Bash 从列名中检索列号
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Bash retrieve column number from column name
提问by Rob
Is there a better way (such as a one liner in AWK) where I can get the column number in a table with headings from a column name? I want to be able to process a column independent of what the column number actually is (such as when another column is added the script will not need to change).
有没有更好的方法（例如 AWK 中的单行），我可以在表格中获取带有列名标题的列号？我希望能够独立于列号实际处理一列（例如，当添加另一列时，脚本将不需要更改）。
For example, given the following table in "table.tsv":
例如，给定“table.tsv”中的下表：
ID Value Target Not Used
1 5 9 11
2 4 8 12
3 6 7 10
I can do a sort on the "Target" column using:
我可以使用以下方法对“目标”列进行排序：
#!/bin/bash
(IFS=$'\t'; read -r; printf "%s\n" "$REPLY"; i=0; for col in $REPLY; do
((++i))
[ "$col" == "Target" ] && break
done; sort -t$'\t' "-k$i,${i}n") < table.tsv
Is there a way to do it without the for loop (or at least clean it up a little)?
有没有办法在没有 for 循环的情况下做到这一点（或者至少稍微清理一下）？
The expected output of the given script is:
给定脚本的预期输出是：
ID Value Target Not Used
3 6 7 10
2 4 8 12
1 5 9 11
However, I was trying to give an example of what I was trying to do. I want to pass/filter my table through several programs so the headings and all columns should be preserved: just have processing occur at each step. In pseudo code, what I would like to do is:
但是，我试图举例说明我正在尝试做的事情。我想通过几个程序传递/过滤我的表格，这样标题和所有列都应该被保留：只需在每一步进行处理。在伪代码中，我想做的是：
print headings from stdin
i=$(magic to determine column position given "Target")
sort -t$'\t' "-k$i,${i}n" # or whatever processing is required on that column
回答by karakfa
another alternative with a lot of pipes
有很多管道的另一种选择
$ head -1 table | tr -s ' ' '\n' | nl -nln | grep "Target" | cut -f1
extract first row, transpose, number lines, find column name, extract number
提取第一行，转置，编号行，查找列名，提取编号
Or, awkto the rescue!
或者，awk救命！
$ awk -v RS='\t' '/Target/{print NR; exit}' file.tsv
3
回答by anubhava
Here is an awk alternative:
这是一个 awk 替代方案：
awk -F '\t' -v col='Target' 'NR==1{for (i=1; i<=NF; i++) if ($i == col){c=i; break}}
{print $c}' file
EDIT:To print column number only:
编辑：仅打印列号：
awk -F '\t' -v col='Target' 'NR==1{for (i=1; i<=NF; i++) if ($i==col) {print i;exit}}' file
3
回答by Ed Morton
$ awk -v name='Target' '{for (i=1;i<=NF;i++) if ($i==name) print i; exit}' file
3
