能的數量可以大於/小於6. – neversaint 2011-01-11 07:55:53

@neversaint：將上面的文件名抽象出來放到shell變量中，並將其全部放入Bash腳本中應該很容易。只需將data.txt替換爲$（FILENAME），並確保在調用awk之前設置FILENAME。 – unwind 2011-01-11 07:58:29

你應該移動打開/關閉輸出文件while循環之外。

來源

2011-01-11 08:48:28 Toto

創建文件迅速從單個列

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