Hi
I have no experience in Unix so any help would be appreciated
I have the flowing text
235543
123
45654
199
225
578
45654
199
225
I need to find this sequence from A file
45654
199
225
and replaced it with in B file
45654
258
so the new file B will be
235543
123
45654
258
578
45654
258
any help?
Here is a solution using awk:
awk '
{
A[++c] = $1
}
END {
for ( i = 1; i <= c; i++ )
{
if ( A == 45654 && A[i+1] == 199 && A[i+2] == 225 )
{
A[i+1] = 258
A[i+2] = 0
}
if ( A )
print A
}
}
' file
Thanks Yoda
but what if I want to search for a variable sequence instead of known. for example
"variable number"
199
225
will be
"variable number"
258
Thanks
I didn't quite understand what you mean by variable sequence.
The program that I posted replaces 45654 199 225 to 45654 258
You just have to modify it as per your requirement.
Thanx again Yoda
sorry for being not clear
what I meant was
what if I want to find sequence that followed by "xxx"
XXX
199
225
and replace it with
XXX
258
xxx could be any number between (1 to 260)
so, every time replace all the sequence followed by that XXX
Thank you
So I assume that you are going to define starting sequence in a variable.
In that case you can pass whatever variable to awk, assign it and use it.
You can code something like:
SEQ=XXX
awk -v S="$SEQ" '
{
A[++c] = $1
}
END {
for ( i = 1; i <= c; i++ )
{
if ( A == S && A[i+1] == 199 && A[i+2] == 225 )
{
A[i+1] = 258
A[i+2] = 0
}
if ( A )
print A
}
}
' file
I used shell variable: SEQ , replace value XXX with the number of your choice. I hope this helps.
Khaled79,
Check this out:
# v=45654;perl -0777 -pe 's/$ENV{v}\n199\n225/$ENV{v}\n258/igs' file
235543
123
45654
258
578
45654
258
rveri
it wont work !
# v=45654;perl -0777 -pe 's/$ENV{v}\n199\n225/$ENV{v}\n258/igs' file
---------- Post updated at 08:14 PM ---------- Previous update was at 07:45 PM ----------
Yoda
for small files awk working with no problem
but, with large files the awk shows this error
awk: cmd. line:3: (FILENAME=a.txt FNR=18498251) fatal: more_nodes: nextfree: can 't allocate 4000 bytes of memory (Cannot allocate memory)
and this error too printed in Cygwin terminal
line 3: 7488 Aborted (core dumped) awk -v S="$SEQ" '
{
A[++c] = $1
}
END {
for ( i = 1; i <= c; i++ )
{
if ( A == S && A[i+1] == 199 && A[i+2] == 225 )
{
A[i+1] = 258
A[i+2] = 0
}
if ( A )
print A
}
}
' ascii.txt >pre.txt
any help about it?
Thanks a lot
How about this awk code?
SEQ=45654
awk -v S="$SEQ" '
$0 == S {
V = $0
getline
if ( $0 == 199 )
{
getline
if ( $0 == 225 )
{
print V RS "258"
next
}
else
{
print V RS "199" RS $0
next
}
}
else
{
print V RS $0
next
}
}
$0 != S {
print $0
}
' file
Dear Yoda
I will test it now for large files I have
Thanks
---------- Post updated at 09:25 PM ---------- Previous update was at 08:58 PM ----------
Thanks Yoda
it works well for large and small files as well
Thank you very much.
could you please tell me what is the different between two codes?
Khaled
The first awk program that I posted loads all the records in your file into an Indexed Array A[++c] = $1 and in the end it performs the required operation.
This caused the program to throw Cannot allocate memory error for large files.
But in the second awk program, entire records are not loaded into any variable or array but instead checking record by record to perform the required operation.
Hence it is not a memory intensive program and works for large files.
