Question

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对应一个由操作员多次分析的数据集。即J1和J2是运营商J的第一次和第二次尝试。措施a，b，c和d略微使用4用于测量值的唯一算法，其值的真值位于列true

中

我想要做的是创建3个新文本文件，比较J1与J2，S1 vs S2和J1 vs的结果S1。 J1与J2的示例输出：

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

其中a1是a的衡量标准J1等等。

另一个例子是S1 vs 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不是按字母数字顺序排列的，也不会为相同的ID对运算符进行聚类。我不确定如何最好地完成这项任务 - 使用linux工具或像perl / python这样的脚本语言。

我最初尝试使用linux快速打砖墙

首先找到所有唯一ID（已排序）

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

循环浏览这些ID并对J1，J2：

进行排序

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。我尝试在awk循环foreach

中向awk 'BEGIN {RS="\n\n"} {print $1, $3,$10,$4,$11,$5,$12,$6,$13,$7}'添加其他管道

有什么想法吗？我确信这可以使用awk以不那么麻烦的方式完成，尽管使用适当的脚本语言可能会更好。

Answer 1

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，做任何你想要的比较都应该是微不足道的。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
}

Answer 2

Python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

O / P：

{'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']}

Answer 3

我没有像TLP那样使用Text :: CSV。如果你需要，但是对于这个例子，我想因为字段中没有嵌入的逗号，我对'，'进行了简单的拆分。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

生成的输出文件位于

之下

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：要获得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test添加了'defined'。因为它是正确的方式（而不仅仅是测试'$ _'，它可能是0并且被grep错误地排除在列表中）。

将线条与匹配的键组合在一起

3 个答案: