3

我有一个具有以下结构的文本文件

ID,operator,a,b,c,d,true
WCBP12236,J1,75.7,80.6,65.9,83.2,82.1
WCBP12236,J2,76.3,79.6,61.7,81.9,82.1
WCBP12236,S1,77.2,81.5,69.4,84.1,82.1
WCBP12236,S2,68.0,68.0,53.2,68.5,82.1
WCBP12234,J1,63.7,67.7,72.2,71.6,75.3
WCBP12234,J2,68.6,68.4,41.4,68.9,75.3
WCBP12234,S1,81.8,82.7,67.0,87.5,75.3
WCBP12234,S2,66.6,67.9,53.0,70.7,75.3
WCBP12238,J1,78.6,79.0,56.2,82.1,84.1
WCBP12239,J2,66.6,72.9,79.5,76.6,82.1
WCBP12239,S1,86.6,87.8,23.0,23.0,82.1
WCBP12239,S2,86.0,86.9,62.3,89.7,82.1
WCBP12239,J1,70.9,71.3,66.0,73.7,82.1
WCBP12238,J2,75.1,75.2,54.3,76.4,84.1
WCBP12238,S1,65.9,66.0,40.2,66.5,84.1
WCBP12238,S2,72.7,73.2,52.6,73.9,84.1

每个ID对应于一个数据集，该数据集由操作员多次分析。即J1和J2是操作员 J 的第一次和第二次尝试。措施a, b,c, 和d种略有不同的算法来测量其真值位于列中的值true

我想做的是创建 3 个新的文本文件来比较J1vs J2、S1vsS2和J1vs的结果S1。J1vs的示例输出J2：

ID,operator,a1,a2,b1,b2,c1,c2,d1,d2,true
WCBP12236,75.7,76.3,80.6,79.6,65.9,61.7,83.2,81.9,82.1
WCBP12234,63.7,68.6,67.7,68.4,72.2,41.4,71.6,68.9,75.3

a1测量a在哪里J1等

另一个例子是S1vs S2：

ID,operator,a1,a2,b1,b2,c1,c2,d1,d2,true
WCBP12236,77.2,68.0,81.5,68.0,69.4,53.2,84.1,68.5,82.1
WCBP12234,81.8,66.6,82.7,67.9,67.0,53,87.5,70.7,75.3

ID 不会按字母数字顺序排列，操作员也不会为相同的 ID 聚集在一起。我不确定如何最好地完成这项任务 - 使用 linux 工具或 perl/python 之类的脚本语言。

我最初使用 linux 的尝试很快就碰壁了

首先找到所有唯一ID（排序）

awk -F, '/^WCBP/ {print $1}' file | uniq | sort -k 1.5n > unique_ids

遍历这些 ID 并排序J1，J2：

foreach i (`more unique_ids`)
    grep $i test.txt | egrep 'J[1-2]' | sort -t',' -k2
end

这给了我排序的数据

WCBP12234,J1,63.7,67.7,72.2,71.6,75.3
WCBP12234,J2,68.6,68.4,41.4,68.9,80.4
WCBP12236,J1,75.7,80.6,65.9,83.2,82.1
WCBP12236,J2,76.3,79.6,61.7,81.9,82.1
WCBP12238,J1,78.6,79.0,56.2,82.1,82.1
WCBP12238,J2,75.1,75.2,54.3,76.4,82.1
WCBP12239,J1,70.9,71.3,66.0,73.7,75.3
WCBP12239,J2,66.6,72.9,79.5,76.6,75.3

我不确定如何重新排列这些数据以获得所需的结构。我尝试awk在foreach循环中添加一个额外的管道awk 'BEGIN {RS="\n\n"} {print $1, $3,$10,$4,$11,$5,$12,$6,$13,$7}'

有任何想法吗？我确信这可以使用 以不那么繁琐的方式完成awk，尽管使用适当的脚本语言可能会更好。

4

3 回答 3

4

您可以使用 Perl csv 模块Text::CSV提取字段，然后将它们存储在哈希中，其中 ID 是主键，第二个字段是辅助键，所有字段都存储为值。然后，进行任何您想要的比较应该是微不足道的。如果要保留行的原始顺序，可以在第一个循环中使用数组。

use strict;
use warnings;
use Text::CSV;

my %data;
my $csv = Text::CSV->new({
            binary => 1,      # safety precaution
            eol    => $/,     # important when using $csv->print()
    });
while ( my $row = $csv->getline(*ARGV) ) {
    my ($id, $J) = @$row;   # first two fields
    $data{$id}{$J} = $row;  # store line
}
于 2013-06-24T10:10:34.833 回答
1

蟒蛇方式：

import os,sys, re, itertools
info=["WCBP12236,J1,75.7,80.6,65.9,83.2,82.1",
  "WCBP12236,J2,76.3,79.6,61.7,81.9,82.1",
  "WCBP12236,S1,77.2,81.5,69.4,84.1,82.1",
  "WCBP12236,S2,68.0,68.0,53.2,68.5,82.1",
  "WCBP12234,J1,63.7,67.7,72.2,71.6,75.3",
  "WCBP12234,J2,68.6,68.4,41.4,68.9,80.4",
  "WCBP12234,S1,81.8,82.7,67.0,87.5,75.3",
  "WCBP12234,S2,66.6,67.9,53.0,70.7,72.7",
  "WCBP12238,J1,78.6,79.0,56.2,82.1,82.1",
  "WCBP12239,J2,66.6,72.9,79.5,76.6,75.3",
  "WCBP12239,S1,86.6,87.8,23.0,23.0,82.1",
  "WCBP12239,S2,86.0,86.9,62.3,89.7,82.1",
  "WCBP12239,J1,70.9,71.3,66.0,73.7,75.3",
  "WCBP12238,J2,75.1,75.2,54.3,76.4,82.1",
  "WCBP12238,S1,65.9,66.0,40.2,66.5,80.4",
  "WCBP12238,S2,72.7,73.2,52.6,73.9,72.7" ]

def extract_data(operator_1, operator_2):
    operator_index=1
    id_index=0
    data={}
    result=[]
    ret=[]
    for line in info:
        conv_list=line.split(",")
        if len(conv_list) > operator_index and ((operator_1.strip().upper() == conv_list[operator_index].strip().upper()) or (operator_2.strip().upper() == conv_list[operator_index].strip().upper()) ):
            if data.has_key(conv_list[id_index]):
                iters = [iter(conv_list[int(operator_index)+1:]), iter(data[conv_list[id_index]])]
                data[conv_list[id_index]]=list(it.next() for it in itertools.cycle(iters))
                continue
            data[conv_list[id_index]]=conv_list[int(operator_index)+1:]
    return data

ret=extract_data("j1", "s2")
print ret

输出/输出：

{'WCBP12239'：['70.9'，'86.0'，'71.3'，'86.9'，'66.0'，'62.3'，'73.7'，'89.7'，'75.3'，'82.1']，'WCBP12238' ：['72.7'，'78.6'，'73.2'，'79.0'，'52.6'，'56.2'，'73.9'，'82.1'，'72.7'，'82.1']，'WCBP12234'：['66.6 '，'63.7'，'67.9'，'67.7'，'53.0'，'72.2'，'70.7'，'71.6'，'72.7'，'75.3']，'WCBP12236'：['68.0'，'75.7 '，'68.0'，'80.6'，'53.2'，'65.9'，'68.5'，'83.2'，'82.1'，'82.1']}

于 2013-06-24T12:07:24.883 回答
1

我没有像 TLP 那样使用 Text::CSV。如果你需要你可以，但对于这个例子，我认为由于字段中没有嵌入的逗号，我对“，”做了一个简单的拆分。此外，列出了两个运算符的真实字段（而不仅仅是 1），因为我认为最后一个值的特殊情况会使解决方案复杂化。

