AWK - 发现每行的最小值与任意大小 [英] AWK - find min value of each row with arbitrary size

问题描述

我与行的文件为：

5 3 6 4 2 3 5
1 4 3 2 6 5 8
..

我想获得每行的分，因此，例如与上面给出的投入，我应该得到：

I want to get the min on each line, so for example with the input given above, I should get:

min of first line:  2
min of second line: 1
..

我如何使用awk来为列的每一行任意数量做到这一点？

How can I use awk to do this for any arbitrary number of columns in each line?

推荐答案

如果你不使用数字代替文字介意输出您可以使用此单行：

If you don't mind the output using digits instead of words you can use this one liner:

$ awk '{m=$1;for(i=1;i<=NF;i++)if($i<m)m=$i;print "min of line",NR": ",m}' file
min of line 1:  2
min of line 2:  1

如果你确实想在序数数：

If you really do want to count in ordinal numbers:

BEGIN {
    split("first second third fourth",count," ")
}
{
    min=$1
    for(i=1;i<=NF;i++)
    if($i<min)
        min=$i

    print "min of",count[NR],"line: \t",min
}

保存这 script.awk 并运行，如：

$ awk -f script.awk file
min of first line:    2
min of second line:   1

显然，这将只适用于高达4行的文件工作，但只是增加了序数列表，你认为你需要的最大数量。你应该能够很容易地找到一个列表的在线pretty。

Obviously this will only work for files with upto 4 lines but just increase the ordinal numbers list to the maximum number you think you will need. You should be able to find a list online pretty easily.

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