基於相同的鍵摺疊行

Question

我想根據第一列的相等性摺疊行。然後將第二列的內容添加到新的摺疊表中，以逗號分隔並添加額外空間。另外，如果第二列的內容相同，則摺疊它們，也就是說，如果輸出文件中出現兩次「非劇毒」，則只顯示一次。

我在這裏很新，請解釋如何運行它。希望任何人都可以幫助我！

輸入（製表符分隔）：

HS372_01446 non-virulent 
HS372_01446 non-virulent 
HS372_01446 lung 
HS372_00498 non-virulent 
HS372_00498 non-virulent 
HS372_00498 non-virulent 
HS372_00498 lung 
HS372_00498 lung 
HS372_00954 jointlungCNS 
HS372_00954 non-virulent 
HS372_00954 non-virulent 
HS372_00954 moderadamentevirulenta(nose) 
HS372_00954 lung 

希望的輸出（製表符分隔）：

HS372_01446 non-virulent, lung 
HS372_00498 non-virulent, lung 
HS372_00954 jointlungCNS, non-virulent, moderadamentevirulenta(nose), lung 

Answer 1

的Perl

從命令行，

perl -lane' 
    ($n, $p) =@F; 
    $s{$n}++ or push @r, $n; 
    $c{$n}{$p}++ or push @{$h{$n}}, $p; 
    END { 
    $" = ",\t"; 
    print "$_\t@{$h{$_}}" for @r; 
    } 
' file 

輸出

HS372_01446  non-virulent, lung 
HS372_00498  non-virulent, lung 
HS372_00954  jointlungCNS, non-virulent, moderadamentevirulenta(nose), lung 

Answer 2

from collections import defaultdict 

a = """HS372_01446 non-virulent 
HS372_01446 non-virulent 
HS372_01446 lung 
HS372_00498 non-virulent 
HS372_00498 non-virulent 
HS372_00498 non-virulent 
HS372_00498 lung 
HS372_00498 lung 
HS372_00954 jointlungCNS 
HS372_00954 non-virulent 
HS372_00954 non-virulent 
HS372_00954 moderadamentevirulenta(nose) 
HS372_00954 lung""".split("\n") 

stuff = defaultdict(set) 

for line in a: 
    uid, symp = line.split(" ") 
    stuff[uid].add(symp) 

for uid, symps in stuff.iteritems(): 
    print "%s %s" % (uid, ", ".join(list(symps))) 

Answer 3

在perl的：

use warnings; 
use strict; 

open my $input, '<', 'in.txt'; 

my %hash; 
while (<$input>){ 
    chomp; 
    my @split = split(' '); 
    $hash{$split[0]}{$split[1]} = 1; 
} 

for my $key (keys %hash){ 
    print "$key\t"; 
     for my $info (keys $hash{$key}){ 
      print "$info\t"; 
     } 
    print "\n"; 
} 

哪個打印：

HS372_01446 non-virulent lung  
HS372_00954 non-virulent moderadamentevirulenta(nose) jointlungCNS lung  
HS372_00498 non-virulent lung 

Answer 4

另一個Perl的溶液：

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

my %hash; 
while (<DATA>) 
{ 
    chomp; 
    my ($key, $value) = split; 
    push @{$hash{$key}}, $value; 
} 

while (my ($key, $values) = each %hash) 
{ 
    print "$key\t", join ', ', uniq @$values, "\n"; 
} 

__DATA__ 
HS372_01446 non-virulent 
HS372_01446 non-virulent 
HS372_01446 lung 
HS372_00498 non-virulent 
HS372_00498 non-virulent 
HS372_00498 non-virulent 
HS372_00498 lung 
HS372_00498 lung 
HS372_00954 jointlungCNS 
HS372_00954 non-virulent 
HS372_00954 non-virulent 
HS372_00954 moderadamentevirulenta(nose) 
HS372_00954 lung 

Answer 5

爪哇：

javac的Collapse.java

的Java收起input.txt中

import java.io.*; 
import java.util.*; 

public class Collapse { 

    public static void main(String[] args) throws Exception { 
     BufferedReader br = new BufferedReader(new InputStreamReader(new FileInputStream(args[0]))); 

     Map<String, Set<String>> output = new HashMap<String, Set<String>>(); 
     String line; 
     while ((line = br.readLine()) != null) { 
      StringTokenizer st = new StringTokenizer(line, "\t"); 
      String key = st.nextToken(); 
      Set<String> set = output.get(key); 
      if (set == null) { 
       output.put(key, set = new LinkedHashSet<String>()); 
      } 
      set.add(st.nextToken()); 
     } 

     for (String key : output.keySet()) { 
      StringBuilder sb = new StringBuilder(); 
      for (String value : output.get(key)) { 
       if (sb.length() != 0) sb.append(", "); 
       sb.append(value); 
      } 
      System.out.println(key + "\t" + sb); 
     } 
    } 
} 

Answer 6

如果您的數據來自一個MySQL數據庫（你可以將它導入一個），你可以使用group_concat操作。

看到這個答案 Can I concatenate multiple MySQL rows into one field?

這目前標有431個upvotes，所以你的問題是一個非常普遍的問題，答案顯示出非常優雅的解決方案。

Answer 7

這確實你問什麼，除了保持在相同的順序，它們出現在文件中的ID和描述中，如果該事項：

use strict; 
use warnings; 

open my $fh, '<', 'diseases.txt'; 

my %diseases; 
my @ids; 

while (<$fh>) { 
    my ($id, $desc) = split; 
    if (not $diseases{$id}) { 
    $diseases{$id}{list} = [$desc]; 
    $diseases{$id}{seen}{$desc} = 1; 
    push @ids, $id; 
    } 
    elsif (not $diseases{$id}{seen}{$desc}) { 
    push @{ $diseases{$id}{list} }, $desc; 
    $diseases{$id}{seen}{$desc} = 1; 
    } 
} 

for my $id (@ids) { 
    printf "%s %s\n", $id, join ', ', @{ $diseases{$id}{list} }; 
} 

輸出

HS372_01446 non-virulent, lung 
HS372_00498 non-virulent, lung 
HS372_00954 jointlungCNS, non-virulent, moderadamentevirulenta(nose), lung 

Answer 8

解析文本文件的標準UNIX工具AWK：

$ awk '!seen[$1,$2]++{a[$1]=(a[$1] ? a[$1]", " : "\t") $2} END{for (i in a) print i a[i]}' file 
HS372_00498  non-virulent, lung 
HS372_00954  jointlungCNS, non-virulent, moderadamentevirulenta(nose), lung 
HS372_01446  non-virulent, lung