Linux 加入多个文件
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join multiple files
提问by prathmesh.kallurkar
I am using the standard join command to join two sorted files based on column1. The command is simple join file1 file2 > output_file.
我正在使用标准 join 命令连接基于 column1 的两个排序文件。该命令是简单的 join file1 file2 > output_file。
But how do I join 3 or more files using the same technique ? join file1 file2 file3 > output_file Above command gave me an empty file. I think sed can help me but I am not too sure how ?
但是如何使用相同的技术连接 3 个或更多文件？join file1 file2 file3 > output_file 上面的命令给了我一个空文件。我认为 sed 可以帮助我，但我不太确定如何？
采纳答案by mata
man join:
man join：
NAME
join - join lines of two files on a common field
SYNOPSIS
join [OPTION]... FILE1 FILE2
it only works with twofiles.
它只适用于两个文件。
if you need to join three, maybe you can first join the first two, then join the third.
如果你需要加入三个，也许你可以先加入前两个，然后加入第三个。
try:
尝试：
join file1 file2 | join - file3 > output
that should join the three files without creating an intermediate temp file. -tells the join command to read the first input stream from stdin
应该在不创建中间临时文件的情况下加入三个文件。-告诉 join 命令从中读取第一个输入流stdin
回答by prathmesh.kallurkar
Join joins lines of two fileson a common field. If you want to join more - do it in pairs. Join first two files first, then join the result with a third file etc.
Join在一个公共字段上连接两个文件的行。如果你想加入更多 - 成对进行。先连接前两个文件，然后将结果与第三个文件连接，依此类推。
回答by Gnosophilon
The manpage of joinstates that it only works for two files. So you need to create and intermediate file, which you delete afterwards, i.e.:
该man页面join声明它仅适用于两个文件。因此，您需要创建中间文件，然后将其删除，即：
> join file1 file2 > temp
> join temp file3 > output
> rm temp
回答by ack
One can join multiple files (N>=2) by constructing a pipeline of joins recursively:
可以通过join递归构造s的管道来连接多个文件 (N>=2) ：
#!/bin/sh
# multijoin - join multiple files
join_rec() {
if [ $# -eq 1 ]; then
join - ""
else
f=; shift
join - "$f" | join_rec "$@"
fi
}
if [ $# -le 2 ]; then
join "$@"
else
f1=; f2=; shift 2
join "$f1" "$f2" | join_rec "$@"
fi
回答by rsz
I know this is an old question but for future reference.
If you know that the files you want to join have a pattern like in the question here e.g. file1 file2 file3 ... fileNThen you can simply join them with this command
我知道这是一个老问题，但供将来参考。如果您知道要加入的文件具有与此处问题类似的模式，例如，file1 file2 file3 ... fileN那么您可以简单地使用此命令加入它们
cat file* > output
Where output will be the series of the joined files which were joined in alphabetical order.
其中输出将是按字母顺序连接的连接文件系列。
回答by gmargari
I created a function for this. First argument is the output file, rest arguments are the files to be joined.
我为此创建了一个函数。第一个参数是输出文件，其余参数是要加入的文件。
function multijoin() {
out=
shift 1
cat | awk '{print }' > $out
for f in $*; do join $out $f > tmp; mv tmp $out; done
}
Usage:
用法：
multijoin output_file file*
回答by kvantour
While a bit an old question, this is how you can do it with a single awk:
虽然有点老问题，但这是您如何使用单个awk：
awk -v j=<field_number> '{key=$j; $j=""} # get key and delete field j
(NR==FNR){order[FNR]=key;} # store the key-order
{entry[key]=entry[key] OFS ~$ cat A.txt
x1 2
x2 3
x4 5
x5 8
~$ cat B.txt
x1 5
x2 7
x3 4
x4 6
~$ cat C.txt
x2 1
x3 1
x4 1
x5 1
~$ cat D.txt
x1 1
} # update key-entry
END { for(i=1;i<=FNR;++i) {
key=order[i]; print key entry[key] # print
}
}' file1 ... filen
This script assumes:
该脚本假设：
- all files have the same amount of lines
- the order of the output is the same order of the first file.
- files do not need to be sorted in field
<field_number> <field_number>is a valid integer.
- 所有文件的行数相同
- 输出的顺序与第一个文件的顺序相同。
- 文件不需要在字段中排序
<field_number> <field_number>是一个有效的整数。
回答by user3200815
Assuming you have four files A.txt, B.txt, C.txt and D.txt as:
假设您有四个文件 A.txt、B.txt、C.txt 和 D.txt，分别为：
firstOutput='0,1.2'; secondOutput='2.2'; myoutput="$firstOutput,$secondOutput"; outputCount=3; join -a 1 -a 2 -e 0 -o "$myoutput" A.txt B.txt > tmp.tmp; for f in C.txt D.txt; do firstOutput="$firstOutput,1.$outputCount"; myoutput="$firstOutput,$secondOutput"; join -a 1 -a 2 -e 0 -o "$myoutput" tmp.tmp $f > tempf; mv tempf tmp.tmp; outputCount=$(($outputCount+1)); done; mv tmp.tmp files_join.txt
Join the files with:
加入文件：
~$ cat files_join.txt
x1 2 5 0 1
x2 3 7 1 0
x3 0 4 1 0
x4 5 6 1 0
x5 8 0 1 0
Results:
结果：
##代码##