BASH - 使用Loop和If语句汇总来自唯一字段中多个字段的信息 [英] BASH - Summarising information from several fields in unique field using Loop and If statements
问题描述
我有以下标签分隔文件：
A1 A1 0 0 2 1 1 1 1 1 1 1 2 1 1 1
A2 A2 0 0 2 1 1 1 1 1 1 1 1 1 1
A3 A3 0 0 2 2 1 1 2 2 1 1 1 1 1
A5 A5 0 0 2 2 1 1 1 1 1 1 1 2 1 1
想法是总结列之间的信息7（包含）和在文件末尾添加的新列中的结尾。
为此，这些是规则：
-
如果行中（第7列和结尾之间）的2总数为 0 ：将1 1添加到新的最后一列
-
如果行中的2总数（第7列和结尾之间）为 1 ：将1 2添加到新的最后一列
-
如果总数为行中2的r（第7列和结尾之间） 2或更多：将2 2添加到新的最后一列
我开始使用命令提取我想要处理的列：
awk'{for（i = 7; i< = NF; i ++）printf $ i; print}'myfile.ped> tmp_myfile.txt
然后我使用以下方法计算每行中的出现次数：
sed's / [^ 2] // g'tmp_myfile.txtt | awk'{print NR，length}'>
tmp_occurences.txt
哪些输出：
1 1
2 0
3 2
4 1
然后我的想法是编写一个循环遍历行的for循环来添加新的汇总列。
我正在考虑这种结构，基于我在这里找到的东西： http://www.thegeekstuff.com/2010/06/bash-if-statement-examples ：
<$读取行时p $ p>
;
do
set $ line
如果[$ 2== 0]
则
$ 3 ==1 1
elif [$ 2== 1]
然后
$ 3 ==1 2
elif [$ 2> = 2]
然后
$ 3 ==2 2
其他
打印[错误]
fi
完成< tmp_occurences.txt
但是我被困在这里。我是否必须在开始循环之前创建新列？我是朝着正确的方向前进？
理想情况下，最终输出（在合并初始文件和摘要列的前6列之后）将是：
A1 A1 0 0 2 1 1 2
A2 A2 0 0 2 1 1 1
A3 A3 0 0 2 2 2 2
A5 A5 0 0 2 2 1 2
感谢您的帮助！
使用gnu-aw你可以这样做：
awk -v OFS ='\ t''{
c = 0;
for（i = 7; i< = NF; i ++）
if（$ i == 2）
c ++
if（c == 0）
s = 1 1
否则if（c == 1）
s =1 2
else
s =2 2
NF = 6
打印$ 0，s
}'档案
A1 A1 0 0 2 1 1 2
A2 A2 0 0 2 1 1 1
A3 A3 0 0 2 2 2 2
A5 A5 0 0 2 2 1 2
PS：如果不使用gnu-awk你可以使用：
awk -v OFS ='\ t''{c = 0; for（i = 7; i< = NF; i ++）{if（$ i == 2）c ++; $ i =} if（c == 0）s =1 1;否则如果（c == 1）s =1 2;否则s =2 2; NF = 6;打印$ 0，s}'文件
I have the following tab-separated file:
A1 A1 0 0 2 1 1 1 1 1 1 1 2 1 1 1
A2 A2 0 0 2 1 1 1 1 1 1 1 1 1 1 1
A3 A3 0 0 2 2 1 1 2 2 1 1 1 1 1 1
A5 A5 0 0 2 2 1 1 1 1 1 1 1 2 1 1
The idea is to summarise the information between column 7 (included) and the end in a new column that is added at the end of the file.
To do so, these are the rules:
If the total number of "2"s in the row (between column 7 and the end) is 0: add "1 1" to the new last column
If the total number of "2"s in the row (between column 7 and the end) is 1: add "1 2" to the new last column
If the total number of "2"s in the row (between column 7 and the end) is 2 or more: add "2 2" to the new last column
I started to extract the columns I want to work on using the command:
awk '{for (i = 7; i <= NF; i++) printf $i " "; print ""}' myfile.ped > tmp_myfile.txt
Then I count the number of occurrence in each row using:
sed 's/[^2]//g' tmp_myfile.txtt | awk '{print NR, length }' > tmp_occurences.txt
Which outputs:
1 1
2 0
3 2
4 1
Then my idea was to write a for loop that loops through the lines to add the new summary column. I was thinking in this kind of structure, based on what I found here: http://www.thegeekstuff.com/2010/06/bash-if-statement-examples:
while read line ;
do
set $line
If ["$2"==0]
then
$3=="1 1"
elif ["$2"==1 ]
then
$3=="1 2"
elif ["$2">=2 ]
then
$3=="2 2"
else
print ["error"]
fi
done < tmp_occurences.txt
But I am stuck here. Do I have to create the new column before starting the loop? Am I going in the right direction?
Ideally, the final output (after merging the first 6 columns from the initial file and the summary column) would be:
A1 A1 0 0 2 1 1 2
A2 A2 0 0 2 1 1 1
A3 A3 0 0 2 2 2 2
A5 A5 0 0 2 2 1 2
Thank you for your help!
Using gnu-awk you can do:
awk -v OFS='\t' '{
c=0;
for (i=7; i<=NF; i++)
if ($i==2)
c++
if (c==0)
s="1 1"
else if (c==1)
s="1 2"
else
s="2 2"
NF=6
print $0, s
}' file
A1 A1 0 0 2 1 1 2
A2 A2 0 0 2 1 1 1
A3 A3 0 0 2 2 2 2
A5 A5 0 0 2 2 1 2
PS: If not using gnu-awk you can use:
awk -v OFS='\t' '{c=0; for (i=7; i<=NF; i++) {if ($i==2) c++; $i=""} if (c==0) s="1 1"; else if (c==1) s="1 2"; else s="2 2"; NF=6; print $0, s}' file
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