Yoda
can I make it search for any of tow or more numbers if found then replaced it like for example
SEQ=45654 | 234567 |57899
awk -v S="$SEQ" '
$0 == S {
V = $0
getline
if ( $0 == 199 )
{
getline
if ( $0 == 225 )
{
print V RS "258"
next
}
else
{
print V RS "199" RS $0
next
}
}
else
{
print V RS $0
next
}
}
$0 != S {
print $0
}
' file
Thanks a lot
Khaled
Yes you can. For implementing this change use regular expression comparison operators ~ and !~ instead:
SEQ="45654|234567|57899"
awk -v S="$SEQ" '
$0 ~ S {
V = $0
getline
if ( $0 == 199 )
{
getline
if ( $0 == 225 )
{
print V RS 258
next
}
else
{
print V RS "199" RS $0
next
}
}
else
{
print V RS $0
next
}
}
$0 !~ S {
print $0
F = 0
}
' file
Yoda
I have used this code to print the number that comes before and after specific character
SEQ=200
awk -v S="$SEQ" '
{
A[++c] = $1
}
END {
for ( i = 1; i <= c; i++ )
{
if ( A == S )
{
print " the letter is " A
print " followed by " A[i+1]
print " comes after " A[i-1]
}
}
}
' file
result was like the following
the letter is 200
followed by 202
comes after 211
the letter is 200
followed by 223
comes after 212
the letter is 200
followed by 202
comes after 211
I need it to print counter of times that happened if it repeated rather than print it many time so the result should be something like this
2 times
the letter is 200
followed by 202
comes after 211
1 times
the letter is 200
followed by 223
comes after 212
how I can do it?
Thanks a lot
Khaled
You can code something like:
SEQ=200
awk -v S="$SEQ" '
{
A[++c] = $1
}
END {
for ( i = 1; i <= c; i++ )
{
if ( A == S )
{
C[A,A[i+1],A[i-1]]++
V[A,A[i+1],A[i-1]] = "the letter is " A RS "followed by " A[i+1] RS "comes after " A[i-1]
}
}
for ( k in V )
{
print C[k] " times"
print V[k]
}
}
' file
Thanks Yoda
You are awesome!
how I can sort the output descending ?
Thanks
By default, the order in which a for (i in array) loop scans an array is not defined; it is generally based upon the internal implementation of arrays inside awk.
So you have to use an Indexed Array to help preserve the order, try this modified code:
SEQ=200
awk -v S="$SEQ" '
{
A[++c] = $1
}
END {
for ( i = 1; i <= c; i++ )
{
if ( A == S )
{
if ( !(V[A A[i+1] A[i-1]]) )
T[++j] = A A[i+1] A[i-1]
C[A A[i+1] A[i-1]]++
V[A A[i+1] A[i-1]] = "the letter is " A RS "followed by " A[i+1] RS "comes after " A[i-1]
}
}
for ( i = 1; i <= j; i++ )
{
print C[T] " times"
print V[T]
}
}
' file
The result wasn't ordered descending based on times
25 times
the letter is 200
followed by 202
comes after 211
36 times
the letter is 200
followed by 223
comes after 212
it should print the 36 times result followed by 25 result
Thanks
I'm sorry. I misread your requirement. I thought you want to print records in the order that you have them in your input file.
If you have gawk then you can use below code to print in descending order:
SEQ=200
gawk -v S="$SEQ" '
{
A[++c] = $1
}
END {
for ( i = 1; i <= c; i++ )
{
if ( A == S )
{
C[A,A[i+1],A[i-1]]++
V[A,A[i+1],A[i-1]] = "the letter is " A RS "followed by " A[i+1] RS "comes after " A[i-1]
}
}
for ( k in V )
{
T[++j] = C[k]
}
n = asort(T)
for ( i = n; i >= 1; i-- )
{
for ( k in V )
{
if ( C[k] == T )
{
print C[k] " times"
print V[k]
}
}
}
}
' file
Thanks Yoda
Suppose that I have texts in different languages
how I could print the frequent n words
like frequent 10 or 100 ?
It is easy to do it for A-Z which is English
but, regardless the input language text how I can do it?
Thanks