如何在不排序的情况下删除两个文件之间的公共行? [英] How to remove common lines between two files without sorting?

问题描述

我有两个未排序的文件，它们有一些共同点.

I have two files not sortered which have some lines in common.

file1.txt

file1.txt

Z
B
A
H
L

file2.txt

file2.txt

S
L
W
Q
A

我用于删除常见行的方法如下:

The way I'm using to remove common lines is the following:

sort -u file1.txt > file1_sorted.txt
sort -u file2.txt > file2_sorted.txt

comm -23 file1_sorted.txt file2_sorted.txt > file_final.txt

输出:

B
H
Z

问题是我想保持file1.txt的顺序，我的意思是:

The problem is that I want to keep the order of file1.txt, I mean:

所需的输出:

Z
B
H

我最坚决的一种解决方案是循环读取file2.txt的所有行，以及:

One solution I tought is doing a loop to read all the lines of file2.txt and:

sed -i '/^${line_file2}$/d' file1.txt

但是，如果文件很大，性能可能会降低.

But if files are big the performance may suck.

• 你喜欢我的主意吗?
• 您有其他选择吗?

推荐答案

grep或awk:

awk 'NR==FNR{a[$0]=1;next}!a[$0]' file2 file1

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