基于价值观在一个特定的列值分裂 [英] splitting values based on values in a specific column

问题描述

我有我想分解成多个文件的uniq值第一列的文件。例如，这里是一个文件：

I have a file that I would like to break up into multiple files with uniq values for the first column. For example, here is a file:

fileA.txt

fileA.txt

1    Cat
1    Dog
1    Frog
2    Boy
2    Girl
3    Tree
3    Leaf
3    Branch
3    Trunk

我想我的输出是这个样子：

I would like my output to look something like this:

FILE1.TXT

file1.txt

1    Cat
2    Boy
3    Tree

FILE2.TXT

file2.txt

1    Dog
2    Girl
3    Leaf

file3.txt

file3.txt

1    Frog
3    Branch

file4.txt

file4.txt

3    Trunk

如果一个值不存在，我希望它被跳过。我试图寻找类似情况下的矿井，但我已经拿出短。有没有人对如何做到这一点的想法？

If a value does not exist, I want it to be skipped. I have tried to search for similar situations to mine, but I've come up short. Does anyone have idea of how to do this?

从理论上讲，这awk命令应该工作：的awk'{打印＆GT; 文件++中的[1 $]名.txt}'输入。但是，我无法得到它的工作适当地（最可能是由于这样的事实，我在Mac上工作）没有人知道的另一种方式？

Theoretically, this awk command should work: awk '{print > "file" ++a[$1] ".txt"}' input. However, I can't get it to work appropriately (most likely due to the fact that I work on a mac) Does anyone know of an alternative way?

推荐答案

在输出重定向右侧的一个前加括号的pression是不确定的行为。尝试的awk'{打印＆GT; （文件++ a [$ 1]名.txt）}'输入。

An unparenthesized expression on the right side of output redirection is undefined behavior. Try awk '{print > ("file" ++a[$1] ".txt")}' input.

如果有太多的文件，同时打开是一个问题，然后得到GNU awk的，但如果你不能

If having too many files open concurrently is an issue then get GNU awk, but if you cant:

$ ls
 fileA.txt

$ awk '{f="file" ++a[$1] ".txt"; print >> f; close(f)}' fileA.txt

$ ls
file1.txt  file2.txt  file3.txt  file4.txt  fileA.txt

$ cat file1.txt
1    Cat
2    Boy
3    Tree

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