庆典通过匹配列合并文件 [英] bash merge files by matching columns

问题描述

我有两个文件：

 File1

 12    abc
 34    cde
 42    dfg
 11    df
 9     e   


 File2

 23    abc
 24    gjr
 12    dfg
 8     df

我想通过柱合并文件列（如果塔2是相同的）为这样的输出：

I want to merge files column by column (if column 2 is the same) for the output like this:

  File1  File2
   12    23    abc
   42    12    dfg
   11    8     df
   34    NA    cde
   9     NA    e
   NA    24    gjr

我怎样才能做到这一点？

How can I do this?

我想它是这样的：

 cat File* >> tmp; sort tmp | uniq -c | awk '{print $2}' > column2; for i in
 $(cat column2); do grep -w "$i" File*

但是，这是我在哪里卡住了...结果
不知道怎么greping我应该列＆放文件合并后列;写NA值在哪里丢失。

But this is where I am stuck...
Don't know how after greping I should combine files column by column & write NA where value is missing.

希望有人能够帮助我。

Hope someone could help me with this.

推荐答案

由于我是用庆典 3.2运行为 SH （它没有进程替换为 SH ），我用了两个临时文件来准备使用数据与加入：

Since I was testing with bash 3.2 running as sh (which does not have process substitution as sh), I used two temporary files to get the data ready for use with join:

$ sort -k2b File2 > f2.sort
$ sort -k2b File1 > f1.sort
$ cat f1.sort
12    abc
34    cde
11    df
42    dfg
9     e  
$ cat f2.sort
23    abc
8     df
12    dfg
24    gjr
$ join -1 2 -2 2 -o 1.1,2.1,0 -a 1 -a 2 -e NA f1.sort f2.sort
12 23 abc
34 NA cde
11 8 df
42 12 dfg
9 NA e
NA 24 gjr
$

使用过程中替换，你可以写：

With process substitution, you could write:

join -1 2 -2 2 -o 1.1,2.1,0 -a 1 -a 2 -e NA <(sort -k2b File1) <(sort -k2b File2)

如果你想以不同格式的数据，使用 AWK 来后处理的输出：

If you want the data formatted differently, use awk to post-process the output:

$ join -1 2 -2 2 -o 1.1,2.1,0 -a 1 -a 2 -e NA f1.sort f2.sort |
> awk '{ printf "%-5s %-5s %s\n", $1, $2, $3 }'
12    23    abc
34    NA    cde
11    8     df
42    12    dfg
9     NA    e
NA    24    gjr
$

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