[英]How to use awk to print counter+1 if line does not match previous line
I have been working in bash trying print text based on an input file. 我一直在bash中尝试根据输入文件打印文本。 My file contains the following text:
我的文件包含以下文本:
a
a
a
b
c
c
c
d
d
e
e
For each line that does not match the previous line, I want to add 1 to a counter and print the counter next to original entry. 对于与上一行不匹配的每一行,我想在计数器上加1并在原始条目旁边打印计数器。 For lines that are equal, I would like to print the counter as is, like so:
对于相等的行,我想按原样打印计数器,如下所示:
a 1
a 1
a 1
b 2
c 3
c 3
c 3
d 4
d 4
e 5
e 5
I have tried the code below, but this only provides me with information for the lines that are equal to the previous lines and does not print the counter. 我已经尝试了下面的代码,但这仅向我提供了与前几行相同的信息,并且不打印计数器。
f=a
counter=1
awk '{
if ($0==f && NR>1) {print f, counter} {f=$0} ;
next
elif ($0!=f && NR>1)
{print f, ++counter} {f=$0}
}' file.txt
output: 输出:
a
a
c
c
d
e
This is all you need 这就是你所需要的
awk '
$1 != prev {c++; prev=$1}
{print $0, c}
' file
On the first line of the file, both prev
and c
will be undefined. 在文件的第一行,
prev
和c
均未定义。 The $1 != prev
test will be true. $1 != prev
测试将为true。 awk will treat undefined c
as the number zero in arithmetic context, so the ++
operator properly sets c to 1. 在算术上下文中,awk将未定义的
c
视为数字零,因此++
运算符将c正确设置为1。
Two place you're going wrong: 您出错的两个地方:
if
command: you have way too many braces. if
命令的语法:大括号过多。 This is your code restructured so that it works: 这是对您的代码进行重组,以使其起作用:
awk -v f=a -v counter=1 '{
if ($0==f && NR>1) {print f, counter; f=$0 ; next}
if ($0!=f && NR>1) {print f, ++counter; f=$0}
}' file
But it's written in a non-idiomatic way, and is pretty repetitive. 但是它是以非惯用的方式编写的,并且非常重复。 Plus you need to know the first line of the file.
另外,您需要知道文件的第一行。
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