[SOLVED] Use variable instead of text file with grep -f (file mode)?

Jason_25 · 12-04-2016, 01:43 PM

This does not work:

Code:

#dictation words variable
INPUT="i fine"

#convert to variable with new lines instead of space
PROCESSED_INPUT=$(echo $INPUT | tr ' ' '\n')

#searches multiple lines for the most words in the variable and orders them
#ai dictionary can be a zeroed file for these purposes
cat AI_DICTIONARY | grep -Ff $PROCESSED_INPUT

This works:

Code:

#dictation words variable
INPUT="i fine"

#convert to file with new lines instead of space
echo $INPUT | tr ' ' '\n' > INPUT_TEST

#searches multiple lines for the most words in the input file and orders them
#ai dictionary can be a zeroed file for these purposes
cat AI_DICTIONARY | grep -Ff INPUT_TEST

How can I do this in memory instead of with files?

grail · 12-04-2016, 02:25 PM

Try quoting your variables to maintain the spacing.

danielbmartin · 12-04-2016, 03:10 PM

With this AI_DICTIONARY ...

Code:

i fine
you fine
he fail
i good
she fine
Herman happy
Melvin euphoric

... this code ...

Code:

#dictation words variable
INPUT="i fine"

#convert to file with new lines instead of space
tr ' ' '\n' <$INPUT >$INPUT_TEST

#searches multiple lines for the most words in the input file and orders them
#ai dictionary can be a zeroed file for these purposes
grep -wFf $INPUT_TEST <$AI_DICTIONARY >$OutFile

... produced this OutFile ...

Code:

i fine
you fine
i good
she fine

With the same AI_DICTIONARY this code ...

Code:

#dictation words variable
INPUT="i fine"

#convert to variable with new lines instead of space
PROCESSED_INPUT=$(tr ' ' '|' <<<$INPUT)

#searches multiple lines for the most words in the variable and orders them
#ai dictionary can be a zeroed file for these purposes
egrep -w "($PROCESSED_INPUT)" <$AI_DICTIONARY >$OutFile

... produced the same OutFile.

If you care for a further refinement it is possible to eliminate the variable PROCESSED_INPUT.
With the same AI_DICTIONARY this code ...

Code:

#dictation words variable
INPUT="i fine"

#convert to variable with new lines instead of space
#searches multiple lines for the most words in the variable and orders them
#ai dictionary can be a zeroed file for these purposes
egrep -w "($(tr ' ' '|' <<<$INPUT))" <$AI_DICTIONARY >$OutFile

... produced the same OutFile.

Daniel B. Martin

Jason_25 · 12-05-2016, 10:17 AM

I still have not figured this out. It may not be possible. The problem appears to actually be with grep -f not grep -F. The list feature appears to not take a variable.

The error is

Code:

grep: i: no such file or directory

There are ways to do what I want without providing a list but I prefer how simple it is.

grail - quoting variables helped but was not the primary problem.

danielbmartin - I always get

Code:

$Outfile: ambiguous redirect

with that code.

edit:
I think i got it to work. I am just still in shock because it was so simple. Use e instead of f.

Code:

cat AI_DICTIONARY | grep -Fe $PROCESSED_INPUT

grail · 12-05-2016, 10:46 AM

Well, that would make sense as your data is no longer coming from a file

Don't forget to mark as SOLVED now you have a solution.

michaelk · 12-05-2016, 11:34 AM

You should not use -fF. -f expects a file. -F is a list.

Code:

#dictation words variable
INPUT="i fine"

#convert to variable with new lines instead of space
PROCESSED_INPUT=$(echo $INPUT | tr ' ' '\n')

#searches multiple lines for the most words in the variable and orders them
#ai dictionary can be a zeroed file for these purposes
grep -F $PROCESSED_INPUT AI_DICTIONARY

It isn't necessary to use cat...

MadeInGermany · 12-05-2016, 12:33 PM

Code:

INPUT="i fine
i good"

grep -F "$INPUT" AI_DICTIONARY

Note that $INPUT must be in "quotes"! Previous posts forgot that.