Awk 搜索图案并打印计数
我有如下输入表Awk 搜索图案并打印计数,awk,sed,Awk,Sed,我有如下输入表 module input output type func AOIO A1,A2,B Z comb ((!A1+!A2) B) AN2 A1,A2 ZN comb (A1 A2) AIO A1,A2,A3 Z comb ((A1 A2)+A3) NOR
module input output type func
AOIO A1,A2,B Z comb ((!A1+!A2) B)
AN2 A1,A2 ZN comb (A1 A2)
AIO A1,A2,A3 Z comb ((A1 A2)+A3)
NOR A1,A2 ,B Z comb (!((A1 A2)+B))
BUF I Q seq I
CLK I QN seq IN
BND
INV IN Z comb (!I)
module input output type func path
AOIO A1,A2,B Z comb ((!A1+!A2) B) 3
AN2 A1,A2 ZN comb (A1 A2) 1
AIO A1,A2,A3 Z comb ((A1 A2)+A3) 2
NOR A1,A2 ,B Z comb (!((A1 A2)+B)) 3
BUF I Q seq I 1
CLK I QN seq IN 1
BND 0
INV IN Z comb (!I) 1
（！计数+1！！！（！I）sed 's/[^(]//g' file | awk '{ print length} | awk '{ if ($2~/^(*!/ ) print $3 length+1 ; elseif($2~/^( / ) print $3 length; elseif ($2 !~ /(/ print $3 "1"; elseif ($2~/(!I)/ ) print $3 "1"
awk '
FNR==1{
OFS="\t"
print $0,"path"
next
}
{
line=$0
if($0~/!I/){
print $0,1
next
}
sub(/[^(]*/,"")
sub(/[a-zA-Z0-9].*/,"")
if($0~/!$/){
print line,length($0)
}
else if(!NF){
print line,length($0)+1
}
else if($0 ~ /!/ && $0 !~ /!$/){
print line,length($0)-1
}
else{
print line,length($0)
}
}' Input_file | column -t -s $'\t'
module input output type func path
AOIO A1,A2,B Z comb ((!A1+!A2) B) 3
AN2 A1,A2 ZN comb (A1 A2) 1
AIO A1,A2,A3 Z comb ((A1 A2)+A3) 2
NOR A1,A2 ,B Z comb (!((A1 A2)+B)) 3
BUF I Q seq I 1
CLK I QN seq IN 1
BND 1
INV IN Z comb (!I) 1
module func path
AO ((!A1+!A2) B) 3
AN (A1 A2) 1
AI ((A1 A2)+A3) 2
NOR (!((A1 A2)+B)) 3
BUF I 1
CLK IQ 1
BND 0
INV (!I) 1
基于仅在GNU
awkawk '
FNR==1{
OFS="\t"
print $0,"path"
next
}
{
line=$0
if($0~/!I/){
print $0,1
next
}
$1=""
sub(/[^\t]*\t+/,"")
sub(/[a-zA-Z0-9].*/,"")
if($0~/!$/){
print line,length($0)
}
else if(!NF){
print line,length($0)+1
}
else if($0 ~ /!/ && $0 !~ /!$/){
print line,length($0)-1
}
else{
print line,length($0)
}
}' Input_file | column -t -s $'\t'
module input output type func path
AOIO A1,A2,B Z comb ((!A1+!A2) B) 3
AN2 A1,A2 ZN comb (A1 A2) 1
AIO A1,A2,A3 Z comb ((A1 A2)+A3) 2
NOR A1,A2 ,B Z comb (!((A1 A2)+B)) 3
BUF I Q seq I 1
CLK I QN seq IN 1
BND 1
INV IN Z comb (!I) 1
module func path
AO ((!A1+!A2) B) 3
AN (A1 A2) 1
AI ((A1 A2)+A3) 2
NOR (!((A1 A2)+B)) 3
BUF I 1
CLK IQ 1
BND 0
INV (!I) 1
awk ' ##Starting awk program from here.
FNR==1{ ##Checking condition if this is first line then do following.
OFS="\t" ##Setting OFS as tab here.
print $0,"path" ##Printing current line and path string here.
next ##next will skip all further statements from here.
}
{
line=$0 ##Setting variable line to current line.
if($0~/!I/){ ##Checking condition if line is equal to !I then do following.
print $0,1 ##Printing current line and 1 here.
next ##next will skip all further statements from here.
}
$1="" ##nullifying first field here.
sub(/[^\t]*\t+/,"") ##Substituting everything till tab and tab occurrences with NULL here.
sub(/[a-zA-Z0-9].*/,"") ##Substituting everything till last in current line.
if($0~/!$/){ ##Checking if line do not end with ! then do following.
print line,length($0) ##Printing line and length of current line.
}
else if(!NF){ ##else if number of fields is NULL then do following.
print line,length($0)+1 ##Printing line and length of current line +1 here.
}
else if($0 ~ /!/ && $0 !~ /!$/){ ##else if line is NOT equal to ! AND not ending with !
print line,length($0)-1 ##Printing line and length of current line-1 here.
}
else{ ##else do following.
print line,length($0) ##Printing line and length of current line.
}
}' Input_file | column -t -s $'\t' ##Passing awk output to column command to separate it tab wise.
您也可以尝试此awk：
cat cpath.awk
开始{
FS=“{3，}”
}
NR==1{
printf“%-30s%s\n”$0，“路径”
下一个
}
{
n=0
如果（$2~/！I |^[^（+]+$/）
n=1
else if（匹配（$2，/^\（+！[^（]/））
n=r长度-1
else if（匹配（$2，/^\（[！（]*/））{
s=substr（$2，1，RLENGTH）
n=gsub（/！$\（/，“”，s）
}
printf“%-30s%s\n”，$0，n
}
awk -f cpath.awk file
请在您的问题中以代码的形式添加您的努力，然后让我们知道。您是否可以让我们知道计数如何在第一行变为
325（（！5（(((（！5（（6（（（）((（4$1$51$5（（a1a2）+（B1）+C）的函数34B1（！（！A1！A2））不计算A1的！数（我们需要先计算jst！
即A1的，而不是A2的），并给出路径为2，但路径应该是3。@先生，是的。谢谢您的逻辑