组文件和管道awk命令 [英] Group files and pipe to awk command

问题描述

我有一个目录中的文件;他们使用的是YYYY_MM_DD命名为：

I have files in a directory; they are named using YYYY_MM_DD:

-rw-r--r-- 1 root root 497186 Apr 21 13:17 2012_03_25
-rw-r--r-- 1 root root 490558 Apr 21 13:17 2012_03_26
-rw-r--r-- 1 root root 488797 Apr 21 13:17 2012_03_27
-rw-r--r-- 1 root root 316290 Apr 21 13:17 2012_03_28
-rw-r--r-- 1 root root 490081 Apr 21 13:17 2012_03_29
-rw-r--r-- 1 root root 486621 Apr 21 13:17 2012_03_30
-rw-r--r-- 1 root root 490904 Apr 21 13:17 2012_03_31
-rw-r--r-- 1 root root 491788 Apr 21 13:17 2012_04_01
-rw-r--r-- 1 root root 488630 Apr 21 13:17 2012_04_02

在文件中的第一列是一个数字，我用下面的 AWK 命令采取的第一列的平均值。

The first column within the file is a number, and I am using the following awk command to take an average of that first column.

awk -F, '{ x += $1 } END { print x/NR }' MyFile

使用相同的命令，我可以通过两个文件awk来获得这两个文件的总平均作为一个整体。

Using the same command i can pass two files to awk to get the total average of both files as a whole.

awk -F, '{ x += $1 } END { print x/NR }' File1 File2

我想要做的是...

What I want to do is this...

我想在我的目录中的所有文件，并每月将它们分组，那么当月的所有文件传递给awk命令。

I want to get all the files in my directory, and group them per month, then pass all the files for the month to the awk command.

所以，按照同样的数据，还有3月7日的文件，我希望所有的7个文件要传递给我的 AWK 命令是这样的：

So as per the same data, there are 7 files in March, I would want all 7 files to be passed to my awk command like this:

awk -F, '{ x += $1 } END { print x/NR }' File1 File2 File3 File4 File5 File6 File7

然后同样为四月份的集中。

Then likewise for April's set.

推荐答案

您想以某种方式与单纯的awk做到这一点，也可以使用文件寻找？例如：

Are you wanting to somehow accomplish this with awk alone, or can you use file globbing? For example:

awk -F, '{ #Do stuff }' 2012_03_[0-3][0-9]

将得到所有的三月文件。

will get all the March files.

您也可以使用 2012_03 * 但这是在比大于1的文件名匹配模式不太具体。

You could also use 2012_03* but that's less specific in its globbing pattern than the above one.

的 修改的

Edit

您可以使用shell脚本是这样的：

You can use a shell script like this:

DIR="/tmp/tmp"
for month in $(find "$DIR" -maxdepth 1 -type f | sed 's/.*\/\([0-9]\{4\}_[0-9]\{2\}\).*/\1/' | sort -u); do
  awk -F, '#dostuff' "$DIR/${month}"_[0-3][0-9] > output/dir/SUM_"${month}"
done

一如往常，有几个注意事项。带有空格的文件将打破它。如果有不符合该目录中的YYYY_MM_DD格式的文件，你会得到错误，但它不应该影响性能。让我知道，如果这些限制是不能接受的，我会觉得它多一点。

As always, there are a few caveats. Files with spaces will break it. You'll get errors if there are files that don't conform to the YYYY_MM_DD format in the directory, but it shouldn't affect performance. Let me know if those constraints are not acceptable and I'll think on it a little more.

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