使用awk按列名获取特定的列号 [英] Get a specific column number by column name using awk
问题描述
我有n个文件,在这些文件中,每个文件中的不同列号都有一个名为"thrudate"的特定列.
i have n number of file, in these files a specific column named "thrudate" is given at different column number in every files.
i我只想一次从所有文件中提取此列的值.所以我尝试使用awk.在这里,我只考虑一个文件,并提取出thrudate的值
i Just want to extract the value of this column from all files in one go. So i tried with using awk. Here i'm considering only one file, and extracting the values of thrudate
awk -F, -v header=1,head="" '{for(j=1;j<=2;j++){if($header==1){for(i=1;i<=$NF;i++){if($i=="thrudate"){$head=$i;$header=0;break}}} elif($header==0){print $0}}}' file | head -10
我的处理方式:
- 使用find命令查找所有相似文件,然后对每个文件执行第二步
- 循环第一行中的所有字段,检查标头值为1的列名(将其初始化为1以仅检查第一行),一旦它与"thrudate"匹配,我将标头设置为0,然后从此循环中断
- 一旦获得列号,然后为每一行打印一次.
推荐答案
您可以使用以下awk脚本:
You can use the following awk script:
print_col.awk :
# Find the column number in the first line of a file
FNR==1{
for(n=1;n<=NF;n++) {
if($n == header) {
next
}
}
}
# Print that column on all other lines
{
print $n
}
然后使用find
在每个文件上执行此脚本:
Then use find
to execute this script on every file:
find ... -exec awk -v header="foo" -f print_col.awk {} +
在注释中,您要求提供一个可以根据其标题名称打印多个列的版本.您可以为此使用以下脚本:
In comments you've asked for a version that could print multiple columns based on their header names. You may use the following script for that:
print_cols.awk :
BEGIN {
# Parse headers into an assoc array h
split(header, a, ",")
for(i in a) {
h[a[i]]=1
}
}
# Find the column numbers in the first line of a file
FNR==1{
split("", cols) # This will re-init cols
for(i=1;i<=NF;i++) {
if($i in h) {
cols[i]=1
}
}
next
}
# Print those columns on all other lines
{
res = ""
for(i=1;i<=NF;i++) {
if(i in cols) {
s = res ? OFS : ""
res = res "" s "" $i
}
}
if (res) {
print res
}
}
这样称呼:
find ... -exec awk -v header="foo,bar,test" -f print_cols.awk {} +
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