#!/usr/bin/perl
use strict;
use warnings;
use List::MoreUtils qw/ mesh /;

my %data;

while (<DATA>) {
    chomp;
    my ($id, $op, @vals) = split /,/;
    $data{$id}{$op} = \@vals;
}

my @ops = ([qw/J1 J2/], [qw/S1 S2/], [qw/J1 S1/]);

for my $id (sort keys %data) {
    for my $comb (@ops) {
        open my $fh, ">>", "@$comb.txt" or die $!;
        my $a1 = $data{$id}{ $comb->[0] };
        my $a2 = $data{$id}{ $comb->[1] };
        print $fh join(",", $id, mesh(@$a1, @$a2)), "\n";
        close $fh or die $!;
    }   
}

__DATA__
WCBP12236,J1,75.7,80.6,65.9,83.2,82.1
WCBP12236,J2,76.3,79.6,61.7,81.9,82.1
WCBP12236,S1,77.2,81.5,69.4,84.1,82.1
WCBP12236,S2,68.0,68.0,53.2,68.5,82.1
WCBP12234,J1,63.7,67.7,72.2,71.6,75.3
WCBP12234,J2,68.6,68.4,41.4,68.9,75.3
WCBP12234,S1,81.8,82.7,67.0,87.5,75.3
WCBP12234,S2,66.6,67.9,53.0,70.7,75.3
WCBP12239,J1,78.6,79.0,56.2,82.1,82.1
WCBP12239,J2,66.6,72.9,79.5,76.6,82.1
WCBP12239,S1,86.6,87.8,23.0,23.0,82.1
WCBP12239,S2,86.0,86.9,62.3,89.7,82.1
WCBP12238,J1,70.9,71.3,66.0,73.7,84.1
WCBP12238,J2,75.1,75.2,54.3,76.4,84.1
WCBP12238,S1,65.9,66.0,40.2,66.5,84.1
WCBP12238,S2,72.7,73.2,52.6,73.9,84.1

产生的输出文件如下

J1 J2.txt

WCBP12234,63.7,68.6,67.7,68.4,72.2,41.4,71.6,68.9,75.3,75.3
WCBP12236,75.7,76.3,80.6,79.6,65.9,61.7,83.2,81.9,82.1,82.1
WCBP12238,70.9,75.1,71.3,75.2,66.0,54.3,73.7,76.4,84.1,84.1
WCBP12239,78.6,66.6,79.0,72.9,56.2,79.5,82.1,76.6,82.1,82.1

S1 S2.txt

WCBP12234,81.8,66.6,82.7,67.9,67.0,53.0,87.5,70.7,75.3,75.3
WCBP12236,77.2,68.0,81.5,68.0,69.4,53.2,84.1,68.5,82.1,82.1
WCBP12238,65.9,72.7,66.0,73.2,40.2,52.6,66.5,73.9,84.1,84.1
WCBP12239,86.6,86.0,87.8,86.9,23.0,62.3,23.0,89.7,82.1,82.1

J1 S1.txt

WCBP12234,63.7,81.8,67.7,82.7,72.2,67.0,71.6,87.5,75.3,75.3
WCBP12236,75.7,77.2,80.6,81.5,65.9,69.4,83.2,84.1,82.1,82.1
WCBP12238,70.9,65.9,71.3,66.0,66.0,40.2,73.7,66.5,84.1,84.1
WCBP12239,78.6,86.6,79.0,87.8,56.2,23.0,82.1,23.0,82.1,82.1

更新：为了只得到 1 个真值，for 循环可以这样写：

for my $id (sort keys %data) {
    for my $comb (@ops) {
        local $" = '';
        open my $fh, ">>", "@$comb.txt" or die $!;
        my $a1 = $data{$id}{ $comb->[0] };
        my $a2 = $data{$id}{ $comb->[1] };
        pop @$a2;
        my @mesh = grep defined, mesh(@$a1, @$a2);
        print $fh join(",", $id, @mesh), "\n";
        close $fh or die $!;
    }   
}

更新：为 grep expr 中的测试添加了“已定义”。因为它是正确的方法（而不是仅仅测试'$_'，它可能是 0 并且被 grep 错误地排除在列表中）。

于 2013-06-24T19:04:22.467 回答