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Doubt in awk

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# 1  
Old 06-17-2014
Doubt in awk
Hi,

I got a below requirement from this forum, but the solution provided was not clear.

Below is the requirement

Code:
 
Input file
 
A 1 Z
A 1 ZZ
B 2 Y
B 2 AA
 
 
Required output
 
B Y|AA
A Z|ZZ

Solution provided wa
Code:
/usr/xpg4/bin/awk '{a[$1]=a[$1]?a[$1]"|"$3:$3}END{for (i in a){print i,a[i] }}' inputfile

The portion is in red was not understandable.

What I assumption is
Code:
{a[$1]=a[$1]?a[$1]"|"$3:$3}

if a[$1] the first row fields are same for next row then including pipeline and printing field $3 , but I am sure of : (colon)

I tried below to understand the code

Code:
/usr/xpg4/bin/awk -F" " '{a[$1]=a[$1]?a[$1]:$3}END{for (i in a){print i,a[i] }}' inputfile

but getting syntax error.
# 2  
Old 06-17-2014
{a[$1]=a[$1]?a[$1]"|"$3:$3} --> If array a with index $1 contains some element then concatinate existing array element with current line $3 where separator between existing element and new element (column 3 $3 from current line read) is pipe |
else (meaning so far no such index in array a) then a[$1] equal to current line $3
This User Gave Thanks to Akshay Hegde For This Post:
stew
# 3  
Old 06-17-2014
Hi Akshay,

Thanks for your explanation

As you said the pipeline is seprator then it should be in second place (as emntioned below) not in the 4th place, also not sure what colon : plays here

Code:
B |Y AA
A |Z ZZ

# 4  
Old 06-17-2014
Quote:
Originally Posted by stew
Hi Akshay,

Thanks for your explanation

As you said the pipeline is seprator then it should be in second place (as emntioned below) not in the 4th place, also not sure what colon : plays here

Code:
B |Y AA
A |Z ZZ

colon : is like else statement in ternary

Example :
A traditional if-else construct in C, Java and JavaScript is written:
Code:
if (a > b) {
    result = x;
} else {
    result = y;
}

This can be rewritten as the following statement:

Code:
result = a > b ? x : y;

I get result properly see here

Code:
$ awk '{a[$1] = a[$1] ? a[$1] "|" $3: $3}END{for(i in a) print i,a[i]}' <<EOF
A 1 Z
A 1 ZZ
B 2 Y
B 2 AA
EOF

A Z|ZZ
B Y|AA

---------- Post updated at 02:27 PM ---------- Previous update was at 02:17 PM ----------

This is the best and safest way actually I prefer.

Code:
awk '{a[$1] = ( $1 in a ) ? a[$1] "|" $3: $3}END{for(i in a) print i,a[i]}'

# 5  
Old 06-17-2014
Thanks.

I understood a bit now, but how this
Code:
a[$1]

is printing
Code:
A Z

then
Code:
"|" $3: $3

is printing
Code:
|ZZ

my understanding is

Code:
 
a[$1] "|"    $3 :  $3}
|        |    |      |
|        |    |      |
V        V    V      V
A        |    Z     ZZ
B        |    Y     AA
 
But why its printing like below
 
A Z|ZZ
B Y|AA

# 6  
Old 06-17-2014
Hope this will clear your doubts please note we are printing array elements in END block

Code:
When awk reads line number  : 1 Array a[A] = Z           row : A 1 Z                             
When awk reads line number  : 2 Array a[A] = Z|ZZ        row : A 1 ZZ                            
When awk reads line number  : 3 Array a[B] = Y           row : B 2 Y                             
When awk reads line number  : 4 Array a[B] = Y|AA        row : B 2 AA

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