在Linux目录中查找与模式匹配的文件数 [英] Find count of files matching a pattern in a directory in linux

问题描述

我是Linux新手.我在linux中有一个大约有25万个文件的目录 我需要找到与模式匹配的文件数.

I am new to linux. I have a directory in linux with approx 250,000 files I need to find count of number of files matching a pattern.

我尝试使用以下命令:

ls -1 20061101-20131101_kh5x7tte9n_2010_* | wc -l

我收到以下错误消息:

-bash: /bin/ls: Argument list too long
0

请帮助.预先感谢

推荐答案

为此最好使用find:

find . -name "pattern_*" -printf '.' | wc -l

在您的特定情况下:

find . -maxdepth 1 -name "20061101-20131101_kh5x7tte9n_2010_*" -printf '.' | wc -m

find将返回符合条件的文件列表. -maxdepth 1将使搜索仅在路径中完成，没有子目录(

find will return a list of files matching the criteria. -maxdepth 1 will make the search to be done just in the path, no subdirectories (thanks Petesh!). -printf '.' will print a dot for every match, so that names with new lines won't make wc -m break.

然后wc -m将指示与文件数匹配的字符数.

Then wc -m will indicate the number of characters which will match the number of files.

两个可能选项的性能比较:

Performance comparation of two possible options:

让我们用这种模式创建10000个文件:

Let's create 10 000 files with this pattern:

$ for i in {1..10000}; do touch 20061101-20131101_kh5x7tte9n_201_$i; done

，然后用ls -1 ...或find ...比较获得结果所需的时间:

And then compare the time it takes to get the result with ls -1 ... or find ...:

$ time find . -maxdepth 1 -name "20061101-20131101_kh5x7tte9n_201_*" | wc -l
10000

real    0m0.034s
user    0m0.017s
sys     0m0.021s

$ time ls -1 | grep 20061101-20131101_kh5x7tte9n_201 | wc -l
10000

real    0m0.254s
user    0m0.245s
sys     0m0.020s

find快x5倍！但是如果我们使用ls -1f(

find is x5 times faster! But if we use ls -1f (thanks Petesh again!), then ls is even faster than find:

$ time ls -1f | grep 20061101-20131101_kh5x7tte9n_201 | wc -l
10000

real    0m0.023s
user    0m0.020s
sys     0m0.012s

这篇关于在Linux目录中查找与模式匹配的文件数的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！