如何读取.dat文件和 grep 特定值

我正在尝试使用 bash 脚本读取.dat文件。

该值包含文件大小，我想 grep 除第一列外大于 0 的值。大于零的值可以出现在任何行中。

I have awk script to read line by line. 

 1349848860, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349848920, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349848980, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349849040, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349849100, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0
1349849160, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 227.736, 2, 0, 29378, 0, 0, 0, 0, 0
1349849220, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349849280, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349849340, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349851200, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349851260, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
1349851320, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 0, 0, 0, 0, 0
1349851380, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 0, 227.736, 2, 0, 29620, 0, 0, 0, 0, 0
1349851440, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0
#!/bin/bash
FILENAME=$1
awk '{kount++;print  kount, $0}     END{print "nTotal " kount " lines read"}' $FILENAME
awk '{print $13}' $FILENAME

所需输出 -

   227.736, 2,  29378 
   227.736, 2,  29620

感谢您的帮助。Naveen<</p>

div class="one_answers">

如果我了解您的需求：

awk -F"," '{for (i=2;i<=NF;i++){if ($i > 1) {print}}}' file.dat

输出

1349939700, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 2, 0, 69832, 0, 0, 0, 0, 0
1349939700, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 2, 0, 69832, 0, 0, 0, 0, 0
1349939700, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 2, 0, 69832, 0, 0, 0, 0, 0
1349939700, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 2, 0, 69832, 0, 0, 0, 0, 0
1349939700, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 2, 0, 69832, 0, 0, 0, 0, 0
1349939880, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 68552, 0, 0, 0, 0, 0
1349939880, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 68552, 0, 0, 0, 0, 0
1349939880, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 68552, 0, 0, 0, 0, 0
1349939880, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 68552, 0, 0, 0, 0, 0
1349940000, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 73826, 0, 0, 0, 0, 0
1349940000, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 73826, 0, 0, 0, 0, 0
1349940000, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 73826, 0, 0, 0, 0, 0
1349940000, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 73826, 0, 0, 0, 0, 0

据我了解，您希望 grep 所有包含 0 或 1 以外的数字的行，而不计算行首的第一个数字。怎么样

egrep ',.*([2-9]|[0-9.]{4,})' tmp.txt

这将从您的示例中打印出行

1349939700, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 2, 0, 69832, 0, 0, 0, 0, 01349939880, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 68552, 0, 0, 0, 0, 01349940000, 0, 0, 0, 0, 2, 0, 0, 0, 0, 0, 0, 14353, 1, 0, 450.03, 1, 0, 73826, 0, 0, 0, 0, 0

如果要获取此文件中每行的第 13 列（即 14353）、第 16 列（即 450.03）和第 19 列（即 69832）的内容（根据您的输入），请尝试以下操作：

awk < test.dat -F ', ' '{print $13,$16,$19}'

数据文件中test.dat的位置。

这将输出：

14353 450.03 69832
0 0 0
0 0 0
14353 450.03 68552
0 0 0
14353 450.03 73826

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